Tamilnadu Board Class 11 Physics Chapter 8 PDF

UNIT
8 HEAT AND THERMODYNAMICS
Classical thermodynamics…. is the only physical theory of universal content which I am convinced…
will never be overthrown. – Albert Einstein
LEARNING OBJECTIVES
In this unit, a student is exposed to

• meaning of heat, work and temperature
• ideal gas laws
• concept of specific heat capacity
• thermal expansion of solids, liquids and gases
• various states of matter
• Newton’s law of cooling
• Stefan’s law and Wien’s law
• meaning of thermodynamic equilibrium
• meaning of internal energy
• zeroth and first laws of thermodynamics
• various thermodynamic processes
• work done in various thermodynamic processes
• second law of thermodynamics
• working of carnot engine and refrigerator
8.1 is a branch of physics which explains the

HEAT AND TEMPERATURE phenomena of temperature, heat etc. The
concepts presented in this chapter will help
us to understand the terms ‘hot’ and ‘cold’ and
also differentiate heat from temperature. In
8.1.1 Introduction thermodynamics, heat and temperature are
Temperature and heat play very important two different but closely related parameters.
role in everyday life. All species can function
properly only if its body is maintained at a
particular temperature. In fact life on Earth 8.1.2 Meaning of heat
is possible because the Sun maintains its When an object at higher temperature
temperature. Understanding the meaning is placed in contact with another object
of temperature and heat are very crucial to at lower temperature, there will be a
understand the nature. Thermodynamics spontaneous flow of energy from the object
95
UNIT-8(XI-Physics_Vol-2) Folder.indd 95 20-08-2018 16:38:22

at higher temperature to the one at lower b. When heated, a cup of coffee receives
temperature. This energy is called heat. heat from the stove. Once the coffee is
This process of energy transfer from higher taken from the stove, the cup of coffee
temperature object to lower temperature has more internal energy than before.
object is called heating. Due to flow of heat ‘Heat’ is the energy in transit and
sometimes the temperature of the body will which flows from an object at higher
increase or sometimes it may not increase. temperature to an object at lower
temperature. Heat is not a quantity.
There is a misconception So the statement ‘A hot cup of coffee
Note that heat is a quantity of has more heat’ is wrong, instead
energy. People often talk ‘this ‘coffee is hot’ will be appropriate.
water has more heat or less heat’. These
words are meaningless. Heat is not a
quantity. Heat is an energy in transit 8.1.3 Meaning of work
which flows from higher temperature When you rub your hands against each
object to lower temperature object. other the temperature of the hands increases.
Once the heating process is stopped You have done some work on your hands
we cannot use the word heat. When by rubbing. The temperature of the hands
we use the word ‘heat’, it is the energy increases due to this work. Now if you place
in transit but not energy stored in the your hands on the chin, the temperature of
body. the chin increases. This is because the hands
are at higher temperature than the chin. In the
above example, the temperature of hands is
increased due to work and temperature of the
E X A M P L E 8. 1 chin is increased due to heat transfer from the
a. ‘A lake has more rain’. hands to the chin. It is shown in the Figure 8.1
b. ‘A hot cup of coffee has more heat’. By doing work on the system, the
What is wrong in these two statements? temperature in the system will increase and
sometimes may not. Like heat, work is also
Solution
a. When it rains, lake receives water Keeping the hands
A persion rubbing on chin
from the cloud. Once the rain stops, his hands
the lake will have more water than
before raining. Here ‘raining’ is a
process which brings water from the
cloud. Rain is not a quantity rather it
is water in transit. So the statement
‘lake has more rain’ is wrong, instead
the ‘lake has more water’ will be Figure 8.1 Difference between work and
appropriate. heat
96 Un i t 8 H E AT A N D T H E R MODY NA M IC S

not a quantity and through the work energy 8.2
is transferred to the system . So we cannot
THERMAL PROPERTIES OF
use the word ‘the object contains more work’
MATTER
or ‘less work’.
Either the system can transfer energy
to the surrounding by doing work on
8.2.1 Boyle’s law,
surrounding or the surrounding may
transfer energy to the system by doing work Charles’ law and ideal
gas law
on the system. For the transfer of energy
from one body to another body through the For a given gas at low pressure (density) kept
process of work, they need not be at different in a container of volume V, experiments
temperatures. revealed the following information.
� When the gas is kept at constant

8.1.4 Meaning of temperature, the pressure of the gas is
temperature inversely proportional to the volume.
Temperature is the degree of hotness or 1
Pµ . It was discovered by Robert Boyle
coolness of a body. Hotter the body higher V
is its temperature. The temperature will (1627-1691) and is known as Boyle’s law.
determine the direction of heat flow when � When the gas is kept at constant
two bodies are in thermal contact. pressure, the volume of the gas is directly
proportional to absolute temperature.
The SI unit of temperature is kelvin (K).
V µ T . It was discovered by Jacques
In our day to day applications, Celsius Charles (1743-1823) and is known as
(˚C) and Fahrenheit (°F) scales are used. Charles’ law.
� By combining these two equations we
Temperature is measured with a
thermometer. have
The conversion of temperature from one PV = CT. Here C is a positive constant.
scale to other scale is given in Table 8.1 We can infer that C is proportional to the
number of particles in the gas container
Table 8.1 Temperature conversion by considering the following argument. If
we take two containers of same type of gas
Scale To Kelvin From Kelvin
with same volume V, same pressure P and
Celsius K=°C + 273.15 °C = K − 273.15 same temperature T, then the gas in each
Fahrenheit K=(°F + 459.67)÷1.8 °F=(K × 1.8)-459.67 container obeys the above equation. PV =
CT. If the two containers of gas is considered
Scale To Fahrenheit From Fahrenheit as a single system, then the pressure and
Celsius °F=(1.8 × °C) + 32 °C = (°F − 32)÷1.8 temperature of this combined system will be
same but volume will be twice and number
Scale To Celsius From Celsius of particles will also be double as shown in
Fahrenheit °C=(°F − 32)÷1.8 °F = (1.8 × °C) + 32 Figure 8.2
Uni t 8 H EAT A N D TH ER MODYNAM IC S 97

P,V,T,N P,V,T,N P,2V,T,2N
Two Separate systems Single system
Figure 8.2 Ideal gas law
For this combined system, V becomes

N = μ NA(8.2)
2V, so C should also double to match
P (2V )
with the ideal gas equation = 2C . It where NA is Avogadro number (6.023
T
implies that C must depend on the number ×1023mol-1)
of particles in the gas and also should have Substituting for N from equation (8.2), the
 PV  equation (8.1) becomes
the dimension of   = JK −1 . So we can
 T  PV = μ NAkT. Here NAk=R called
write the constant C as k times the number universal gas constant and its value is
of particles N. 8.314 J /mol. K.
Here k is the Boltzmann constant So the ideal gas law can be written for μ
(1.381×10−23 JK−1) and it is found to be a mole of gas as
universal constant.
So the ideal gas law can be stated as follows PV = μRT(8.3)
PV = NkT(8.1)
This is called the equation of state for an
The equation (8.1) can also be expressed in ideal gas. It relates the pressure, volume and
terms of mole. temperature of thermodynamic system at
equilibrium.
Mole is the practical unit to express the
amount of gas. One mole of any substance
is the amount of that substance which E X A M P L E 8. 2
contains Avogadro number (NA) of A student comes to school by a bicycle
particles (such as atoms or molecules). whose tire is filled with air at a pressure 240
The Avogadro’s number NA is defined as kPa at 27°C. She travels 8 km to reach the
the number of carbon atoms contained in
school and the temperature of the bicycle
exactly 12 g of 12C .
tire increases to 39°C. What is the change
Suppose if a gas contains μ mole of particles in pressure in the tire when the student
then the total number of particles can be reaches school?
written as
98 Un i t 8 HE AT A N D TH E R MODY NA MIC S

V = 5.5 L
P = 101 kPa
T = 310 K
Solution
We can take air molecules in the tire as an
ideal gas. The number of molecules and Solution
the volume of tire remain constant. So the
air molecules at 27°C satisfies the ideal We can treat the air inside the lungs as an
gas equation P1V1 = NkT1 and at 39°C it ideal gas. To find the number of molecules,
satisfies P2V2 = NkT2 we can use the ideal gas law.
But we know PV = NkT
V1 = V2 = V Here volume is given in the Litre. 1 Litre is

volume occupied by a cube of side 10 cm.
PV NkT1 1Litre = 10cm × 10cm × 10cm = 10-3 m3
1
=
P2V NkT2
PV 1.01 × 105 Pa × 5.5 × 10 −3 m3
P1 T1 N= = −23 −1
= 1.29 × 1023
= kT 1.38 × 10 JK × 310 K
P2 T2 PV 1.01 × 105 Pa × 5.5 × 10 −3 m3
N= = = 1.29 × 1023 Molecules
T2 kT 1.38 × 10 −23 JK −1 × 310 K
P2 = P1
T1 Only 21% of N are oxygen. The total
312K number of oxygen molecules
P2 = 240 103 Pa = 249.6 kPa
300K
21
= 1.29 × 1023 ×
100
E X A M P L E 8. 3 Number of oxygen molecules

= 2.7 × 1022 molecules
When a person breaths, his lungs can hold
up to 5.5 Litre of air at body temperature
37°C and atmospheric pressure E X A M P L E 8. 4
(1 atm =101 kPa). This Air contains 21%
Calculate the volume of one mole of any
oxygen. Calculate the number of oxygen
gas at STP and at room temperature (300K)
molecules in the lungs.
with the same pressure 1 atm.

Solution: determine the number of mole. At room
Here STP means standard temperature temperature 300K, the volume of a gas
(T=273K or 0°C) and Pressure (P=1 atm or occupied by any gas is equal to 24.6L.
101.3 kPa) 120m3
The number of mole μ=
µRT 24.6×10−3 m3
We can use ideal gas equation V = . ≈4878 mol .
P
Here µ = 1 mol and R =8.314 J/mol.K. Air is the mixture of about 20% oxygen,
79% nitrogen and remaining one percent
By substituting the values are argon, hydrogen, helium, and xenon.
J The molar mass of air is 29 gmol-1.
(1mol )(8.314 K )(273K )
V= mol So the total mass of air in the room
1.013 × 105 Nm −2 m = 4878 × 29 = 141.4kg.
=22.4 × 10-3 m3
We know that 1 Litre (L) = =10-3m3. So we
can conclude that 1 mole of any ideal gas 8.2.2 Heat capacity and
has volume 22.4 L. specific heat capacity
300K
By multiplying 22.4L by we get Take equal amount of water and oil at
273K
the volume of one mole of gas at room temperature 27°C and heat both of them
temperature. It is 24.6 L. till they reach the temperature 50°C. Note
down the time taken by the water and oil
to reach the temperature 50°C. Obviously
E X A M P L E 8. 5
these times are not same. We can see that
Estimate the mass of air in your class water takes more time to reach 50°C than
room at NTP. Here NTP implies normal oil. It implies that water requires more heat
temperature (room temperature) and 1 energy to raise its temperature than oil. Now
atmospheric pressure. take twice the amount of water at 27°C and
heat it up to 50°C , note the time taken for
this rise in temperature. The time taken
by the water is now twice compared to the
previous case.
We can define ‘heat capacity’ as the
amount of heat energy required to raise
the temperature of the given body from
T to T + ∆T .
DQ
Heat capacity S =
DT
Solution
Specific heat capacity of a substance is defined
The average size of a class is 6m length, 5 as the amount of heat energy required to raise
m breadth and 4 m height. The volume of the temperature of 1kg of a substance by 1
the room V = 6 × 5 × 4 = 120m3. We can Kelvin or 1°C

∆Q = m s∆T
The term heat capacity
Therefore, Note
or specific heat capacity
1  ∆Q  does not mean that object
s=   contains a certain amount of heat.
m  ∆T 
Heat is energy transfer from the object
Where s is known as specific heat capacity of at higher temperature to the object at
a substance and its value depends only on lower temperature. The correct usage
the nature of the substance not amount of is ‘internal energy capacity’. But for
substance historical reason the term ‘heat capacity’
or ‘specific heat capacity’ are retained.
ΔQ = Amount of heat energy
ΔT = Change in temperature
m = Mass of the substance
When two objects of
The SI unit for specific heat capacity is
same mass are heated at
J kg-1 K-1. Heat capacity and specific heat
equal rates, the object
capacity are always positive quantities.
with smaller specific heat
capacity will have a faster temperature
Table 8.2 Specific heat capacity of increase.
some common substances at 1 atm (20°C) When two objects of same mass
are left to cool down, the temperature
Specific heat of the object with smaller specific heat
Material
capacity (Jkg−lK−1) capacity will drop faster.
Air 1005
Lead 130 When we study properties of gases, it is

more practical to use molar specific heat
Copper 390 capacity. Molar specific heat capacity is
Iron (steel) 450 defined as heat energy required to increase
the temperature of one mole of substance by
Glass 840 1K or 1°C. It can be written as follows
Aluminium 900
1  ∆Q 
C=  
Human body 3470 µ  ∆T 
Water 4186 Here C is known as molar specific heat

capacity of a substance and μ is number of
From the table it is clear that water has the moles in the substance.
highest value of specific heat capacity. For
this reason it is used as a coolant in power The SI unit for molar specific heat capacity
stations and reactors. is J mol-1 K-1 . It is also a positive quantity.
Un it 8 H E AT A N D T H E R MODY NA M IC S 101

Liquids, have less intermolecular forces
8.2.3 Thermal expansion
than solids and hence they expand more
of solids, liquids and gases
than solids. This is the principle behind the
Thermal expansion is the tendency of matter mercury thermometers.
to change in shape, area, and volume due to a In the case of gas molecules, the
change in temperature. intermolecular forces are almost negligible
All three states of matter (solid, liquid and and hence they expand much more than
gas) expand when heated. When a solid is solids. For example in hot air balloons when
heated, its atoms vibrate with higher amplitude gas particles get heated, they expand and
about their fixed points. The relative change take up more space.
in the size of solids is small. Railway tracks are The increase in dimension of a body due
given small gaps so that in the summer, the to the increase in its temperature is called
tracks expand and do not buckle. Railroad thermal expansion.
tracks and bridges have expansion joints to The expansion in length is called linear
allow them to expand and contract freely with expansion. Similarly the expansion in
temperature changes. It is shown in Figure 8.3 area is termed as area expansion and the
To
A
Lo L
Ao
T = To + T
L
L A
T T
Lo Ao
(a) Linear expansion (b) Area expansion (c) Vo
To
A
Lo L
Ao Vo
T = To + T
L V
L A V
T T T
Lo Ao Vo
(a) Linear expansion (b) Area expansion (c) Volume expansion
To
A
Lo L
Ao Vo
T = To + T
L V
L A V
T T T
Lo Ao Vo
(a) Linear expansion (b) Area expansion (c) Volume expansion
Figure 8.3 Expansion joints for safety Figure 8.4 Thermal expansions

expansion in volume is termed as volume
expansion. It is shown in Figure 8.4
Linear Expansion
In solids, for a small change in temperature
 
ΔT, the fractional change in length  ∆L  is
 L 
directly proportional to ΔT.
DL
= αLΔT
L
DL Solution
Therefore, αL = DL
LDT = αL ΔT
Where, αL = coefficient of linear expansion. L
ΔL = Change in length ΔL = αL L∆T
L = Original length ΔL = 10 × 10−6 × 300 × 23 = 0.69 m=69 cm

ΔT = Change in temperature.
Area Expansion
• When the lid of a For a small change in temperature ΔT
 ∆A 
glass bottle is tight, keep the fractional change in area   of a
  A 
the lid near the hot water
substance is directly proportional to ΔT and
which makes it easier to
it can be written as
open. It is because the lid has higher
thermal expansion than glass. DA
= αAΔT
• When the hot boiled egg is dropped A
in cold water, the egg shell can be DA
Therefore, αA =
removed easily. It is because of the ADT
different thermal expansions of the Where, αA = coefficient of area expansion.
shell and egg. ΔA = Change in area
A = Original area
E X A M P L E 8. 6 ΔT = Change in temperature
Eiffel tower is made up of iron and its height Volume Expansion
is roughly 300 m. During winter season For a small change in temperature ΔT the
(January) in France the temperature is 2°C  ∆V 
fractional change in volume   of a
and in hot summer its average temperature   V 
25°C. Calculate the change in height of substance is directly proportional to ΔT.
Eiffel tower between summer and winter. DV
The linear thermal expansion coefficient = αV ΔT
V
for iron α = 10 ×10−6 per °C ∆V
Therefore, αV =
V ∆T

Where, αV = coefficient of volume expansion.
ΔV = Change in volume y y
Volume of 1kg of water (cm3 )
Density of water (g cm-3)

V = Original volume 1000.35 1.0000
1000.30
ΔT = Change in temperature 1000.25

0.9999
Unit of coefficient of linear, area and

1000.20
0.9998
1000.15
volumetric expansion of solids is ˚C-1 or K-1 1000.10
0.9997
1000.05
1000.00 0.9996
x
For a given specimen, 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6
Note Temperatureº
Temperatureº C C Temperatu
(a)
= αL ΔT (Linear
expansion)
y y
Volume of 1kg of water (cm3 )
≈ 2 αL ΔT (Area expansion ≈ 2 ×
Density of water (g cm-3)

1000.35 1.0000
Linear expansion)
1000.30
0.9999
1000.25
≈ 3 αL ΔT (Volume expansion
1000.20
≈3
× Linear expansion)
1000.15
0.9998
1000.10
0.9997
1000.05
1000.00 0.9996 x
x
0 1 2 3 4 5 6 7 8 9 10
8.2.4 Anomalous expansion
Temperatureº
0 1 2 3 4 5 6 7 8 9 10
Temperatureº C C Temperatureº C
of water (b)
Liquids expand on heating and contract Figure 8.5 Anomalous Expansion of water
on cooling at moderate temperatures. But
water exhibits an anomalous behavior.
It contracts on heating between 0˚C and
4˚C. The volume of the given amount of on the top surface above the liquid water
water decreases as it is cooled from room (ice floats). This is due to the anomalous
temperature, until it reach 4˚C . Below expansion of water. As the water in lakes
4˚C the volume increases and so the density and ponds freeze only at the top the species
decreases. This means that the water has a living in the lakes will be safe at the bottom.
maximum density at 4˚C . This behavior
Summer winter
of water is called anomalous expansion of
water. It is shown in the Figure 8.5
In cold countries during the winter 8°C 0°C
season, the surface of the lakes will be at 7 1
6 2
lower temperature than the bottom as 5 3
4 4
shown in the Figure 8.6. Since the solid
water (ice) has lower density than its liquid Figure 8.6 Anomalous expansion of
form, below 4°C, the frozen water will be water in lakes

8.2.5 Change of state Q=m×L
Q
All matter exists normally in three states Therefore, L=
m
as solids, liquids or gases. Matter can be
changed from one state to another either by Where L = Latent heat capacity of the
heating or cooling. substance
Q = Amount of heat
Examples:
1. Melting (solid to liquid) m = mass of the substance
2. Evaporation (liquid to gas) The SI unit for Latent heat capacity is J kg-1 .
3. Sublimation (solid to gas)
4. Freezing / Solidification (liquid to solid)
Gas
5. Condensation (gas to liquid)
Liquid
Temperature
Solid
Gas
Melting Vaporization
Co
n
tio
nd
Energy
ma
en
Ev
ion
sat
bli
ap
sit
ion
Su
ora
po
Figure 8.8 Temperature versus heat for

De
tio
n
water
Melting
When heat is added or

Note
Solid
Solidification
Liquid removed during a change
of state, the temperature
Figure 8.7 Change of states of matter remains constant.
• The latent heat for a solid - liquid state

Latent heat capacity:
change is called the latent heat of fusion
While boiling a pot of water, the temperature (Lf )
of the water increases until it reaches 100 ˚C • The latent heat for a liquid - gas state
which is the boiling point of water, and then change is called the latent heat of
the temperature remains constant until all vaporization (Lv)
the water changes from liquid to gas. During • The latent heat for a solid - gas state change
this process heat is continuously added to is called the latent heat of sublimation (Ls)
the water. But the temperature of water does
not increase above its boiling point. This is Triple point
the concept of latent heat capacity. the triple point of a substance is the
Latent heat capacity of a substance is temperature and pressure at which the three
defined as the amount of heat energy required phases (gas, liquid and solid) of that substance
to change the state of a unit mass of the coexist in thermodynamic equilibrium.
material. The triple point of water is at 273.1 K and a
partial vapour pressure of 611.657 Pascal.

8.2.6 Calorimetry Thermometer Stirrer
Calorimetry means the measurement of

the amount of heat released or absorbed by
thermodynamic system during the heating
process. When a body at higher temperature Water
is brought in contact with another body at Sample
lower temperature, the heat lost by the hot Insulating

wood
body is equal to the heat gained by the
cold body. No heat is allowed to escape to Air (insulation) Calorimeter cup
the surroundings. It can be mathematically
expressed as
Figure 8.10 Calorimeter with sample
Qgain = −Qlost of block
Qgain + Qlost= 0 Qgain = −Qlost

Note the sign convention. The heat lost is
Heat gained or lost is measured with a denoted by negative sign and heat gained is
calorimeter. Usually the calorimeter is denoted as positive.
an insulated container of water as shown in From the definition of specific heat capacity
Figure 8.9.
Qgain =m2s2 (Tf – T2)
Thermometer Stirrer Qlost= m1s1 (Tf – T1)
Here s1 and s2 specific heat capacity of hot

sample and water respectively.
So we can write
m2s2 (Tf – T2) = − m1s1 (Tf – T1)
Water Insulating
wood m2s2Tf – m2s2T2= − m1s1Tf + m1s1T1
Air (insulation) Calorimeter cup

m2s2Tf + m1s1Tf = m2s2T2 + m1s1T1
The final temperature

Figure 8.9 Calorimeter
m1s1T1 + m2 s2T2
Tf =
m1s1 + m2 s2
A sample is heated at high temperature
(T1) and immersed into water at room
E X A M P L E 8. 7
temperature (T2) in the calorimeter. After
some time both sample and water reach a If 5 L of water at 50°C is mixed with 4L
final equilibrium temperature Tf . Since the of water at 30°C, what will be the final
calorimeter is insulated, heat given by the temperature of water? Take the specific
hot sample is equal to heat gained by the heat capacity of water as 4184 J kg-1 K-1.
water. It is shown in the Figure 8.10

Solution Conduction
We can use the equation Conduction is the process of direct transfer
of heat through matter due to temperature
m1s1T1 + m2 s2T2
Tf = difference. When two objects are in direct
m1s1 + m2 s2
contact with one another, heat will be
m1 = 5L = 5kg and m2= 4L = 4kg, s1 = s2 transferred from the hotter object to the colder
and T1=50°C =323K and T2 = 30°C=303 K. one. The objects which allow heat to travel
So easily through them are called conductors.
m1T1 + m2T2 5×323 + 4 ×303 Thermal conductivity

Tf = = =314.11 K
m1 + m2 5+4 Thermal conductivity is the ability to
Tf = 314.11 K-273K ≈ 41°C. conduct heat.
Suppose if we mix equal amount of water The quantity of heat transferred through a
(m1 = m2) with 50°C and 30°C, then unit length of a material in a direction normal
the final temperature is average of two to unit surface area due to a unit temperature
temperatures. difference under steady state conditions is
known as thermal conductivity of a material.
T1 + T2 323 + 303
Tf = = = 313K = 40°C
2 2 Material having
thermal conductivity K
Suppose if both the water are at 30°C then Area A
the final temperature will also 30°C. It
implies that they are at equilibrium and Q
T2 T1
no heat exchange takes place between each
d
other. T2 > T1
Figure 8.11 Steady state heat flow by

It is important to note
Note conduction.
that the final equilibrium
temperature of mixing of
gas or liquid depends on mass of the In steady state, the rate of flow of heat Q is
substances, their specific heat capacities proportional to the temperature difference
and their temperatures. Only if we mix ΔT and the area of cross section A and is
the same substances at equal amount, inversely proportional to the length L. So
the final temperature will be an average the rate of flow of heat is written as
of the individual temperatures. Q KA∆T
=
t L
Where, K is known as the coefficient of
8.2.7 Heat transfer
thermal conductivity.
As we have seen already heat is a energy in (Not to be confused with Kelvin represented
transit which is transferred from one body to by upper case K)
another body due to temperature difference.
There are three modes of heat transfer: The SI unit of thermal conductivity is J s-1
Conduction, Convection and Radiation. m-1 K-1 or W m-1 K-1.

Table 8.3: Thermal conductivities (in W m−1 K−1) of some materials at 1 atm
Thermal Thermal
Material Material
Conductivity Conductivity
Diamond 2300 Water 0.56
Silver 420 Human tissue 0.2
Copper 380 Wood 0.17
Aluminum 200 Helium 0.152
Steel 40 Cork 0.042
Glass 0.84 Air 0.023
Brick 0.84
Ice 2
aluminum have high thermal conductivities.

Steady state: So they are used to make cooking vessels.
Note
The state at which
Convection
temperature attains constant
value everywhere and there is no Convection is the process in which heat
further transfer of heat anywhere is transfer is by actual movement of molecules
called steady state. in fluids such as liquids and gases. In
convection, molecules move freely from
Thermal conductivity depends on the nature one place to another. It happens naturally or
of the material. For example silver and forcefully.
During the day, sun rays warm up the land more quickly than sea water. It
is because land has less specific heat capacity than water. As a result the air
above the land becomes less dense and rises. At the same time the cooler
air above the sea flows to land and it is called ‘sea breeze’. During the night
time the land gets cooled faster than sea due to the same reason (specific heat). The air
molecules above sea are warmer than air molecules above the land. So air molecules
above the sea are replaced by cooler air molecules from the land. It is called ‘land breeze’.
Day Day Night Night
Sea breeze Sea breeze

land breeze land breeze
Land cooler Land cooler
Land warmer Land warmer
than water than water
than water than water

Boiling water in a cooking pot is an
The parameter temperature
example of convection. Water at the bottom Note is generally thought to be
of the pot receives more heat. Due to
associated with matter
heating, the water expands and the density (solid, liquid and gas).
of water decreases at the bottom. Due to this But radiation is also considered as a
decrease in density, molecules rise to the thermodynamic system which has
top. At the same time the molecules at the well defined temperature and pressure.
top receive less heat and become denser and The visible radiation coming from
come to the bottom of the pot. This process the Sun is at the temperature of 5700 K
goes on continuously. The back and forth and the Earth re emits the radiation in
movement of molecules is called convection the infrared range into space which is
at a temperature of around 300K.
current.
To keep the room warm, we use room
heater. The air molecules near the heater 8.2.8 Newton’s law of
will heat up and expand. As they expand, cooling
the density of air molecules will decrease
Newton’s law of cooling states that the rate of
and rise up while the higher density cold
loss of heat of a body is directly proportional
air will come down. This circulation of air
to the difference in the temperature between
molecules is called convection current.
that body and its surroundings .
Radiation: dQ
∝ −(T− Ts ) (8.4)
When we keep our hands near the hot stove dt
we feel the heat even though our hands The negative sign indicates that the quantity
are not touching the hot stove. Here heat of heat lost by liquid goes on decreasing
transferred from the hot stove to our hands with time. Where,
is in the form of radiation. We receive energy T = Temperature of the object
from the sun in the form of radiations. Ts = Temperature of the surrounding
These radiations travel through vacuum and
reach the Earth. It is the peculiar character T
100
of radiation which requires no medium to 90
transfer energy from one object to another. 80

Temperature(°C)
70
The conduction or convection requires 60
medium to transfer the heat. 50
40
Radiation is a form of energy transfer 30

TS
from one body to another by electromagnetic
20
10
waves. 0
t
30 60 90 120 150 180 210 240 270 300
Time (seconds)
Example:
1. Solar energy from the Sun. Figure 8.12 Cooling of hot water with time
2. Radiation from room heater. From the graph in Figure 8.12 it is clear
that the rate of cooling is high initially and
decreases with falling temperature.

Let us consider an object of mass m and above room temperature. Using equation
specific heat capacity s at temperature T. Let (8.8)
Ts be the temperature of the surroundings. If dT a
= − dt or = dT = − a (T − T )
the temperature falls by a small amount dT T − Ts ms dt ms
s
in time dt, then the amount of heat lost is,

8o C a
= − (61o C )
dQ = msdT (8.5) 3 min ms
Similarly the average temperature of
Dividing both sides of equation (8.5) by dt 65°C and 60°C is 62.5°C. The average
dQ msdT temperature is 35.5°C above the room
= (8.6)
dt dt temperature. Then we can write
From Newton’s law of cooling 5o C a
= − (35.5o C )
dQ dt ms
∝ −(T − Ts )
dt
By diving both the equation, we get
dQ
= - a (T -Ts)(8.7) 8o C a
dt - (61o C )
Where a is some positive constant. 3 min = ms
5o C a
From equation (8.6) and (8.7) - (35.5o C )
dt ms
dT 8×dt 61
- a (T -Ts) = ms =
dt 3×5 35.5
dT a
= − dt(8.8)
T − Ts ms 61×15 915
dt = = = 3.22 min
Integrating equation (8.8) on both sides, 35.5×8 284
∞ dT t a
∫ T − Ts
= −∫
0 ms
dt
8.3
0
ln (T -Ts) = −
a
t + b1 LAWS OF HEAT
ms TRANSFER
Where b1 is the constant of integration.
taking exponential both sides, we get
8.3.1 Prevost theory of heat
a
T = Ts + b2 e
-
ms
t
(8.9) exchange
here b2 = eb1 = constant Every object emits heat radiations at all
finite temperatures (except 0 K) as well as it
E X A M P L E 8. 8 absorbs radiations from the surroundings.
A hot water cools from 92°C to 84°C in For example, if you touch someone, they
3 minutes when the room temperature is might feel your skin as either hot or cold.
27°C. How long will it take for it to cool A body at high temperature radiates
from 65°C to 60°C? more heat to the surroundings than it
The hot water cools 8°C in 3 minutes. The receives from it. Similarly, a body at a lower
average temperature of 92°C and 84°C is temperature receives more heat from the
88°C. This average temperature is 61°C surroundings than it loses to it.

Prevost applied the idea of ‘thermal radiations have different wavelengths and
equilibrium’ to radiation. He suggested that all the emitted wavelengths will not have
all bodies radiate energy but hot bodies equal intensity.
radiate more heat than the cooler bodies. Wien’s law states that, the wavelength of
At one point of time the rate of exchange maximum intensity of emission of a black
of heat from both the bodies will become body radiation is inversely proportional to
the same. Now the bodies are said to be in the absolute temperature of the black body.
‘thermal equilibrium’.
1 b
Only at absolute zero temperature a body λm ∝ (or ) λ m = (8.12)
T T
will stop emitting. Therefore Prevost theory
states that all bodies emit thermal radiation Where, b is known as Wien’s constant. Its
at all temperatures above absolute zero value is 2.898× 10-3 m K
irrespective of the nature of the surroundings. It implies that if temperature of the body
increases, maximal intensity wavelength
( λ m ) shifts towards lower wavelength
8.3.2 Stefan Boltzmann law (higher frequency) of electromagnetic
Stefan Boltzmann law states that, the spectrum. It is shown in Figure 8.13
total amount of heat radiated per second Graphical representation
per unit area of a black body is directly
proportional to the fourth power of its 5500k
absolute temperature. 8 x 107
Radiation intensity (kj/m3 nm)
E ∝ T4 or E = σ T4 (8.10)
Where, σ is known as Stefan’s constant. Its 6 x 107 5000k
value is 5.67 × 10−8 W m−2 k−4

4 x 107 4500k
If a body is not a perfect

Note 4000k
black body, then
2 x 107 3500k
E = e σ T4 (8.11) 0 500 1000 1500 2000
Where ‘e’ is emissivity of surface. Wavelength (nm)

Emissivity is defined as the ratio of the
energy radiated from a material’s surface Figure 8.13 Wien’s displacement law
to that radiated from a perfectly black body
at the same temperature and wavelength. From the graph it is clear that the peak of
the wavelengths is inversely proportional to
temperature. The curve is known as ‘black
8.3.3 Wien’s displacement body radiation curve’.
law
Wien’s law and Vision:
In the universe every object emits radiation.
Why our eye is sensitive to only visible
The wavelengths of these radiations depend
wavelength (in the range 400 nm to 700nm)?
on the object’s absolute temperature. These

The Sun is approximately taken as a black The humans evolved under the Sun by
body. Since any object above 0 K will emit receiving its radiations. The human eye is
radiation, Sun also emits radiation. Its sensitive only in the visible not in infrared
surface temperature is about 5700 K. By or X-ray ranges in the spectrum.
substituting this value in the equation (8.12), Suppose if humans had evolved in a planet
b 2.898×10−8 near the star Sirius (9940K), then they would
λm = = ≈ 508 nm have had the ability to see the Ultraviolet
T 5700
rays!
It is the wavelength at which maximum
intensity is 508nm. Since the Sun’s E X A M P L E 8. 9
temperature is around 5700K, the spectrum
of radiations emitted by Sun lie between 400 The power radiated by a black body A is
nm to 700 nm which is the visible part of the EA and the maximum energy radiated was
spectrum. It is shown in Figure 8.14 at the wavelength λA. The power radiated
by another black body B is EB = N EA and
the radiated energy was at the maximum
400 nm
0.0001 nm 0.01 nm
λ max
Gamma rays
1
wavelength, λA. What is the value of N?
2
According to Wien’s displacement law
8 x 107 λmax T = constant for both object A and B
Radiation intensity (kj/m3 nm)
X-rays
1
VISIBLE SPECTRUM
λA TA = λB TB. Here λB = λA
500 nm
10 nm
2
Ultra-
6 x 107
violet
Visible light
The sun TB λ A
5700 k = = 1 =2
1000 nm
TA λ B 1
Infraed
4 x 107
2
600 nm
0.01 cm
TB = 2TA
2 x 107
From Stefan-Boltzmann law 1 cm
Rader TV FM
Radio waves
4
EB  TB 
1m
700 nm
4
0 400 500 600 700 =   = (2) = 16 = N
Wavelenght (nm) E A TA 
100 m
AM
400 nm
0.0001 nm 0.01 nm
Gamma rays
max
Object B has emitted at lower wavelength
compared to A. So the object B would have
emitted more energetic radiation than A.
X-rays
VISIBLE SPECTRUM
500 nm
10 nm
Ultra-
violet
Visible light
The sun
5700 k 8.4
1000 nm
Infraed
THERMODYNAMICS:
600 nm
0.01 cm
1 cm
Rader TV FM
Radio waves
8.4.1 Introduction
1m
700 nm
0 600 700
nght (nm) In the previous sections we have studied
100 m
AM
Figure 8.14 Wien’s law and Human’s about the heat, temperature and thermal
vision properties of matter. Thermodynamics is

a branch of physics which describes the Examples:
laws governing the process of conversion A thermodynamic system can be liquid,
of work into heat and conversion of heat solid, gas and radiation.
into work. The laws of thermodynamics
are formulated over three centuries of Thermodynamic Surrounding
experimental works of Boyle, Charles, system
Bernoulli, Joule, Clausius, Kelvin, Carnot
Bucket of water Open atmosphere
and Helmholtz. In our day to day life,
the functioning of everything around Air molecules in the Outside air
us and even our body is governed by room
the laws of thermodynamics. Therefore Human body Open atmosphere
thermodynamics is one of the essential
branches of physics. Fish in the sea Sea of water
We can classify thermodynamics system

Thermodynamic system:
into three types: It is given in Figure 8.16
A thermodynamic system is a finite part
of the universe. It is a collection of large
number of particles (atoms and molecules) Thermodynamic system
specified by certain parameters called Open system can exchange both matter
and energy with the environment.
pressure (P), Volume (V) and Temperature Closed system exchange energy but not
(T). The remaining part of the universe is

matter with the environment.
Isolated system can exchange neither

called surrounding. Both are separated by a energy nor matter with the enviroment. Open Closed
Isolated
boundary. It is shown in Figure 8.15

(a) Open (a) Closed (a) Isolated
Matter Matter Matter
Energy
Energy Energy
Surroundings
Figure 8.16 Different types of
thermodynamic systems
System
8.4.2 Thermal equilibrium
When a hot cup of coffee is kept in the room,
heat flows from coffee to the surrounding
Boundary air. After some time the coffee reaches the
same temperature as the surrounding air
and there will be no heat flow from coffee
to air or air to coffee. It implies that the
Figure 8.15 Thermodynamic system
coffee and surrounding air are in thermal
equilibrium with each other.

Two systems are said to be in thermal each other. In a state of thermodynamic
equilibrium with each other if they are at the equilibrium the macroscopic variables such
same temperature, which will not change as pressure, volume and temperature will have
with time. fixed values and do not change with time.
Mechanical equilibrium:
Consider a gas container with piston as
8.4.3 Thermodynamic state
shown in Figure 8.17. When some mass is
variables
placed on the piston, it will move downward
due to downward gravitational force and In mechanics velocity, momentum and
after certain humps and jumps the piston acceleration are used to explain the state of
will come to rest at a new position. When any moving object (which you would have
the downward gravitational force given by realized in Volume 1). In thermodynamics,
the piston is balanced by the upward force the state of a thermodynamic system is
exerted by the gas, the system is said to represented by a set of variables called
be in mechanical equilibrium. A system thermodynamic variables.
is said to be in mechanical equilibrium if Examples: Pressure, temperature, volume
no unbalanced force acts on the thermo and internal energy etc.
dynamic system or on the surrounding by The values of these variables completely
thermodynamic system. describe the equilibrium state of a
thermodynamic system. Heat and work are
Mechanical equilibrium not state variables rather they are process
With out masses With masses
variables.
There are two types of thermodynamic
variables: Extensive and Intensive
Moveable Masses Extensive variable depends on the size or
piston
Moveable mass of the system.
Gas piston
Example: Volume, total mass, entropy,
Gas internal energy, heat capacity etc.
Intensive variables do not depend on the
size or mass of the system.
Figure 8.17 Mechanical equilibrium Example: Temperature, pressure, specific
heat capacity, density etc.
Chemical equilibrium: Equation of state:

If there is no net chemical reaction between The equation which connects the state
two thermodynamic systems in contact with variables in a specific manner is called
each other then it is said to be in chemical equation of state. A thermodynamic
equilibrium. equilibrium is completely specified by these
Thermodynamic equilibrium: state variables by the equation of state. If the
If two systems are set to be in thermodynamic system is not in thermodynamic equilibrium
equilibrium, then the systems are at thermal, then these equations cannot specify the state
mechanical and chemical equilibrium with of the system.

An ideal gas obeys the equation PV = NkT then A and B are in thermal equilibrium
at thermodynamic equilibrium. Since all with each other.
four macroscopic variables (P,V,T and N) Consider three systems A, B and C which are
are connected by this equation, we cannot initially at different temperatures. Assume
change one variable alone. For example, if that A and B are not in thermal contact with
we push the piston of a gas container, the each other as shown in Figure 8.18 (a) but
volume of the gas will decrease but pressure each of them is in thermal contact with a
will increase or if heat is supplied to the gas, third system C. After a lapse of time, system
its temperature will increase, pressure and A will be in thermal equilibrium with C and
volume of the gas may also increase. B also will be in thermal equilibrium with
There is another example of equation of C. In this condition, if the systems A and
state called van der Waals equation. Real B are kept in thermal contact as shown in
gases obey this equation at thermodynamic the Figure 8.18 (b), there is no flow of heat
equilibrium. The air molecules in the room between the systems A and B. It implies
truly obey van der Waals equation of state. that the system A and B are also in thermal
But at room temperature with low density equilibrium with each other. Once the three
we can approximate it into an ideal gas. systems are at thermal equilibrium, there
will be no heat flow between them as they
are at the same temperature. This can be
8.5 mathematically expressed as
ZEROTH LAW OF If TA = TC and TB = TC, it implies that TA = TB,
THERMODYNAMICS where TA, TB and TC are the temperatures
of the systems A, B, and C respectively.
The zeroth law of thermodynamics states
Temperature is the property which
that if two systems, A and B, are in thermal
determines whether the system is in
equilibrium with a third system, C,
A B A B
(a) (b)
Figure 8.18 (a) Two systems A and B in thermal contact with object C separately (b) If
systems A and B are in thermal contact, they are also in thermal equilibrium with each
other.

thermal equilibrium with other systems or
8.6
not. Zeroth law enables us to determine
INTERNAL ENERGY (U)
the temperature. For example, when a
thermometer is kept in contact with a human The internal energy of a thermodynamic
body, it reaches thermal equilibrium with system is the sum of kinetic and potential
the body. At this condition, the temperature energies of all the molecules of the system with
of the thermometer will be same as the respect to the center of mass of the system.
human body. This principle is used in The energy due to molecular motion
finding the body temperature. including translational, rotational and
vibrational motion is called internal kinetic
Activity
energy (EK)
We often associate the temperature as a The energy due to molecular interaction
measure of how hot or cold an object is is called internal potential energy (EP).
while touching it. Can we use our sensory Example: Bond energy.
organs to determine the temperature of U = EK + EP
an object? � Since ideal gas molecules are assumed to
When you stand bare feet with one
have no interaction with each other the
foot on the carpet and the other on the
internal energy consists of only kinetic
tiled floor, your foot on tiled floor feels
energy part (EK) which depends on the
cooler than the foot on the carpet even
temperature, number of particles and is
though both the tiled floor and carpet
independent of volume. However this is not
are at the same room temperature. It is
true for real gases like Van der Waals gases.
because the tiled floor transfers the heat � Internal energy is a state variable. It
energy to your skin at higher rate than
depends only on the initial and final
the carpet. So the skin is not measuring
states of the thermodynamic system. For
the actual temperature of the object;
example, if the temperature of water is
instead it measures the rate of heat energy
raised from 30°C to 40°C either by heating
transfer. But if we place a thermometer
or by stirring, the final internal energy
on the tiled floor or carpet it will show
depends only on the final temperature
the same temperature.
40°C and not the way it is arrived at.
It is very important to note
Note that the internal energy of
a thermodynamic system is
associated with only the kinetic
energy of the individual molecule due
to its random motion and the potential
energy of molecules which depends on
their chemical nature. The bulk kinetic
energy of the entire system or gravitational
Carpet potential energy of the system should not
Tiled floor
be mistaken as a part of internal energy.
For example

(a) Consider two gas containers
Note at the same temperature having Hot object
the same internal energy, one
is kept at rest on the ground Internal energy
and the other is kept in a moving train. Decreases
Even though the gas container in the train
is moving with the speed of the train, the
internal energy of the gas in it will not Thermal Heat flows
increase. contact
(b) Consider two gas containers at the
same temperature having the same
internal energy, one is kept on the Cold object
ground and the other is kept at some
height h. Even though the container at Internal energy
height h is having higher gravitational Increases
potential energy, this has no influence
on internal energy of the gas molecules.
E X A M P L E 8. 1 0 It is to be noted that heat

Note
does not always increases
When you mix a tumbler of hot water with
the internal energy. Later
one bucket of normal water, what will be
we shall see that in ideal gases during
the direction of heat flow? Justify.
isothermal process the internal energy
The water in the tumbler is at a higher
will not increase even though heat
temperature than the bucket of normal
flows in to the system.
water. But the bucket of normal water has
larger internal energy than the hot water
in the tumbler. This is because the internal
8.6.1 Joule’s mechanical
energy is an extensive variable and it
equivalent of heat
depends on the size or mass of the system.
Even though the bucket of normal water The temperature of an object can be increased
has larger internal energy than the tumbler by heating it or by doing some work on it.
of hot water, heat will flow from water in In the eighteenth century, James Prescott
the tumbler to the water in the bucket. Joule showed that mechanical energy can
This is because heat flows from a body at be converted into internal energy and vice
higher temperature to the one at lower versa. In his experiment, two masses were
temperature and is independent of internal attached with a rope and a paddle wheel as
energy of the system. shown in Figure 8.19. When these masses fall
Once the heat is transferred to an object through a distance h due to gravity, both the
it becomes internal energy of the object. masses lose potential energy equal to 2mgh.
The right way to say is ‘object has certain When the masses fall, the paddle wheel
amount of internal energy’. Heat is one of turns. Due to the turning of wheel inside
the ways to increase the internal energy of water, frictional force comes in between
a system as shown in the Figure. the water and the paddle wheel. This causes

mechanical energy, but this is not so with
heat as it is not a quantity. However this
terminology is retained for historical
Thermometer
reasons. A correct name would be ‘Joule’s
m h mechanical equivalence of internal
Measured m mg energy’. Joule Essentially converted
height of mechanical energy to internal energy.
descent
mg
Insulated In his experiment potential energy is
container of water
converted to rotational kinetic energy of
paddle wheel and this rotational kinetic
Figure.8.19 Joule’s experiment for energy is converted to internal energy of
determining the mechanical equivalent water.
of heat energy.
a rise in temperature of the water. This

E X A M P L E 8. 11
implies that gravitational potential energy is
converted to internal energy of water. The A student had a breakfast of 200 food
temperature of water increases due to the calories. He thinks of burning this energy
work done by the masses. In fact, Joule was by drawing water from the well and
able to show that the mechanical work has watering the trees in his school. Depth of
the same effect as giving heat. He found that the well is about 25 m. The pot can hold
to raise 1 g of an object by 1°C , 4.186 J of 25L of water and each tree requires one pot
energy is required. In earlier days the heat of water. How many trees can he water?
was measured in calorie. (Neglect the mass of the pot and the energy
spent by walking. Take g =10 m s-2)
1 cal = 4.186 J
This is called Joule’s mechanical equivalent
of heat.
Before James Prescott Joule,

Note people thought that heat was
a kind of fluid called caloric
fluid which flows from an
object at high temperature to that in Solution
low temperature. According to caloric To draw 25 L of water from the well, the
fluid idea, the hot object contains more
student has to do work against gravity by
caloric fluid and the cold object contains
less caloric fluid since heat was treated burning his energy.
as a quantity. Now we understand that Mass of the water = 25 L = 25 kg (1L=1kg )
heat is not a quantity but it is an energy
in transit. So the word ‘mechanical The work required to draw 25 kg of water
equivalent of heat’ is wrong terminology. = gravitational potential energy gained by
Because mechanical energy is a quantity water.
and any object can contain more or less
W = mgh = 25×10×25 = 6250 J

The total energy gained from the food =
200 food cal =200 kcal. ∆U=Q-W (8.13)
= 200×103 × 4.186 J = 8.37 ×105 J The internal energy of a thermodynamic

If we assume that by using this energy the system can be changed either by heating or
student can drawn ‘n’ pots of water from the by work as shown below.
well, the total energy spent by him = 8.37 ×
Heat flows into the Internal energy
105 J = nmgh
system increases
8.37 ´105 J
n= ≈ 134.
6250 J Heat flows out of the Internal energy
This n is also equal to the number of trees system decreases
that he can water. Work is done on the Internal energy
Is it possible to draw 134 pots of water system increases
from the well just by having breakfast? No.
Actually the human body does not convert Work is done by the Internal energy
entire food energy into work. It is only system decreases
approximately 20% efficient. It implies that
Based on the above table the sign
only 20% of 200 food calories is used to draw
convention is introduced to use first law
water from the well. So 20 % of the 134 is
of thermodynamics appropriately. It is
only 26 pots of water. It is quite meaningful.
shown in the following table and the
So he can water only 26 trees.
Figure 8.20.
The remaining energy is used for blood
circulation and other functions of the body.
It is to be noted that some energy is always System gains heat Q is positive
‘wasted’. Why is it that the body cannot have System loses heat Q is negative
100% efficiency? You will find the answer in
section 8.9 Work done on the W is negative
system
8.6.2 First law of
thermodynamics Work done by the W is positive
The first law of thermodynamics is a system
statement of the law of conservation of energy.
In Newtonian mechanics conservation Q > 0, W > 0 Q >0, W < 0 Q < 0, W < 0 Q < 0, W > 0
of energy involves kinetic and potential
energies of bulk objects. But the first law of
thermodynamics includes heat also. This law
states that ‘Change in internal energy (ΔU)
of the system is equal to heat supplied to the
system (Q) minus the work done by the system
(W) on the surroundings’. Mathematically it Figure 8.20 The Sign convention for
is written as heat and work

Even though we often explain first law of Solution
thermodynamics using gases, this law is
universal and applies to liquids and solids also.
Some book presents the

Note first law of thermodynamics
as ∆U = Q + W. Here the
work done by the system is taken
as negative and work done on the
system is positive. However both the
conventions are correct and we can
follow any one of the convention. Work done by the system (body),
W = +500 kJ
Heat released from the system (body),
E X A M P L E 8. 1 2
Q = –230 kJ
A person does 30 kJ work on 2 kg of water The change in internal energy of a body
by stirring using a paddle wheel. While = ΔU= – 230 kJ – 500 kJ = – 730 kJ
stirring, around 5 kcal of heat is released
from water through its container to the
surface and surroundings by thermal 8.6.3 Quasi-static process
conduction and radiation. What is the Consider a system of an ideal gas kept in
change in internal energy of the system? a cylinder of volume V at pressure P and
temperature T. When the piston attached
Solution
to the cylinder moves outward the volume
Work done on the system (by the person of the gas will change. As a result the
while stirring), W = -30 kJ = -30,000J temperature and pressure will also change
Heat flowing out of the system, because all three variables P,T and V are
Q = -5 kcal = 5 × 4184 J =-20920 J related by the equation of state PV = NkT. If
Using First law of thermodynamics a block of some mass is kept on the piston,
∆U = Q-W it will suddenly push the piston downward.
∆U = -20,920 J-(-30,000) J The pressure near the piston will be larger
than other parts of the system. It implies
∆U = -20,920 J+30,000 J = 9080 J
that the gas is in non-equilibrium state. We
Here, the heat lost is less than the work
cannot determine pressure, temperature or
done on the system, so the change in
internal energy of the system until it reaches
internal energy is positive.
another equilibrium state. But if the piston is
pushed very slowly such that at every stage
E X A M P L E 8. 1 3 it is still in equilibrium with surroundings,
Jogging every day is good for health. Assume we can use the equation of state to calculate
that when you jog a work of 500 kJ is done the internal energy, pressure or temperature.
and 230 kJ of heat is given off. What is the This kind of process is called quasi-static
change in internal energy of your body? process.

A quasi-static process is an infinitely The small work done by the gas on the
slow process in which the system changes piston
its variables (P,V,T) so slowly such that
it remains in thermal, mechanical and dW = Fdx(8.14)
chemical equilibrium with its surroundings
The force exerted by the gas on the piston
throughout. By this infinite slow variation,
F = PA. Here A is area of the piston and P
the system is always almost close to
is pressure exerted by the gas on the piston.
equilibrium state.
E X A M P L E 8. 1 4
Give an example of a quasi-static process.
Consider a container of gas with volume
V, pressure P and temperature T. If we add
sand particles one by one slowly on the top
of the piston, the piston will move inward
very slowly. This can be taken as almost dx
a quasi-static process. It is shown in the
Gas Gas
figure
Sand
Figure 8.21 Work done by the gas
Equation (8.14) can be rewritten as
dW = PA dx(8.15)
Sand particles added slowly- quasi-static
process But Adx = dV= change in volume during
this expansion process.
So the small work done by the gas during
8.6.4 Work done in volume the expansion is given by
changes
dW = PdV(8.16)
Consider a gas contained in the cylinder
fitted with a movable piston. Suppose the Note here that is positive since the volume
gas is expanded quasi-statically by pushing is increased. Here, is positive.
the piston by a small distance dx as shown In general the work done by the gas by
in Figure 8.21. Since the expansion occurs increasing the volume from Vi to Vf is given by
quasi-statically the pressure, temperature
and internal energy will have unique values Vf
at every instant.
W= ò Vi
PdV (8.17)

Suppose if the work is done on the system, E X A M P L E 8. 15
then Vi > Vf . Then, W is negative. A gas expands from volume 1m3 to 2m3 at
Note here the pressure P is inside the integral constant atmospheric pressure.
in equation (8.17). It implies that while the (a) Calculate the work done by the gas.
system is doing work, the pressure need (b) Represent the work done in PV
not be constant. To evaluate the integration diagram.
we need to first express the pressure as a
function of volume and temperature using Solution
the equation of state.
(a)
The pressure P = 1 atm = 101 kPa,
Vf =2 m3 and Vi = 1m3
V f
8.6.5 PV diagram From equation (8.17) W = ò PdV = P

Vi
Vf
PV diagram is a graph between pressure ò dV
Vi
P and volume V of the system. The P-V Since P is constant. It is taken out of the
diagram is used to calculate the amount of integral.
work done by the gas during expansion or
on the gas during compression. In Unit 2, W = P (Vf – Vi) = 101×103 × (2 – 1) = 101 kJ
we have seen that the area under the curve
will give integration of the function from (b)
Since the pressure is kept constant, PV
lower limit to upper limit. The area under diagram is straight line as shown in the
the PV diagram will give the work done figure. The area is equal to work done
during expansion or compression which is by the gas.
shown in Figure 8.22
P
P
(Pi , Vi) 101 kPa
Area = W
W = 101 kJ
(Pf , Vf)
1m3 2m3 V
Vf

W = Area =
Vi
PdV
Note the arrow mark in the curve. Suppose
the work is done on the system, then
volume will decreases and the arrow will
V point in the opposite direction.
Figure 8.22 Work done by the gas during

8.7
expansion
SPECIFIC HEAT CAPACITY
OF A GAS
The shape of PV diagram depends on the
Specific heat capacity of a given system
nature of the thermodynamic process.
plays a very important role in determining

the structure and molecular nature of the is called specific heat capacity at constant
system. Unlike solids and liquids, gases have volume.
two specific heats: specific heat capacity If the volume is kept constant, then the
at constant pressure (sp) and specific heat supplied heat is used to increase only the
capacity at constant volume (sv). internal energy. No work is done by the gas
as shown in Figure 8.24.
8.7.1 Specific heat capacity

Insulator
Specific heat capacity at constant pressure
(sp):
The amount of heat energy required to raise
the temperature of one kg of a substance by
1 K or 1°C by keeping the pressure constant
is called specific heat capacity of at constant Pin
pressure. When the heat energy is supplied
to the gas, it expands to keep the pressure Gas
constant as shown in Figure 8.23 Q
Conductor
Insulator
Gas
Figure 8.24 Specific heat capacity at
Q
Conductor
constant volume
It implies that to increase the temperature

of the gas at constant volume requires less
heat than increasing the temperature of the
Figure 8.23 Specific heat capacity at
gas at constant pressure. In other words sp is
constant pressure
always greater than sv.
In this process a part of the heat energy Molar Specific heat capacities
is used for doing work (expansion) and Sometimes it is useful to calculate the molar
the remaining part is used to increase the heat capacities Cp and Cv. The amount of
internal energy of the gas. heat required to raise the temperature of one
Specific heat capacity at constant volume mole of a substance by 1K or 1°C at constant
(sv): volume is called molar specific heat capacity
The amount of heat energy required to raise at constant volume (Cv). If pressure is kept
the temperature of one kg of a substance by constant, it is called molar specific heat
1 K or 1°C by keeping the volume constant capacity at constant pressure (Cp).

If Q is the heat supplied to mole of a gas Q = µCpdT(8.22)
at constant volume and if the temperature
changes by an amount DT , we have If W is the workdone by the gas in this
process, then
Q = µCvDT.(8.18)
W = PdV (8.23)
By applying the first law of thermodynamics
for this constant volume process (W=0, But from the first law of thermodynamics,
since dV=0), we have
Q = dU + W (8.24)
Q = DU - 0 (8.19)
Substituting equations (8.21), (8.22) and
By comparing the equations (8.18) and (8.23) in (8.24), we get,
(8.19),
µCpdT = µCv dT + PdV (8.25)
1 DU
DU = µCvDT or Cv =
µ DT For mole of ideal gas, the equation of state
If the limit DT goes to zero, we can write is given by
1 dU PV = µRT ⇒ PdV+VdP = µRdT (8.26)

Cv = (8.20)
µ dT
Since the temperature and internal energy Since the pressure is constant, dP=0
are state variables, the above relation holds ∴CpdT = CvdT +RdT
true for any process.
∴CP = Cv +R (or) Cp - Cv = R (8.27)
This relation is called Meyer’s relation

8.7.2 Meyer’s relation
It implies that the molar specific heat
Consider µ mole of an ideal gas in a container capacity of an ideal gas at constant pressure
with volume V, pressure P and temperature is greater than molar specific heat capacity
T. at constant volume.
When the gas is heated at constant The relation shows that specific heat at
volume the temperature increases by dT. constant pressure (sp) is always greater that
As no work is done by the gas, the heat that specific heat at constant volume (sv).
flows into the system will increase only the
internal energy. Let the change in internal
8.8
energy be dU.
THERMODYNAMIC
If Cv is the molar specific heat capacity at
PROCESSES
constant volume, from equation (8.20)
dU = µCvdT(8.21) 8.8.1 Isothermal process

Suppose the gas is heated at constant pressure It is a process in which the temperature
so that the temperature increases by dT. remains constant but the pressure and
If ‘Q’ is the heat supplied in this process and volume of a thermodynamic system will
‘dV’ the change in volume of the gas. change. The ideal gas equation is

PV = µRT For an isothermal process, the first law of
Here, T is constant for this process thermodynamics can be written as follows,
So the equation of state for isothermal Q = W(8.30)

process is given by
From equation (8.30), we infer that the heat
PV = constant(8.28)
supplied to a gas is used to do only external
This implies that if the gas goes from work. It is a common misconception that
one equilibrium state (P1,V1) to another when there is flow of heat to the system, the
equilibrium state (P2,V2) the following temperature will increase. For isothermal
relation holds for this process process this is not true. The isothermal
compression takes place when the piston
P1V1 = P2V2(8.29) of the cylinder is pushed. This will increase
the internal energy which will flow out of
Since PV = constant, P is inversely the system through thermal contact. This is
1
proportional toV (P ∝ ). This implies shown in Figure 8.26.
V
that PV graph is a hyperbola. The pressure-
volume graph for constant temperature is Isothermal expansion
Q > 0, W > 0
Isothermal compression
Q < 0, W < 0
also called isotherm.

Figure 8.25 shows the PV diagram for
quasi-static isothermal expansion and
quasi-static isothermal compression.
Thermal
We know that for an ideal gas the internal contact
energy is a function of temperature only. For

an isothermal process since temperature
is constant, the internal energy is also Figure 8.26 Isothermal expansion and
constant. This implies that dU or DU = 0. isothermal compression
P P
A(Pi , Vf) A(Pf , Vf)
Pi Pf

T

T
Isothermal Isothermal
=
=
co
co
Expansion Compression
ns
ns
t
t
Pf B(Pf , Vf) Pi B(Pi , Vi)
Vi
V Vf Vf V Vi
(a) (b)
Figure 8.25 (a) Quasi-static isothermal expansion (b) Quasi-static isothermal
compression

Examples: Vf
(i) When water is heated, at the boiling Since we have an isothermal expansion,
Vi
V f 
point, even when heat flows to water, the 
> 1, so ln   > 0. As a result the work done
temperature will not increase unless the  Vi 
water completely evaporates. Similarly, by the gas during an isothermal expansion is
at the freezing point, when the ice melts positive.
to water, the temperature of ice will not The above result in equation (8.34) is true
increase even when heat is supplied to ice. for isothermal compression also. But in an
(ii)
All biological processes occur at constant Vf V 
body temperature (37°C). isothermalcompression <1.so ln  f  <0.
Vi  Vi 
Work done in an isothermal process: As a result the work done on the gas in an
Consider an ideal gas which is allowed isothermal compression is negative.
to expand quasi-statically at constant In the PV diagram the work done during
temperature from initial state (Pi,Vi) to the the isothermal expansion is equal to the area
final state (Pf , Vf ). We can calculate the work under the graph as shown in Figure 8.27
done by the gas during this process. From
equation (8.17) the work done by the gas, P P
Pi Pf
Shaded area = work done Shaded are
Vf during isothermal expansion during isot
W = ò PdV(8.31)
T

T
=
Isothermal
=
Vi
co
co
Expansion
ns
ns
t
t
As the process occurs quasi-statically, at
Pf B(Pf , Vf) Pi
every stage the gas is at equilibrium with
the surroundings. Since it is in equilibrium
at every stage the ideal gas law is valid. Vi
V Vf Vf V
Writing pressure in terms of volume and (a) (b)
temperature,
µRT P P
P= P
A(P , V ) (8.32) i f
Pf
A(Pf , Vf)
V i
Shaded area = work done Shaded area = work done
during isothermal expansion during isothermal compression
Substituting equation (8.32) in (8.31) we get

T

T
=
co
co
V µRT Expansion Compression

ns
ns
f
W= ò dV
t
t
Vi V Pf B(Pf , Vf) Pi B(Pi , Vi)

Vf dV
W = µRT ò (8.33)
Vi V
In equation (8.33), we take µRTV
out of the V
V i f
Vf V Vi
(b)
integral, since it is constant throughout
(a) the
isothermal process. Figure 8.27 Work done in an isothermal
By performing the integration in process.
equation (8.33), we get
V f  Similarly for an isothermal compression,
W = µRT In   (8.34)
 Vi  the area under the PV graph is equal to the

work done on the gas which turns out to be Therefore, Q = W = 1.369 kJ. Thus Q is also
the area with a negative sign. positive which implies that heat flows in to
the system.
To calculate the work done (c) For an isothermal process
Note in an isothermal process, we
assume that the process is PiVi = PfVf = µRT
quasi-static. If it is not quasi-static we µRT 8.31J 300K
Pf = = 0.5mol × ×
cannot substitute P = in equation Vf mol.K 6×10−3 m3
(8.31). It is because the ideal gas law =207.75 k Pa
is not valid for non equilibrium states.
But equation (8.34) is valid even
when the isothermal process is not E X A M P L E 8. 17
quasi-static. This is because the state
variables like pressure and volume The following PV curve shows two isothermal
depend on initial and final state alone processes for two different temperatures and.
of an ideal gas and do not depend on Identify the higher temperature of these two.
the way the final state is reached. The P
assumption of ‘quasi-static’ requires to T2 T1
do the integration.
E X A M P L E 8. 1 6
A 0.5 mole of gas at temperature 300
K expands isothermally from an initial
O V
volume of 2 L to 6 L
(a) What is the work done by the gas?
(b) Estimate the heat added to the gas? Solution
(c) What is the final pressure of the gas? To determine the curve corresponding to
(The value of gas constant, R = 8.31 J mol-1 higher temperature, draw a horizontal line
K-1) parallel to x axis as shown in the figure.
This is the constant pressure line. The
Solution volumes V1 and V2 belong to same pressure
(a) We know that work done by the gas in as the vertical lines from V1 and V2 meet
an isothermal expansion the constant pressure line.
Since µ = 0.5
P
T2 T1
8.31J  6L 
W = 0.5mol × × 300 K In  
mol.K  2L 
W = 1.369 kJ
Note that W is positive since the work is
done by the gas.
(b) From the First law of thermodynamics,
in an isothermal process the heat O v2 v1 V
supplied is spent to do work.

At constant pressure, higher the volume (ii)
If the process occurs so quickly that
of the gas, higher will be the temperature. there is no time to exchange heat with
From the figure, as V1 > V2 we conclude surroundings even though there is no
T1 > T2. In general the isothermal curve thermal insulation. A few examples are
closer to the origin, has lower temperature. shown in Figure 8.29.
Examples:
8.8.2 Adiabatic process
This is a process in which no heat flows
into or out of the system (Q=0). But the gas
can expand by spending its internal energy
or gas can be compressed through some
external work. So the pressure, volume and
temperature of the system may change in an
adiabatic process.
For an adiabatic process, the first law Figure 8.29 (a) When the tyre bursts the
becomes ΔU = W. air expands so quickly that there is no time
This implies that the work is done by to exchange heat with the surroundings.
the gas at the expense of internal energy or
work is done on the system which increases
its internal energy.
The adiabatic process can be achieved by the
following methods
(i)
Thermally insulating the system from
surroundings so that no heat flows into
or out of the system; for example, when
thermally insulated cylinder of gas is
compressed (adiabatic compression)
or expanded (adiabatic expansion) as Figure 8.29 (b): When the gas is
shown in the Figure 8.28 compressed or expanded so fast, the gas
cannot exchange heat with surrounding
even though there is no thermal insulation.
Insulation Insulation Insulation Insulation
P increases
P decreases
T increases
T decreases
(a) Adiabatic compression (P and T increases) (b) Adiabatic expansion (P and T decreases)
Figure 8.28 Adiabatic compression and expansion

The PV diagram of an adiabatic expansion
and adiabatic compression process are
shown in Figure 8.30. The PV diagram for
an adiabatic process is also called adiabat
Note that the PV diagram for isothermal
(Figure 8.25) and adiabatic (Figure 8.30)
processes look similar. But actually the
adiabatic curve is steeper than isothermal
curve.
We can also rewrite the equation (8.35)
in terms of T and V. From ideal gas equation,
Figure 8.29 (c): When the warm air µRT
rises from the surface of the Earth, it the pressure P = . Substituting this
V
adiabatically expands. As a result the equation in the equation (8.35), we have
water vapor cools and condenses into
water droplets forming a cloud. µRT γ T constant
V = constant (or) Vγ =
V V μR
Note here that is another constant. So it can
The equation of state for an adiabatic process
be written as
is given by
TVγ-1 = constant.(8.37)
PVγ = constant(8.35)
The equation (8.37) implies that if the gas
Here γ is called adiabatic exponent
goes from an initial equilibrium state (Ti, Vi)
(γ = Cp/Cv) which depends on the nature of
to final equilibrium state (Tf, Vf) adiabatically
the gas.
then it satisfies the relation
The equation (8.35) implies that if the gas goes
from an equilibrium state (Pi,Vi) to another TiViγ-1 = TfVfγ-1(8.38)
equilibrium state (Pf ,Vf) adiabatically then it
satisfies the relation The equation of state for adiabatic process
can also be written in terms of T and P as
PiViγ = PfVfγ(8.36)
TγP1-γ = constant.(8.39)
P P
Pi (Pi ,Vi , Ti ) Pf (Pf ,Vf , Tf )
Adiabatic
Adiabatic
Compression
Expansion
Pf Pi (Pi ,Vi , Ti )
(Pf ,Vf , Tf )
Vi Vf V Vf Vi V
Figure 8.30 PV diagram for adiabatic expansion and adiabatic compression

(The proof of equation (8.39) left as an
exercise). When the piston is
compressed so quickly
E X A M P L E 8. 1 8 that there is no time
to exchange heat to
the surrounding, the
temperature of the gas increases
rapidly. This is shown in the figure. This
principle is used in the diesel engine.
The air-gasoline mixer is compressed
so quickly (adiabatic compression) that
the temperature increases enormously,
which is enough to produce a spark.
We often have the experience of pumping
air into bicycle tyre using hand pump.
Consider the air inside the pump as a
thermodynamic system having volume
V at atmospheric pressure and room
temperature, 27°C. Assume that the nozzle
of the tyre is blocked and you push the
pump to a volume 1/4 of V. Calculate the
final temperature of air in the pump? (For
air , since the nozzle is blocked air will not
flow into tyre and it can be treated as an
adiabatic compression). Work done in an adiabatic process:
Consider μ moles of an ideal gas enclosed in
Solution a cylinder having perfectly non conducting
Here, the process is adiabatic compression. walls and base. A frictionless and insulating
The volume is given and temperature is to piston of cross sectional area A is fitted in
be found. we can use the equation (8.38 ) the cylinder as shown in Figure 8.31.
TiViγ-1 = TfVfγ-1. Insulation Insulation
Ti = 300 K (273+27°C = 300 K)

V
Vi = V &Vf = Vi
γ −1 4 Vf
 V 

Tf =Ti  i  = 300 K × 41.4-1 = 300K×1.741
V f 
Figure 8.31 Work done in an adiabatic
T2 ≈ 522 K or 2490C
process
This temperature is higher than the boiling
point of water. So it is very dangerous to
touch the nozzle of blocked pump when Let W be the work done when the system
you pump air. goes from the initial state (Pi,Vi,Ti) to the
final state (Pf,Vf,Tf ) adiabatically.

Vf
W= ò PdV (8.40) Even though we have
Vi
Note derived equations (8.41)
By assuming that the adiabatic process and (8.42) by assuming
occurs quasi-statically, at every stage the that the adiabatic process
ideal gas law is valid. Under this condition, is quasi-static, both the equations are
the adiabatic equation of state is PV γ = valid even if the process is not quasi-
constant (or) static. This is because P and V are state
variables and are independent of how
constant
P = can be substituted in the the state is arrived.
Vγ In the adiabatic PV diagram shown
equation (8.40), we get
Vf in the Figure 8.32, the area under the
constant adiabatic curve from initial state to
∴ Wadia = ∫ dV
vi
Vγ final state will give the total work done
Vf in adiabatic process.
= constant ∫ V −γ dV
Vi
V
 V −γ +1  f P P
= constant  
 −γ + 1 (Pf , Vf) (Pi , Vi)
  Vi
Adiabatic Adiabatic
constant  1 1  Compression Expansion

= − Isotherm Isotherm
1 − γ  V f γ −1 Vi γ −1  Isotherm
for Tf
Isotherm
for Ti
for Ti for Tf
1  connstant constant 
= −
1 − γ  V f γ −1 Vi γ −1 
W (Pi , Vi) W (Pf , Vf)
Vf Vi V Vi Vf V
But, PiViγ = PfVfγ = constant.
 γ γ 
∴ Wadia =
1  Pf V f − PV
i i 
P P
1− γ  V γ −1 V γ −1  (Pf , Vf) (Pi , Vi)

 f i 
Adiabatic Adiabatic
1 Compression Expansion
Wadia = [Pf V f − PiVi ] Isotherm Isotherm
1− γ (8.41) for Tf for Ti
Isotherm Isotherm
From ideal gas law, for Ti for Tf
PfVf = μRTf and PiVi = μRTi W (Pi , Vi) W (Pf , Vf)
Substituting in equation (8.41), we get Vf Vi V Vi Vf V
µR Figure 8.32 PV diagram -Work done in

∴Wadia = [T -T ] (8.42)
γ -1 i f the adiabatic process
In adiabatic expansion, work is done by the
gas. i.e., Wadia is positive. As Ti>Tf , the gas To differentiate between isothermal
cools during adiabatic expansion. and adiabatic curves in (Figure 8.32)
In adiabatic compression, work is done on the adiabatic curve is drawn along with
the gas. i.e., Wadia is negative. As Ti<Tf, the isothermal curve for Tf and Ti. Note that
temperature of the gas increases during adiabatic curve is steeper than isothermal
adiabatic compression. curve. This is because γ > 1 always.

8.8.3 Isobaric process (ii) Most of the cooking processes in our
This is a thermodynamic process that occurs kitchen are isobaric processes. When
at constant pressure. Even though pressure the food is cooked in an open vessel,
is constant in this process, temperature, the pressure above the food is always at
volume and internal energy are not constant. atmospheric pressure. This is shown in
From the ideal gas equation, we have Figure 8.34
 µR 
V =  T(8.43)
 P 
µR
Here = constant
P
In an isobaric process the temperature is
directly proportional to volume.
V ∝ T (Isobaric process) (8.44)
This implies that for a isobaric process, the

V-T graph is a straight line passing through
Figure 8.34 Isobaric process
the origin.
If a gas goes from a state (Vi ,Ti) to (Vf ,Tf ) at
constant pressure, then the system satisfies The PV diagram for an isobaric process is
the following equation a horizontal line parallel to volume axis as
Tf Ti shown in Figure 8.35.
= (8.45)
Vf Vi Figure 8.35 (a) represents isobaric process
Examples for Isobaric process: where volume decreases
(i)
When the gas is heated and pushes the Figure 8.35 (b) represents isobaric process
piston so that it exerts a force equivalent where volume increases
to atmospheric pressure plus the force P P Isobaric expansion
Isobaric compression
due to gravity then this process is
isobaric. This is shown in Figure 8.33
The masses maintain constant
pressure in the cylinder
Vf Vi V Vi Vf V
(a) (b)
Figure 8.35 PV diagram for an isobaric

Vf process
Initial Final
Vi
The work done in an isobaric process:

Work done by the gas
Vf
Figure 8.33 Isobaric process W= ò PdV(8.46)

Vi

In an isobaric process, the pressure is DU = Q - PDV(8.50)
constant, so P comes out of the integral,
W = Pò
Vf
dV(8.47) E X A M P L E 8. 19
Vi
The following graph shows a V-T graph
W = P[Vf – Vi] = PΔV(8.48) for isobaric processes at two different
pressures. Identify which one occurs at
Where ΔV denotes change in the volume. higher pressure.
If ΔV is negative, W is also negative. This
T
implies that the work is done on the gas. If
P1
ΔV is positive, W is also positive, implying
that work is done by the gas.
The equation (8.48) can also be rewritten P2
using the ideal gas equation.

From ideal gas equation
µRT V
PV = μRT and V =
P
Substituting this in equation (8.48) we get Solution
 
From the ideal gas equation, V =  µR  T
 Ti   P 
W = μRTf 1− (8.49) V-T graph is a straight line passing the
 T f 
origin.
In the PV diagram, area under the isobaric µR
The slope =
curve is equal to the work done in isobaric P
The slope of V-T graph is inversely
process. The shaded area in the following
proportional to the pressure. If the slope is
Figure8.36 is equal to the work done by the
greater, lower is the pressure.
gas.
Here P1 has larger slope than P2. So P2 > P1.
P
Suppose the graph is drawn between T
P and V (Temperature along the x-axis and
Volume along the y-axis) then will we
still have P2 > P1?
E X A M P L E 8. 20
O Vi Vf V
One mole of an ideal gas initially kept in a
cylinder at pressure 1 MPa and temperature
Figure 8.36 Work done in an isobaric 27°C is made to expand until its volume is
process doubled.
How much work is done if the
(a)
The first law of thermodynamics for isobaric expansion is (i) adiabatic (ii) isobaric
process is given by (iii) isothermal?

Identify the processes in which change
(b) (b) Comparing all three processes, we
in internal energy is least and is see that the work done in the isobaric
maximum. process is the greatest, and work done
Show each process on a PV diagram.
(c) in the adiabatic process is the least.
Name the processes in which the heat
(d) (c) The PV diagram is shown in the Figure.
transfer is maximum and minimum.
(Take γ =
5
and R=8.3 J mol-1 K-1) P
3
Isobar
Solution A B
(a) (i) In an adiabatic process the work Isotherm
done by the system is
µR C Adiabat
Wadia = [T – Tf ]
γ -1 i D
To find the final temperature Tf, we can use
adiabatic equation of state.
Vi 2Vi V
TfVfγ-1 = TiViγ-1
γ −1
 V  2
  1 3
Tf=Ti  i  = 300×  
V f   2 The area under the curve AB = Work done
= 0.63 × 300K = 189.8K during the isobaric process
3 The area under the curve AC = Work done
W = 1 × 8.3 × (300 – 189.8) = 1.37kJ during the isothermal process
2
(ii)
In an isobaric process the work The area under the curve AD= Work done
done by the system during the adiabatic process
W = PΔV = P(Vf – Vi) From the PV diagram the area under the
curve AB is more, implying that the work
and Vf = 2Vi so W = 2PVi done in isobaric process is highest and
To find Vi, we can use the ideal gas law for work done in adiabatic process is least.
initial state. PiVi = RTi In an adiabatic process no heat enters
(d)
RTi 300 into the system or leaves from the
Vi = = 8.3 × × 10−6 = 24.9×10−4m3
Pi 1 system. In an isobaric process the work
The work done during isobaric process, done is more so heat supplied should
W = 2 × 106 × 24.9 × 10−4 = 4.9 kJ be more compared to an isothermal
process.
In an isothermal process the work
(iii)
done by the system, 8.8.4 Isochoric process
V 
W = μRT ln  f  This is a thermodynamic process in which
 Vi 
In an isothermal process the initial room the volume of the system is kept constant.
temperature is constant. But pressure, temperature and internal
energy continue to be variables.
W = 1 × 8.3 × 300 × ln(2) = 1.7kJ

The pressure - volume graph for an isochoric Pi P
process is a vertical line parallel to pressure = f (8.52)
Ti Tf
axis as shown in Figure 8.37.
For an isochoric processes, ΔV=0 and W=0.
Then the first law becomes
P
Pf ΔU = Q(8.53)
Implying that the heat supplied is used to

increase only the internal energy. As a result
Pi the temperature increases and pressure also
increases. This is shown in Figure 8.38
Constant volume
V container
(a) P1 P2
P
Pi Gas Gas
Pf Initial Final
Figure 8.38 Isochoric process
V
(b) Suppose a system loses heat to the
surroundings through conducting walls
Figure 8.37 Isochoric process with by keeping the volume constant, then its
(a) increased pressure and (b) decreased
internal energy decreases. As a result the
pressure
temperature decreases; the pressure also
decreases.
The equation of state for an isochoric process
Examples:
is given by
1. When food is cooked by closing with a
 µR 
P =   T(8.51) lid as shown in figure.
 V 
 µR 
where   = constant
 V 
We can infer that the pressure is directly
proportional to temperature. This implies
that the P-T graph for an isochoric process
is a straight line passing through origin.
If a gas goes from state (Pi,Ti) to (Pf,Tf ) at
constant volume, then the system satisfies
the following equation

Table 8.4: Summary of various thermodynamic processes
Temperature
Heat
S. No. Process & internal Pressure
Q
energy
Expansion Q>0 Constant decreases
1 Isothermal
Compression Q<0 Constant Increases
Expansion Q>0 increases Constant
2 Isobaric
Compression Q<0 decreases Constant
Q>0 Increases increases
3 Isochoric
Q<0 Decreases decreases
expansion Q=0 Decreases Decreases
4 Adiabatic
Compression Q=0 Increases increases

Equation of Work done
Volume Indicator diagram (PV diagram)
state (ideal gas)
P P
increases
V  Pi Pf
W=μRT ln  f  >0
T

 V 
T
=
=
co
co
ns
ns
i
t
t
Vi
V Vf Vf V Vi
PV = Constant P P
Decreases
V  Pi Pf
W=μRT ln  f  <0
T
T
 V 
=
=
co
co
ns
ns
i
t
t
Vi
V Vf Vf V Vi
P Isobaric compression P Isobaric expansion
increases
W=P [Vf−Vi]=PΔV
>0
Vf Vi V Vi Vf V
V
= Constant P Isobaric compression P Isobaric expansion
decreases T
W=P [Vf−Vi] =PΔV < 0
Vf Vi V Vi Vf
V
P
P
P
Pf f
P
Pi i
V
V
Constant P Zero
= Constant P
P
T Pi
Pi
P
Pf f
V
V
P P
increases µR Pi (Pi ,Vi , Ti ) Pf (Pf ,Vf , Tf )
W= (T −T ) > 0
γ -1 i f Adiabatic
Expansion
Adiabatic
Compression
Pf Pi (Pi ,Vi , Ti )
(Pf ,Vf , Tf )
Vi Vf V Vf Vi V
PVγ = Constant P P
Pi (Pi ,Vi , Ti ) Pf (Pf ,Vf , Tf )
Adiabatic
Decreases µR Adiabatic
Compression
W= (Ti−T ) < 0 (P V T )
Expansion
γ -1 P f f
f, f, f
Pi (Pi ,Vi , Ti )
Vi Vf V Vf Vi V

When food is being cooked in this closed Solution
position, after a certain time you can When the water is heated from 30°C to
observe the lid is being pushed upwards by 60°C,there is only a slight change in its
the water steam. This is because when the volume. So we can treat this process as
lid is closed, the volume is kept constant. As isochoric. In an isochoric process the
the heat continuously supplied, the pressure work done by the system is zero. The given
increases and water steam tries to push the heat supplied is used to increase only the
lid upwards. internal energy.
2. In automobiles the petrol engine
undergoes four processes. First the ∆U = Q = msv ∆T
piston is adiabatically compressed to
The mass of water = 500 g =0.5 kg
some volume as shown in the Figure (a).
In the second process (Figure (b)), the The change in temperature = 30K
volume of the air-fuel mixture is kept
constant and heat is being added. As a The heat Q = 0.5×4184×30 = 62.76 kJ
result the temperature and pressure are
increased. This is an isochoric process.
For a third stroke (Figure (c)) there will 8.8.5 Cyclic processes
be an adiabatic expansion, and fourth This is a thermodynamic process in which
stroke again isochoric process by keeping the thermodynamic system returns to its
the piston immoveable (Figure (d)). initial state after undergoing a series of
changes. Since the system comes back to the
initial state, the change in the internal energy
(a) (b) Qin (c) (d) Qout is zero. In cyclic process, heat can flow in
Air
to system and heat flow out of the system.
From the first law of thermodynamics, the
net heat transferred to the system is equal to
work done by the gas.
Adibatic Const Adibatic Const
Compression volume heat
addition
Expansion volume heat Qnet = Qin − Qout = W (for a cyclic process)
rejection
(8.54)
The summary of various thermodynamic 8.8.6 PV diagram for a cyclic

processes is give in the Table 8.4. process
E X A M P L E 8. 2 1 In the PV diagram the cyclic process is
represented by a closed curve.
500 g of water is heated from 30°C to
Let the gas undergo a cyclic process in
60°C. Ignoring the slight expansion of
which it returns to the initial stage after an
water, calculate the change in internal
expansion and compression as shown in
energy of the water? (specific heat of water
Figure 8.39
4184 J/kg.K)

P (P1 ,V1) Let W2 be the work done on the gas during
B
P1
C compression from volume V2 to volume V1.
It is equal to the area under the graph ADC
as shown in Figure 8.40 (b)
D A (P2 ,V2)
P2
The total work done in this cyclic process
= W1 - W2 = Green shaded area inside the
loop, as shown in Figure 8.41.
V
P
V1 V2 (P1 ,V1)
B
P1
C
Figure 8.39 PV diagram for cyclic
process
D
Let W1 be the work done by the gas during P2 A (P2 ,V2)
expansion from volume V1 to volume V2.

It is equal to area under the graph CBA as
shown in Figure 8.40 (a) .
P
(P1 ,V1)
B
P1
P C V1 V2 V
(P1 ,V1)
B
P1
C
Figure 8.41 Net work done in a cyclic
D
P2 A (P2 ,V2) process
D
P2 A (P2 ,V2)
Thus the net work done during the cyclic
process shown above is not zero. In general
V
the net work done can be positive or negative.
V1 V2
If the net work done is positive, then work
V
V1 V2 done by the system is greater than the work
Figure 8.40 (a) W for path CBA done on the system. If the net work done is
P
(P1 ,V1)
B
negative then the work done by the system
P1
P C
(P1 ,V1)
is less than the work done on the system.
B
P1
C
D
A (P2 ,V2)
P2
D
Further, in a cyclic process
Note
P2 A (P2 ,V2) the net work done is positive
if the process goes clockwise
and net work done is negative if the
V1 V2 V process goes anti-clockwise. In Figure
V
8.41the process goes clockwise.
V1 V2
Figure 8.40 (b) W for path ADC

E X A M P L E 8. 2 2 Area under the curve BC = rectangle area
= 1 × 2 = − 2J
The PV diagrams for a thermodynamical Network done in the cyclic process = 1 J,
system is given in the figure below. which is positive.
Calculate the total work done in each of In the case (c) the closed curve is
the cyclic processes shown. anticlockwise. So the net work done is
P
(Pa)
P
(Pa)
P negative, implying that the work done on
A (Pa)
the system is greater than work done by

C 4 B A
4 B 4
2 A 2
C
B 2
C
the system. The area under the curve AB
D
1 2 V(m3) 1 2 V(m3) 1 2 V(m3)

will give the work done on the gas (isobaric
(a) (b) (c)
compression) and area under the curve CA
Solution (work done by the system) will give the total
work done by the system.
In the case (a) the closed curve is The area under the curve AB =Rectangle of
anticlockwise. So the net work done is area = 4 × 1 = - 4J
negative, implying that the work done on The area under the curve CA = Rectangle
the system is greater than the work done 1
by the system. The area under the curve area + triangle area = (1×2) + × 1×2 = +3J
2
BC will give work done on the gas (isobaric The total work in the cyclic process = -1 J. It
compression) and area under the curve DA is negative
(work done by the system) will give the
total work done by the system.
8.8.7 Limitations of first law
Area under the curve BC = Area of of thermodynamics
rectangle BC12 = 1 × 4= − 4J The first law of thermodynamics explains
well the inter convertibility of heat and
Area under the curve DA = 1 × 2= + 2J
work. But it does not indicate the direction
Net work done in cyclic process of change.
= −4 + 2= −2 J For example,
a. When a hot object is in contact with a
In the case (b) the closed curve is clockwise. cold object, heat always flows from the
So the net work done is positive, implying hot object to cold object but not in the
that the work done on the system is less than reverse direction. According to first
the work done by the system. Area under law, it is possible for the energy to flow
the curve BC will give work done on the gas from hot object to cold object or from
(isobaric compression) and area under the cold object to hot object. But in nature
curve AB will give the total work done by the direction of heat flow is always from
the system. higher temperature to lower temperature.
b. When brakes are applied, a car stops due
Area under the curve AB = rectangle area+ to friction and the work done against
1 friction is converted into heat. But this
triangle area = (1×2) + × 1×2 = +3J
2

heat is not reconverted to the kinetic The first law of thermodynamics is the
energy of the car. statement about conservation of energy in
So the first law is not sufficient to explain a thermodynamic process. For example, if
many of natural phenomena. a hotter object is placed on a colder object,
heat flows from hotter to colder object.
8.8.8 Reversible process Why does heat not flow from the colder
object to hotter object? Even if energy flows
A thermodynamic process can be considered
from colder object to hotter object, the first
reversible only if it possible to retrace the
law of thermodynamics is not violated. For
path in the opposite direction in such a
example, if 5 J of heat flows form hotter
way that the system and surroundings pass
to colder or from colder to hotter objects
through the same states as in the initial,direct
the total internal energy of this combined
process.
system remains the same. But 5 J of heat
Example: A quasi–static isothermal
never flows from the colder object to hotter
expansion of gas, slow compression and
object. In nature all such process occur
expansion of a spring.
only in one direction but not in the reverse
Conditions for reversible process:
direction, even if the energy is conserved
1. The process should proceed at an in both the processes. Thus the first law of
extremely slow rate. thermodynamics has no explanation for
2. The system should remain in mechanical, this irreversibility. When the scientists of
thermal and chemical equilibrium state the eighteenth century tried to explain this
at all the times with the surroundings, irreversibility, they discovered a new law
during the process. of nature. This is called the second law of
3. No dissipative forces such as friction, thermodynamics. According to second law
viscosity, electrical resistance should be of thermodynamics
present. “Heat always flows from hotter object
to colder object spontaneously”. This is
All reversible processes are known as the Clausius form of second law
Note quasi-static but all quasi- of thermodynamics.
static processes need not
be reversible. For example
when we push the piston very slowly, if E X A M P L E 8. 23
there is friction between cylinder wall
and piston some amount of energy is Give some examples of irreversible
lost to surroundings, which cannot be processes.
retrieved back. All naturally occuring processes are
irreversible. Here we give some interesting
Irreversible process:
examples.
All natural processes are irreversible.
(a) When we open a gas bottle, the gas
Irreversible process cannot be plotted in a
molecules slowly spread into the entire
PV diagram, because these processes cannot
room. These gas molecules can never
have unique values of pressure, temperature
get back in to the bottle.
at every stage of the process.

Natural process Never occurs puts a fundamental restriction on efficiency
of engines. Therefore understanding heat
engines is very important.
Reservoir:
It is defined as a thermodynamic system which
has very large heat capacity. By taking in heat
from reservoir or giving heat to reservoir, the
(b) Suppose one drop of an ink is dropped
reservoir’s temperature does not change.
in water, the ink droplet slowly spreads
Example: Pouring a tumbler of hot water
in the water. It is impossible to get the
in to lake will not increase the temperature
ink droplet back.
of the lake. Here the lake can be treated as a
(c)
When an object falls from some
reservoir.
height, as soon as it hits the earth it
When a hot cup of coffee attains equilibrium
comes to rest. All the kinetic energy
with the open atmosphere, the temperature
of the object is converted to kinetic
of the atmosphere will not appreciably
energy of molecules of the earth
change. The atmosphere can be taken as a
surface, molecules of the object and
reservoir.
small amount goes as sound energy.
We can define heat engine as follows.
The spreaded kinetic energy to the
Heat engine is a device which takes heat
molecules never collected back and
as input and converts this heat in to work by
object never goes up by itself.
undergoing a cyclic process.
Note that according to first law of
A heat engine has three parts:
thermodynamics all the above processes are
possible in both directions. But second law (a) Hot reservoir
of thermodynamics forbids The processes (b) Working substance
to occur in the reverse direction. The second (c) Cold reservoir
law of thermodynamics is one of the very A Schematic diagram for heat engine is
important laws of nature. It controls the given below in the figure 8.42.
way the natural processes occur.
Hot reservoir
TH
8.9
HEAT ENGINE QH Work done by
heat engine
W
In the modern technological world, the Heat engine
role of automobile engines plays a vital

role in for transportation. In motor bikes
QL
and cars there are engines which take in
petrol or diesel as input, and do work by
Cold reservoir
rotating wheels. Most of these automobile TL
engines have efficiency not greater than
40%. The second law of thermodynamics Figure 8.42 Heat Engine

1. Hot reservoir (or) Source: It supplies heat
to the engine. It is always maintained at a Hot reservoir at temperature TH
high temperature TH
QH
2. Working substance: It is a substance like
gas or water, which converts the heat
supplied into work. Heat engine
W
A simple example of a heat engine is
a steam engine. In olden days steam
engines were used to drive trains. The QL
working substance in these is water
which absorbs heat from the burning Cold reservoir at temperature TL
of coal. The heat converts the water
into steam. This steam is does work by
rotating the wheels of the train, thus Figure 8.43 Heat engine
making the train move.
3. Cold reservoir (or) Sink: The heat We can write
engine ejects some amount of heat (QL) Input heat = Work done + ejected heat
in to cold reservoir after it doing work. QH = W + QL
It is always maintained at a low
temperature TL. W = QH - QL
For example, in the automobile engine, Then the efficiency of heat engine
the cold reservoir is the surroundings at
room temperature. The automobile ejects output W QH − QL
η= = =
heat to these surroundings through a input QH QH
silencer. output W Q
η= = = 1 − L (8.55)
input QH QH
The heat engine works in a cyclic process.
After a cyclic process it returns to the same Note here that QH, QL and W all are taken as
state. Since the heat engine returns to the positive, a sign convention followed in this
same state after it ejects heat, the change in expression.
the internal energy of the heat engine is zero. Since QL < QH, the efficiency (η) always less
The efficiency of the heat engine is than 1. This implies that heat absorbed is not
defined as the ratio of the work done (out completely converted into work. The second
put) to the heat absorbed (input) in one law of thermodynamics placed fundamental
cylic process. restrictions on converting heat completely
Let the working substance absorb heat into work.
QH units from the source and reject QL units We can state the heat engine statement of
to the sink after doing work W units, as second law of thermodynamics. This is also
shown in the Figure 8.43. called Kelvin-Planck’s statement.

Kelvin-Planck statement: 8.9.1 Carnot’s ideal heat
It is impossible to construct a heat engine engine
that operates in a cycle, whose sole effect is
to convert the heat completely into work. This In the previous section we have seen that
implies that no heat engine in the universe the heat engine cannot have 100% efficiency.
can have 100% efficiency. What is the maximum possible efficiency
can a heat engine have?. In the year 1824 a
young French engineer Sadi Carnot proved
According to first law of
Note that a certain reversible engine operated in
thermodynamics, in an
isothermal process the given cycle between hot and cold reservoir can
heat is completely converted into work have maximum efficiency. This engine is
(Q = W). Is it a violation of the second called Carnot engine.
law of thermodynamics? No. For
A reversible heat engine operating in
non-cyclic process like an isothermal
expansion, the heat can be completely a cycle between two temperatures in a
converted into work. But Second law particular way is called a Carnot Engine.
of thermodynamics implies that ‘In The carnot engine has four parts which
a cyclic process only a portion of the are given below.
heat absorbed is converted into work’.
All heat engines operate in a cyclic i Source: It is the source of heat maintained
process. at constant high temperature TH.
Any amount of heat can be extracted from
it, without changing its temperature.
E X A M P L E 8. 2 4 ii Sink: It is a cold body maintained at
During a cyclic process, a heat engine a constant low temperature TL. It can
absorbs 500 J of heat from a hot reservoir, absorb any amount of heat.
does work and ejects an amount of heat 300 J iii Insulating stand: It is made of perfectly
into the surroundings (cold reservoir). non-conducting material. Heat is not
Calculate the efficiency of the heat engine? conducted through this stand.
Solution iv Working substance: It is an ideal gas
The efficiency of heat engine is given by enclosed in a cylinder with perfectly
non-conducting walls and perfectly
QL
η=1− conducting bottom. A non-conducting
QH
and frictionless piston is fitted in it.
300 3
η=1− =1− The four parts are shown in the following
500 5
Figure 8.44
η = 1 – 0.6 = 0.4
The heat engine has 40% efficiency,

implying that this heat engine converts
only 40% of the input heat into work.

P
A(V1, P1,TH)
Iso
the
Insulating stand rm
B(V2, P2,TH)
A dia b a
Insulating
wall
Source at TH Sink at TL
A dia
Idealgas
ba
D (V
,P C(V3, P3,TL)
4
4, T) Isother m
L
Conducting bottom
V
O E F G H
Insulating stand Figure 8.45 PV diagram for Carnot cycle

Source at TH Sink at TL
Then the work done by the gas (working
substance) is given by
V2
∴QH = WA→B = ò PdV

Figure 8.44 Carnot engine V1
Since the process occurs quasi-statically,

the gas is in equilibrium with the source
Carnot’s cycle: till it reaches the final state. The work done
The working substance is subjected to four in the isothermal expansion is given by the
successive reversible processes forming equation (8.34)
what is called Carnot’s cycle. V 
Let the initial pressure, volume of the WA→B = μRTHIn  2  = Area under the
 V1 
working substance be P1,V1. curve AB
Step A to B: Quasi-static isothermal (8.56)

expansion from (P1,V1,TH) to (P2,V2,TH):
This is shown in Figure 8.46 (a)
The cylinder is placed on the source. The P P
P
heat (QH) flows from source to the working A(V , P ,T ) 1
A(V , P ,T )
1 H 1 1 H
A(V1, P1,TH)
substance (ideal gas) through the bottom of B(V , P ,T ) B(V , P ,T ) 2 2 H 2 2 H

B(V2, P2,TH)
the cylinder. Since the process is isothermal, D (V D (V D (V

C(V , P ,T ) C(V , P ,T ) ,P C(V3, P3,T
,P ,P
the internal energy of the working substance
3 3 L
T) 4
T)
4,
3 3 L 4
4,
4
4, T)
L L L
will not change. The input heat increases the O E F G H V O E F G H V O E F G H V
volume of the gas. The piston is allowed to (a) (b) (c)
P P P
move outP very slowly(quasi-statically).
A(V , P ,T ) A(V , P ,T )
It is 1 1 H
A(V , P ,T ) A(V , P ,T )
1 1 H 1 1 H
1 1 H
shown in the figure 8.47(a).B(V , P ,T ) 2 2 H

B(V , P ,T ) 2 2 H
B(V , P ,T ) B(V , P ,T ) 2 2 H 2 2 H
W1 is the work done by the gas in

D (V D (V D (V D (V
expanding from , P volume
T) 4
4,
V1 to volume
C(V , P ,T )
L
3 3,P L
T ) V2
C(V , P ,T )
4
4,
L
,P
3 3
T)
L
C(V , P ,T )
4 ,P
4,T)
C(V , P ,T )
L
3 3 L
4
4,
L
3 3 L
with a decrease of pressure from P1 to P2.

O E F G H V O E F G H V O E F G H V O E F G H V
This is represented (a) by the P-V diagram (b) along (c) (d)
the path AB as shown in the Figure 8.45. Figure 8.46 Work done in Carnot cycle

Idealgas
QH
(a) Source at TH
A→B
Isothermal expansion
(d) (b)
Idealgas Carnot Cycle Idealgas
Thermal insulation Thermal insulation
D→A B→C
Adiabatic compression Adiabatic expansion
Idealgas
QL
Sink at TL
(c)
C→D
Isothermal compression
Figure 8.47 Carnot cycle

Step B to C: Quasi-static adiabatic expansion The cylinder is placed on the insulating
from (P2,V2,TH) to (P3,V3,TL) stand again and the gas is compressed
The cylinder is placed on the insulating adiabatically till it attains the initial pressure
stand and the piston is allowed to move P1, volume V1 and temperature TH. This is
out. As the gas expands adiabatically from shown by the curve DA in the P-V diagram.
volume V2 to volume V3 the pressure falls V1
µR
from P2 to P3. The temperature falls to TL. ∴ WD→A = ∫ PdV = (TL −TH ) = − Area
V4
γ −1
This adiabatic expansion is represented by
under the curve DA (8.59)
curve BC in the P-V diagram. This adiabatic
process also occurs quasi-statically and In the adiabatic compression also work is
implying that this process is reversible and done on the gas so it is negative, as is shown
the ideal gas is in equilibrium throughout in Figure 8.46 (d)
the process. It is shown in the figure 8.47(b). Let ‘W’ be the net work done by the working
From the equation (8.42) substance in one cycle
The work done by the gas in an adiabatic ∴W=Work done by the gas – work done on
expansion is given by, the gas
V3 µR = WA→B + WB→C − WC→D − WD→A

WB→C = ò PdV = [TH −TL ] =
V2 γ −1 since WB→C = WD→A
Area under the curve BC (8.57) = WA→B − WC→D
This is shown in Figure 8.46 (b) The net work done by the Carnot engine in
Step C → D: Quasi-static isothermal one cycle W=WA→B − WC→D(8.60)
compression from (P3,V3,TL) to (P4,V4,TL): It
is shown in the figure 8.47(c) Equation (8.60) shows that the net work
The cylinder is placed on the sink and done by the working substance in one cycle
the gas is isothermally compressed until is equal to the area (enclosed by ABCD) of
the pressure and volume become P4 and V4 the P-V diagram (Figure 8.48)
respectively. This is represented by the curve
P
CD in the PV diagram as shown in Figure
A(V1, P1,TH)
8.45. Let WC→D be the work done on the gas.
According to first law of thermodynamics
B(V2, P2,TH)
V4
V  V 
∴ WC→D = ∫ PdV = µRTL ln  4  = −µRTL ln  3 
V3
 V3  V 
4
D (V
. = - Area under the curve CD (8.58) 4 ,P
4, T)
L
C(V3, P3,TL)
This is shown in Figure 8.46 (c)

Here V3 is greater than V4. So the work done
V
is negative, implying work is done on the gas. O E F G H
Step D→A: Quasi-static adiabatic
compression from (P4,V4,TL) to (P1,V1,TH):
Figure 8.48 Net work done in Carnot cycle
It is shown in the figure 8.47(d)

It is very important to note that after one V2 V3
Which implies that = (8.65)
cycle the working substance returns to the V1 V4
initial temperature TH. This implies that the
change in internal energy of the working Substituting equation (8.65) in (8.64), we get
substance after one cycle is zero. QL TL
= (8.66)
QH TH
8.9.2 Efficiency of a Carnot TL
engine ∴The efficiency η = 1 − (8.67)
TH
Efficiency is defined as the ratio of work Note : TL and TH should be expressed in
done by the working substance in one cycle Kelvin scale.
to the amount of heat extracted from the Important results:
source. 1. η is always less than 1 because TL is
less than TH. This implies the efficiency
work done W
η= = (8.61) cannot be 100%. It can be 1 or 100%
Heat extracted QH only when TL = 0K (absolute zero of
temperature) which is impossible to
From the first law of thermodynamics,
attain practically.
W = QH − QL
2. The efficiency of the Carnot’s engine is
QH − QL Q independent of the working substance.
∴ η= = 1 − L (8.62)
QH QH It depends only on the temperatures of
the source and the sink. The greater the
Applying isothermal conditions, we get,
difference between the two temperatures,
V  higher the efficiency.
QH = µRTH ln 2 
 V1  3. When TH=TL the efficiency η =0. No
(8.63)
V3  engine can work having source and sink
QL = µRTL ln 
 V4  at the same temperature.
Here we omit the negative sign. Since we are 4. The entire process is reversible in the
interested in only the amount of heat (QL) Carnot engine cycle. So Carnot engine
ejected into the sink, we have is itself a reversible engine and has
V  maximum efficiency. But all practical
TL ln  3 
Q  V4  heat engines like diesel engine, petrol
∴ L= (8.64)
QH T ln V2  engine and steam engine have cycles
H  V 
1 which are not perfectly reversible. So
By applying adiabatic conditions, we get, their efficiency is always less than the
THV2γ −1 = TLV3γ −1 Carnot efficiency. This can be stated in
the form of the Carnot theorem. It is
THV1γ −1 = TLV4γ −1
stated as follows ‘Between two constant
By dividing the above two equations, we get temperatures reservoirs, only Carnot
V 
γ −1
V 
γ −1
engine can have maximum efficiency. All
 2 
 V  =  3  real heat engines will have efficiency less
1
V 
4
than the Carnot engine’

E X A M P L E 8. 2 5 Even though the differences between the
(a)
A steam engine boiler is maintained temperature of hot and cold reservoirs in
at 250°C and water is converted into both engines is same, the efficiency is not
steam. This steam is used to do work same. The efficiency depends on the ratio
and heat is ejected to the surrounding of the two temperature and not on the
air at temperature 300K. Calculate the difference in the temperature. The engine
maximum efficiency it can have? which operates in lower temperature has
highest efficiency.
Solution
The steam engine is not a Carnot engine, Diesel engines used in
because all the process involved in the cars and petrol engines
steam engine are not perfectly reversible. used in our motor bikes
But we can calculate the maximum are all real heat engines.
possible efficiency of the steam engine by The efficiency of diesel engines has
considering it as a Carnot engine. maximum up to 44% and the efficiency
300 K of petrol engines are maximum up to
TL
η = 1− = 1 − 523 K = 0.43 30%. Since these engines are not ideal
TH
heat engines (Carnot engine), their
The steam engine can have maximum efficiency is limited by the second law
possible 43% of efficiency, implying this of thermodynamics. Now a days typical
steam engine can convert 43% of input bikes give a mileage of 50 km per Liter of
heat into useful work and remaining 57% petrol. This implies only 30% of 1 Liter
is ejected as heat. In practice the efficiency of petrol is converted into mechanical
is even less than 43%. work and the remaining 70% goes out
as wasted heat and ejected into the
E X A M P L E 8. 2 6 surrounding atmosphere!
There are two Carnot engines A and B
operating in two different temperature
regions. For Engine A the temperatures of 8.9.3 Entropy and second
the two reservoirs are 150°C and 100°C. For law of thermodynamics
engine B the temperatures of the reservoirs We have seen in the equation (8.66) that the
are 350°C and 300°C. Which engine has Q Q
quantity H is equal to L . The quantity
TH TL
lesser efficiency? Q
is called entropy. It is a very important
T
Solution thermodynamic property of a system. It
373 Q
The efficiency for engine A = 1 − = is also a state variable. H is the entropy
423 TH
0.11. Engine A has 11% efficiency received by the Carnot engine from hot
QL
573 reservoir and is entropy given out by
The efficiency for engine B = 1 - = 0.08 TL
623 the Carnot engine to the cold reservoir. For
Engine B has only 8% efficiency. reversible engines (Carnot Engine) both
entropies should be same, so that the change

in entropy of the Carnot engine in one cycle processes occur in such a way that entropy
is zero. This is proved in equation (8.66). should increase for all irreversible process.
But for all practical engines like diesel and
petrol engines which are not reversible 8.10
Q
engines, they satisfy the relation L > H .
Q REFRIGERATOR
TL TH
In fact we can reformulate the second law of
thermodynamics as follows A refrigerator is a Carnot’s engine working
in the reverse order. It is shown in the
“For all the processes that occur in nature
figure 8.49.
(irreversible process), the entropy always
increases. For reversible process entropy will Working Principle:
not change”. Entropy determines the direction
in which natural process should occur. (a)
Hot reservoir
We now come back to the question: Why (room temperature
of kitchen)
does heat always flows from a state of higher
temperature to one of lower temperature
QH
and not in the opposite direction? Because
entropy increases when heat flows from hot
W
object to cold object. If heat were to flow
Refrigerator
from a cold to a hot object, entropy will
decrease leading to violation of second law
thermodynamics. QL
Entropy is also called ‘measure of disorder’.
All natural process occur such that the disorder
Cold reservoir
should always increases. (inside of refrigerator)
Consider a bottle with a gas inside. When

the gas molecules are inside the bottle it has (b)
less disorder. Once it spreads into the entire High
room it leads to more disorder. In other pressure
words when the gas is inside the bottle the (liquid) Condenser
entropy is less and once the gas spreads into coils
QH
entire room, the entropy increases. From (to outside)
the second law of thermodynamics, entropy
always increases. If the air molecules go
back in to the bottle, the entropy should
decrease, which is not allowed by the
Compressor
second law of thermodynamics. The same
explanation applies to a drop of ink diffusing
into water. Once the drop of ink spreads, its low pressure
entropy is increased. The diffused ink can Fig 8.49 (a) Schematic diagram of a
never become a drop again. So the natural refrigerator (b) Actual refrigerator

The working substance (gas) absorbs a Substituting this equation into equation
quantity of heat QL from the cold body (sink) (8.70) we get
at a lower temperature TL. A certain amount 1 TL
of work W is done on the working substance β= =
TH TH − TL
−1
by the compressor and a quantity of heat QH TL
is rejected to the hot body (source) ie, the Inferences:
atmosphere at TH. When you stand beneath 1. The greater the COP, the better is the
of refrigerator, you can feel warmth air. condition of the refrigerator. A typical
From the first law of thermodynamics , we refrigerator has COP around 5 to 6.
have 2. Lesser the difference in the temperatures
of the cooling chamber and the
QL +W = QH (8.68)
atmosphere, higher is the COP of a
As a result the cold reservoir (refrigerator) refrigerator.
further cools down and the surroundings 3. In the refrigerator the heat is taken
(kitchen or atmosphere) gets hotter. from cold object to hot object by doing
external work. Without external work
Coefficient of performance (COP) (β): heat cannot flow from cold object to hot
COP is a measure of the efficiency of a object. It is not a violation of second law
refrigerator. It is defined as the ratio of heat of thermodynamics, because the heat
extracted from the cold body (sink) to the is ejected to surrounding air and total
external work done by the compressor W. entropy of (refrigerator + surrounding)
QL is always increased.
COP= β = (8.69)
W
E X A M P L E 8. 27
From the equation (8.68)
A refrigerator has COP of 3. How much
QL
β= work must be supplied to the refrigerator
QH − QL
in order to remove 200 J of heat from its
1 interion?
β= (8.70)
QH QL
−1 COP = β =
QL W
QH TH
But we know that = Qc 200
QL TL W= = = 66.67 J
COP 3

Green House Effect
The presence of atmosphere in the earth plays very important role in human lives. Top
of the atmosphere is at -19°C and bottom of the atmosphere is at +14°C. The increase in
33°C from top to bottom is due to some gases present in the atmosphere. These gases are
called Greenhouse gases and this effect is called Greenhouse effect.
The greenhouse gases are mainly CO2, water vapour, Ne, He, NO2, CH4, Xe, Kr, ozone
and NH3. Except CO2 and water vapor, all others are present only in very small amount
in the atmosphere. The radiation from the Sun is mainly in the visible region of the
spectrum. The earth absorbs these radiations and reradiate in the infrared region. Carbon
dioxide and water Vapour are good absorbers of infrared radiation since they have more
vibrational degree of freedom compared to nitrogen and oxygen (you will learn in unit 9)
which keeps earth warmer as shown in Figure.
The amount of CO2 present in the atmosphere is increased from 20% to 40% due to
human activities since 1900s. The major emission of CO2 comes from burning of fossil
fuels. The increase in automobile usage worldwide causes this damage. Due to this increase
in the CO2 content in the atmosphere, the average temperature of the earth increases by
1°C. This effect is called global warming. It has serious influence and alarming effect on
ice glaciers on Arctic and Antarctic regions. In addition, the CO2 content is also increasing
in ocean which is very dangerous to species in the oceans.
In addition to CO2, another very important greenhouse gas is Chloro flouro
carbon(CFC) which is used as coolant in refrigerators worldwide. In the human made
greenhouse gases CO2 is 55%, CFCs are 24%. Nitrogen oxide is 6% and methane is 15%.
CFCs also has made huge damage to ozone layer.
Lot of efforts are taken internationally to reduce the emission of CO2 and CFCs in
various countries. Nowadays a lot of research is going to replace non fossil fuels to replace
the fossil-fuels in automobile industry. The major emission of CO2 comes from developed
countries like USA and European countries. Various treaties are formed between countries
to reduce the emission of CO2 to considerable level before 2020s. But still global warming
is not taken seriously in various countries.
The greenhouse effect
CO2 and other gases

in the atmosphere
trap IR radiation, keeping
the earth warm.
phere
Atmos

In hot summer, we use earthern pots to drink cold water.
The pot reduces the temperature of water inside it. Does
the earthern pot act as a refrigerator? No. cyclic process
is the basic necessity for heat engine or refrigerator. In
earthern pot, the cooling process is not due to any cyclic process. The
cooling occurs due to evaporation of water molecules which oozes out
through pores of the pot. Once the water molecules evaporate, they
never come back to the pot. Even though the heat flows from cold water
to open atmosphere, it is not a violation of second law of thermodynamics. The water
inside the pot is an open thermodynamic system, so the entropy of water + surrounding
always increases.
SUMMARY
Heat is energy in transit which flows from hot object to cold object. However it is
not a quantity.
Work is a process to transfer energy from one object to another object.
Temperature is a measure of hotness of the object. It determines the direction of the
flow of heat.
The ideal gas law is PV = NkT or PV = μRT. The Ideal gas law holds for only at
thermodynamic equilibrium. For non-equilibrium process, it is not valid.
Heat capacity is the amount of heat energy required to increase the object’s
temperature by 1°C or 1K. It is denoted by S.
Specific heat capacity is the amount of heat energy required to increase the 1 kg of
object’s temperature by 1°C or 1K. It is denoted by s.
Molar specific heat capacity is the amount of heat energy requires to increase the 1
mole of substance’s temperature by 1°C or 1K. It is denoted by C.
Thermal expansion is a tendency of an object to change its shape, area, and volume
due to change in temperature.
Water has an anomalous behavior of expansion.
Latent heat capacity is the amount of heat energy required to change the phase of
the substance.
Calorimetry is the measurement of the amount of heat energy released or absorbed
by a thermodynamic system during the heating process.
Heat transfers in three different modes: conduction, convection and radiation
Stefan-Boltmann law: E = σ T4 and Wien’s law: λmax T = b
Thermodynamic equilibrium: thermal, mechanical and chemical equilibrium
Thermodynamic variables : Pressure, temperature, volume, internal energy and
entropy

Zeroth law of thermodynamics: If two objects are separately in thermal equilibrium
with the third object, then these two are in thermal equilibrium. Temperature is a
property which is the same for both the systems.
Internal energy is the sum of kinetic and potential energies of molecules in the
thermodynamic system.
Joule converted mechanical energy to internal energy of the thermodynamic system
First law of thermodynamics is a statement of conservation of energy. It included
heat energy of the thermodynamic system.
A quasi-static process is an infinitely slow process in which the system is always at
equilibrium with the surrounding.
When the volume of the system changes, the work done W = ∫P dV
The area under the PV diagram gives the work done by the system or work done on
the system.
Specific heat capacity at constant volume is always less than specific heat capacity at
constant pressure.
Isothermal process: T = constant, Isobaric process: P=constant, Isochoric process:
V= constant, Adiabatic process Q = 0
Work done in the isobaric process is most and work done in the adiabatic process
is least
In a cyclic process, change in internal energy is zero.
The total work done in the cyclic process is given by a closed area in PV diagram
A reversible process is an ideal process.
All natural processes are irreversible.
Heat engine takes input from the hot reservoir, performs work and rejects some
amount of heat energy into sink.
Carnot engine is a reversible engine. It has the highest efficiency. No real heat engine
can have the efficiency of that of a Carnot engine.
A refrigerator is reverse of a Carnot engine. COP (coefficient of performance) of the
practical refrigerator is always less than ideal refrigerator.

CONCEPT MAP
Thermodynamics
Heat Work
Temperature
Heat capacity Thermal properties
Ideal gas law Thermal expansion
Internal energy
First law of thermodynamics
Isobaric process Adiabatic process Isochoric process Isothermal process
Second law of thermodynamics
Carnot engine Refrigerator

EVALUATION
I. Multiple choice questions:

1. In hot summer after a bath, the body’s a) Q > 0, W > 0,
a) internal energy decreases b) Q < 0, W > 0,
b) internal energy increases c) Q > 0, W < 0,
c) heat decreases d) Q < 0, W < 0,
d) no change in internal energy and 7. When you exercise in the morning,
heat by considering your body as
2. The graph between volume and thermodynamic system, which of the
temperature in Charles’ law is following is true?
a) an ellipse b) a circle a) ΔU > 0, W > 0,
c)a straight line d) a parabola b) ΔU < 0, W > 0,
c) ΔU < 0, W < 0,
3. When a cycle tyre suddenly bursts, the
air inside the tyre expands. This process d) ΔU = 0, W > 0,
is 8. A hot cup of coffee is kept on the table.
a) isothermal b) adiabatic After some time it attains a thermal
equilibrium with the surroundings.
c) isobaric d) isochoric
By considering the air molecules in
4. An ideal gas passes from one the room as a thermodynamic system,
equilibrium state (P1, V1, T1, N) to which of the following is true
another equilibrium state (2P1, 3V1, T2,
a) ΔU > 0, Q = 0
N). Then
T2 b) ΔU > 0, W < 0
a) T1 = T2 b)T1 =
6 c) ΔU > 0, Q > 0
c) T1 = 6T2 d) T1 = 3T2
d) ΔU = 0, Q > 0
5. When a uniform rod is heated, which
9. An ideal gas is taken from (Pi,Vi) to
of the following quantity of the rod will
(Pf,Vf ) in three different ways. Identify
increase
the process in which the work done on
a) mass the gas the most.
b) weight
P P P
c) center of mass Pf f f f
Pf Pf
d) moment of inertia
6. When food is cooked in a vessel Pi i i i
Pi Pi
V V V
by keeping the lid closed, after
Vf Vi Vf Vi Vf Vi
Process A Process B Process C
some time the steam pushes the lid

outward. By considering the steam as
a thermodynamic system, then in the
cooking process

a) Process A
P
b)Process B
D C
c) Process C
d)
Equal work is done in Process
A B
A,B &C
10. The V-T diagram of an ideal gas d)
V
which goes through a reversible

11. A distant star emits radiation with
cycle A→B→C→D is shown below.
maximum intensity at 350 nm. The
(Processes D→A and B→C are
temperature of the star is
adiabatic)
a) 8280 K b) 5000K
V C
c) 7260 K d) 9044 K
B
12. Identify the state variables given here?
D
a) Q, T, W b) P, T, U
A
c) Q, W d) P, T, Q
T 13. In an isochoric process, we have
a) W = 0 b) Q = 0
The corresponding PV diagram for the
c) ∆U = 0 d) ∆T = 0
process is (all figures are schematic)
14. The efficiency of a heat engine working
P
between the freezing point and boiling
A B point of water is
(NEET 2018)
D C a) 6.25% b)20%
c) 26.8% d)12.5%
V
a) 15. An ideal refrigerator has a freezer at
P temperature −12°C. The coefficient of
D C performance of the engine is 5. The
temperature of the air (to which the
heat ejected) is
A B
a) 50°C (b)45.2°C
V c) 40.2°C (d)37.5°C
b)
Answers:
P
1) a 2) c 3) b 4) b
A B
5) d 6) a 7) b 8) c
9) b 10) b 11) a 12) b
D C 13) a 14) b 15) c
V
c)

II. Short answer questions: 23. Did joule converted mechanical energy
1. ‘An object contains more heat’- is it a to heat energy? Explain.
right statement? If not why? 24. State the first law of thermodynamics.
2. Obtain an ideal gas law from Boyle’s 25. Can we measure the temperature of the
and Charles’ law. object by touching it?
3. Define one mole. 26. Give the sign convention for Q and W.
4. Define specific heat capacity and give 27. Define the quasi-static process.
its unit. 28. Give the expression for work done by
5. Define molar specific heat capacity. the gas.
6. What is a thermal expansion? 29. What is PV diagram?
7. Give the expressions for linear, area 30. Explain why the specific heat capacity
and volume thermal expansions. at constant pressure is greater than
8. Define latent heat capacity. Give its the specific heat capacity at constant
unit. volume.
9. State Stefan-Boltzmann law. 31. Give the equation of state for an

isothermal process.
10. What is Wien’s law?
32. Give an expression for work done in an
11. Define thermal conductivity. Give its
isothermal process.
unit.
33. Express the change in internal energy
12. What is a black body?
in terms of molar specific heat capacity.
13. What is a thermodynamic system?
34. Apply first law for (a) an isothermal (b)
Give examples.
adiabatic (c) isobaric processes.
14. What are the different types of
35. Give the equation of state for an
thermodynamic systems?
adiabatic process.
15. What is meant by ‘thermal equilibrium’?
36. Give an equation state for an isochoric
16. What is mean by state variable? Give process.
example.
37. If the piston of a container is pushed
17. What are intensive and extensive fast inward. Will the ideal gas equation
variables? Give examples. be valid in the intermediate stage? If
18. What is an equation of state? Give an not, why?
example. 38. Draw the PV diagram for
19. State Zeroth law of thermodynamics. a. Isothermal process
20. Define the internal energy of the b. Adiabatic process
system.
c. isobaric process
21. Are internal energy and heat energy
d. Isochoric process
the same? Explain.
39. What is a cyclic process?
22. Define one calorie.

40. What is meant by a reversible and 9. Discuss the
irreversible processes? a. thermal equilibrium
41. State Clausius form of the second law b. mechanical equilibrium
of thermodynamics c. Chemical equilibrium
42. State Kelvin-Planck statement of d. thermodynamic equilibrium.
second law of thermodynamics.
10. Explain Joule’s Experiment of the
43. Define heat engine. mechanical equivalent of heat.
44. What are processes involves in a Carnot 11. Derive the expression for the work
engine? done in a volume change in a
45. Can the given heat energy be completely thermodynamic system.
converted to work in a cyclic process? If 12. Derive Mayer’s relation for an ideal gas.
not, when can the heat can completely
13. Explain in detail the isothermal
converted to work?
process.
46. State the second law of thermodynamics
14. Derive the work done in an isothermal
in terms of entropy.
process
47. Why does heat flow from a hot object
15. Explain in detail an adiabatic process.
to a cold object?
16. Derive the work done in an adiabatic
48. Define the coefficient of performance.
process
III. Long answer Questions: 17. Explain the isobaric process and derive
the work done in this process
1. Explain the meaning of heat and work
18. Explain in detail the isochoric process.
with suitable examples.
19. What are the limitations of the first law
2. Discuss the ideal gas laws.
of thermodynamics?
3. Explain in detail the thermal expansion.
20. Explain the heat engine and obtain its
4. Describe the anomalous expansion of efficiency.
water. How is it helpful in our lives?
21. Explain in detail Carnot heat engine.
5. Explain Calorimetry and derive an
22. Derive the expression for Carnot
expression for final temperature when
engine efficiency.
two thermodynamic systems are mixed.
23. Explain the second law of
6. Discuss various modes of heat transfer.
thermodynamics in terms of entropy.
7. Explain in detail Newton’s law of
24. Explain in detail the working of a
cooling.
refrigerator.
8. Explain Wien’s law and why our eyes
are sensitive only to visible rays?

TT
1 2 (PV
1 1 + P2V2 )
IV. Numerical Problems Answer: T =
PV
1 1T2 + P2V2T1
1. Calculate the number of moles of 4. The temperature of a uniform rod
air is in the inflated balloon at room of length L having a coefficient of
temperature as shown in the figure. linear expansion αL is changed by ∆T.
Calculate the new moment of inertia
of the uniform rod about axis passing
through its center and perpendicular
to an axis of the rod.
Answer: I' = I (1 + αL ∆T)2
5. Draw the TP diagram (P-x axis, T-y

axis), VT(T-x axis, V-y axis) diagram
for
a. Isochoric process
b. Isothermal process
c. isobaric process
The radius of the balloon is 10 cm, and 6. A man starts bicycling in the morning
pressure inside the balloon is 180 kPa. at a temperature around 25°C, he
Answer: μ ≅ 0.3 mol checked the pressure of tire which is
equal to be 500 kPa. Afternoon he found
2. In the planet Mars, the average
that the absolute pressure in the tyre is
temperature is around -53°C and
increased to 520 kPa. By assuming the
atmospheric pressure is 0.9 kPa.
expansion of tyre is negligible, what is
Calculate the number of moles of the
the temperature of tyre at afternoon?
molecules in unit volume in the planet
Mars? Is this greater than that in earth? Answer: T= 36.9°C
Answer: μMars = 0.49 mol 7. Normal human body of the temperature
is 98.6°F. During high fever if the
μEarth ≅ 40 mol temperature increases to 104°F, what
3. An insulated container of gas has two is the change in peak wavelength that
chambers separated by an insulating emitted by our body? (Assume human
partition. One of the chambers has body is a black body)
volume V1 and contains ideal gas at Answer: (a) λmax ≈ 9348 nm at 98.6°F
pressure P1 and temperature T1. The (b) λmax ≈ 9258 nm at 104°F
other chamber has volume V2 and
8. In an adiabatic expansion of the air,
contains ideal gas at pressure P2 and
the volume is increased by 4%, what is
temperature T2. If the partition is
percentage change in pressure?
removed without doing any work on
the gases, calculate the final equilibrium (For air γ = 1.4)
temperature of the container. Answer: 5.6%

9. In a petrol engine, (internal combustion 11. An ideal gas is taken in a cyclic process
engine) air at atmospheric pressure as shown in the figure. Calculate
and temperature of 20°C is compressed (a) work done by the gas.
in the cylinder by the piston to 1/8 (b) work done on the gas
of its original volume. Calculate the
(c) Net work done in the process
temperature of the compressed air.
P (Pa)
(For air γ = 1.4) 600
A
Answer: T ≅ 400°C
10. Consider the following cyclic process
400 B
consist of isotherm, isochoric and C
isobar which is given in the figure.

0 3 6
P V
V (m)
1 3 2
3
1 Answer: (a) W = +1.5kJ

2
V (b)
T W = −1.2kJ
(c) W = +300J.
Draw the same cyclic process 12. For a given ideal gas 6 × 105J heat
qualitatively in the V-T diagram where energy is supplied and the volume of
T is taken along x direction and V is gas is increased from 4 m3 to 6 m3 at
taken along y-direction. Analyze the atmospheric pressure. Calculate (a)
nature of heat exchange in each process. the work done by the gas (b) change in
Answer: T= 36.9°C internal energy of the gas (c) graph this
process in PV and TV diagram.
P V
2 3
1 3 Answer: (a) W = +202.6 kJ
(b) dU = 397.4 kJ
1
2
P T
V T
tm

1a
P=
1 atm
Process 1 to 2 = increase in volume. So
heat must be added. 4 6 V (m) 4 6 V (m)
Process 2 to 3 = Volume remains (c)

constant. Increase in temperature. 13. Suppose a person wants to increase the
The given heat is used to increase the efficiency of the reversible heat engine
internal energy. that is operating between 100°C and
Process 3 to 1 : Pressure remains 300°C. He had two ways to increase
constant. Volume and Temperature are the efficiency. (a) By decreasing the
reduced. Heat flows out of the system. cold reservoir temperature from 100°C
It is an isobaric compression where the to 50°C and keeping the hot reservoir
work is done on the system. temperature constant (b) by increasing

the temperature of the hot reservoir engine of efficiency 60% what must be
from 300°C to 350°C by keeping the the intake temperature for the same
cold reservoir temperature constant. exhaust (sink) temperature?
Which is the suitable method?
Answer: 552°C
Answer: Initial efficiency = 44.5% 15. An ideal refrigerator keeps its content
Efficiency in method (a) =52 % at 0°C while the room temperature
Efficiency in method (b) =48 % is 27°C. Calculate its coefficient of
Method (a) is more efficient. performance.
14. A Carnot engine whose efficiency is 45% Answer: β=10.11
takes heat from a source maintained
at a temperature of 327°C. To have an
BOOKS FOR REFERENCE
1. Serway and Jewett, Physics for scientist and Engineers with modern physics, Brook/Coole
publishers, Eighth edition
2. Paul Tipler and Gene Mosca, Physics for scientist and engineers with modern physics,
Sixth edition, W.H.Freeman and Company
3. James Walker, Physics, Addison-Wesley publishers, 4th Edition
4. Douglas C Giancoli, Physics for scientist & Engineers with modern physics, Pearson
Prentice Hall, 4th edition.
5. H.C.Verma, Concepts of physics volume 1 and Volume 2, Bharati Bhawan Publishers
6. Tarasov and Tarasova, Question and problems in school physics, Mir Publishers

ICT CORNER
Heat and Thermodynamics
Through this activity you will be able

to learn the PV diagrams for various
thermodynamic process.
STEPS:
• Use the URL or scan the QR code to open interactive simulation on ‘Pressure and Volume
Diagram ”.
• At selected temperature, change the “Volume” given below the graph and click play button.
• Now select a different temperature, change the ‘Volume’ again and find the change in the
pressure both in the left image and graph.
• Repeat the same with different values and try drawing the graph accordingly. This also
helps to understand isothermal process.
Step1 Step2
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Tamilnadu Board Class 11 Physics Chapter 8 PDF

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Tamilnadu Board Class 11 Physics Chapter 8 PDF

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UNIT

8 HEAT AND THERMODYNAMICS

In this unit, a student is exposed to

8.1 is a branch of physics which explains the

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� When the gas is kept at constant

Uni t 8 H EAT A N D TH ER MODYNAM IC S 97

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Two Separate systems Single system

Figure 8.2 Ideal gas law

For this combined system, V becomes

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But we know PV = NkT

V1 = V2 = V Here volume is given in the Litre. 1 Litre is

E X A M P L E 8. 3 Number of oxygen molecules

Uni t 8 H EAT A N D TH ER MODYNAM IC S 99

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100 Un i t 8 HE AT A N D TH E R MODY NA MIC S

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Lead 130 When we study properties of gases, it is

Water 4186 Here C is known as molar specific heat

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(a) Linear expansion (b) Area expansion (c) Vo

(a) Linear expansion (b) Area expansion (c) Volume expansion

(a) Linear expansion (b) Area expansion (c) Volume expansion

102 Un i t 8 HE AT A N D TH E R MODY NA MIC S

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L = Original length ΔL = 10 × 10−6 × 300 × 23 = 0.69 m=69 cm

Uni t 8 H EAT A N D TH ER MODYNAM IC S 103

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Volume of 1kg of water (cm3 )

Density of water (g cm-3)

ΔT = Change in temperature 1000.25

Unit of coefficient of linear, area and

Density of water (g cm-3)

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Figure 8.8 Temperature versus heat for

When heat is added or

• The latent heat for a solid - liquid state

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Calorimetry means the measurement of

lower temperature, the heat lost by the hot Insulating

Qgain + Qlost= 0 Qgain = −Qlost

Here s1 and s2 specific heat capacity of hot

Air (insulation) Calorimeter cup

The final temperature

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m1T1 + m2T2 5×323 + 4 ×303 Thermal conductivity

Figure 8.11 Steady state heat flow by

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aluminum have high thermal conductivities.

Day Day Night Night

Sea breeze Sea breeze

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transfer energy from one object to another. 80

medium to transfer the heat. 50

Radiation is a form of energy transfer 30

1 dU PV = µRT ⇒ PdV+VdP = µRdT (8.26)

dU = µCvdT(8.21) 8.8.1 Isothermal process

So the equation of state for isothermal Q = W(8.30)