PHY 103 Lecture Note

PHY 103: Basic Principle of Physics II
Heat and Thermodynamics

KINETIC THEORY OF GASES Contd.
• Molar Specific Heat of an Ideal Gas
We have seen that energy required to raise the temperature of n
moles of gas from Ti to Tf depends on the path taken between the
initial and final states
• Consider the case of an ideal gas
taken from one isotherm at T to
another isotherm at T+T.
• In this case Q that takes the gas
from T to T+T at constant volume
is different from the one that takes
it from T to T+T at constant
pressure.
• But But , since n and T are
constants. Then the only quantity
that can differentiate the two paths
is , the molar specific heat.
Kinetic Theory of Gases
At a constant volume,
molar specific heat capacity at constant volume.

At constant pressure
molar specific heat capacity at constant pressure.

We shall see later that Q at a constant pressure > Q at constant
volume
Relationship between CV and CP

Consider a monoatomic gas.
• If energy is added to the gas at constant volume, from 1st law of
thermodynamics ( = Q – W) we know that all the energy is used to increase
the internal energy.
• In this case all the added energy goes to increase translational kinetic energy
of the molecules.
• Therefore, Since at constant volume,
,
Using
This equation is valid for all ideal gas

• So that .
In the limit of infinitesimal change,
recall that diffentiating with respect to T, we have

• Now consider the path when the gas is taken from isotherm T to isotherm
T+T, at a constant pressure.
• Energy transferred to the gas in this process is given by
The change in internal energy for the process , however is equal to that for the
process because internal energy depends only on temperature for an ideal gas
and is the same for both processes. So, in the two cases.
• In this path, work PdV is done and this can be obtained using PV=nRT,
differentiating this we have,
• since, the process is an isobaric,

• From first law equation, = Q – W, we have
This is valid for any ideal gas. It predicts that the molar specific heat of an ideal
gas at constant pressure is greater than the molar specific heat at constant
volume by an amount R.
• For a monoatomic gas where and ,

• Let be the ratio of molar specific heat at a constant pressure to molar specific
heat at a constant volume, i.e.,
Adiabatic Processes for an Ideal Gas
PV Diagram for an adiabatic process
Adiabatic expansion process,

from state a to state b
Kinetic Theory of Gases: Adiabatic Processes for an Ideal Gas contd.
• As noted before in an adiabatic process, no energy is transferred by heat between a

system and its surrounding.
• In an adiabatic process, pressure and volume are related by the expression

• *
Where c is a constant
Prove of equation (*)
• In adiabatic process, recall Q = 0, W = PdV, the change in internal energy will be the
same as for isovolumetric process between the same temperatures,
(because change in internal energy depends only on temperature).
• Therefore the 1st law equation, (dEint = Q – W),
(**)
Taking the total differential of , we have

(***)
making the subject of the relation in equation (**),
we have,
and substituting in equation(***) we have,

which can be written as;
,
(++)
Recall that , so equation (++) becomes
Divide both sides by PV we have
Opening the bracket and rearranging, we arrive at

By integrating we get
where K is the constant of integration.

() 1.0
i.e.
1.1
Which gives the relationship between pressure and volume in an adiabatic
process.
• Relationship between temperature and volume in adiabatic processes
Using so that in equation 1.0, we have
This becomes,
Since n and R are constants, we have

1.3
Where
i.e.
1.4
Which gives the relationship between temperature and volume in adiabatic
processes.
• Relationship between temperature and pressure in adiabatic processes
• From equation of state of an ideal gas, we have
• substitute this expression in equation (1.0)
Since n and R are constant, we have,
1.5
Where (constant)
i.e.
1.6
Work done in adiabatic processes
• Recall that (constant) so that
Also, recall that,
Integrating to have,
Using the fact that,
1.7
Equation (1.7) is the formula to calculate work done of an adiabatic process.
In Summary, in adiabatic process,
Relationship between pressure Relationship between Pressure and
and volume is given by: Temperature is written as:
Relationship between The work done in adiabatic process is

temperature and volume is: given by:
WORKED EXAMPLES
1. Consider the adiabatic process
represented by the curve on the
right side.
(a) Determine the pressure in stage
2 in the process
(b) If the temperatute at stage 1 is
270C, what will the temperature
at stage 2 be?
(c) Calculate the sum of work done
of all the processes where work
is done (take .
SOLUTION
P1 = 2 x 105N/m2 P2 = ?
V1 = V3 =0.05m3 V2 = 0.1m3
b) In an adiabatic process, temperature and pressure are related by:
This shows that adiabatic expansion leads to decrease in temperature and hence,
decrease in the internal energy of a system as said in the previous classes.
•
c) There are just two processes where work is done in all the three: Stage 1 to 2 and stage 2 to 3.
For state 1 to state 2,
(Using area under the curve will still work here perfectly)
• For state 2 to state 3 (isobaric),
Total work done

Question SOLUTION
2. A diatomic gas is at an initial • a) For an isovolumetric process,
temperature of 300 K. The gas
undergoes an isovolumetric process,
acquiring 500 J of energy by heat. It
then undergoes an isobaric process, n = 1, recall that,
losing this same amount of energy by
heat. Take the mole of the gas to be
1mol, determine:
• (a) the new temperature of the gas
and
• (b) the work done on the gas. For isobaric process,
n=1
24.1
b) Using the first law equation, = Q – W

and since Q absorbed by the system during the first stage (isovolumetric
process) has been lost during the second process (isobaric process),
consequently, Q = 0, we now have,
In this case, ,
Question
• Two moles of an ideal gas ( = 1.40) expands slowly and adiabatically

from a pressure of 5.00 atm and a volume of 12.0 L to a final volume
of 30.0 L. (a) What is the final pressure of the gas? (b) What are the
initial and final temperatures? (c) Find Q, W, and Eint .
• 1atm= 1.013x105N/m2.
Solution
a) We use the relationship between pressure and volume in adiabatic
process,
So that,
b) we need to determine the initial temperature, using
Using PV = nRT
Pi = 5 x 1.013 x 105 N/m2 = 5.065 x 105 N/m2

Vi = 12 L = 12 x 10-3 m3.
Ti = 5.065 x 105 x 12 x 10-3 /(2*8.31)= 365.7 K
• Similarly, PfVf = nRTf  Tf = PfVf/nR
Question
Pf = 1.39 x 1.013 x 105 N/m2 = 1.408 x 105 N/m2
Vf = 30 L = 30 x 10-3 m3.
Tf = 1.408 x 105 x 30 x 10-3 /(2*8.31)= 254.2 K
OR you can use to determine
• (C) In adiabatic process, Q = 0 J
Eint = (DO IT IN CLASS)

The figure above shows an adiabatic
expansion in 2.5mol of monoatomic
ideal gas from 22.5 litres at 1 bar and
00C (point A) to 45.5 litres (at point C)
and the pressure decreases to 0.3bar.
a. Determine the temperature at
point C.
b. How much work is done in the
adiabatic expansion?
c. How much work is done in the
isothermal expansion at the final
volume.
1bar =105N/m2 or Pascal, Take  = 2.0
b) Work done in adiabatic process,
a) Using the relationship between
pressure and temperature in adiabatic
process
,
,
c)Work done in isothermal process,
So,
TO BE CONTINUED……….………..
ALL THE BEST

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PHY 103: Basic Principle of Physics II

Heat and Thermodynamics

molar specific heat capacity at constant volume.

molar specific heat capacity at constant pressure.

Relationship between CV and CP

This equation is valid for all ideal gas

recall that diffentiating with respect to T, we have

• since, the process is an isobaric,

• For a monoatomic gas where and ,

Adiabatic expansion process,

• As noted before in an adiabatic process, no energy is transferred by heat between a

• In an adiabatic process, pressure and volume are related by the expression

Taking the total differential of , we have

and substituting in equation(***) we have,

Recall that , so equation (++) becomes

Divide both sides by PV we have

Opening the bracket and rearranging, we arrive at

where K is the constant of integration.

Since n and R are constants, we have

Since n and R are constant, we have,

Also, recall that,

Using the fact that,

Relationship between The work done in adiabatic process is

b) In an adiabatic process, temperature and pressure are related by:

Total work done

b) Using the first law equation, = Q – W

• Two moles of an ideal gas ( = 1.40) expands slowly and adiabatically

Pi = 5 x 1.013 x 105 N/m2 = 5.065 x 105 N/m2

Eint = (DO IT IN CLASS)

ALL THE BEST

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