Chapter14 Thermodynamics

Chapter 14
Thermodynamics
Reference:chap.15
Alan Giambattista, Betty McCarthy Richardson and Robert C. Richardson,
2009, “College Physics”, 3rd ed., McGraw-Hill.
Only for education
•The First Law of Thermodynamics
•Thermodynamic Processes (isobaric, isochoric, isothermal, adiabatic)
•Reversible and Irreversible Processes
•Heat Engines
•Refrigerators and Heat Pumps
•The Carnot Cycle
•Entropy (The Second Law of Thermodynamics)
•The Third Law of Thermodynamics
普物AB—海大輪機系 1
張宏宜
The First Law of Thermodynamics
The first law of thermodynamics says

the change in internal energy of a system
is equal to the heat flow into the system
plus the work done on the system.
ΔU = Q + W
張宏宜
Thermodynamic Processes
A state variable describes the state of a system at time t,

but it does not reveal how the system was put into that state.
Examples of state variables: pressure, temperature, volume,
number of moles, and internal energy.
A PV diagram can be used to represent the state changes
of a system, provided the system is always near
equilibrium.
The area under a PV curve

gives the magnitude of the
work done on a system.
W>0 for compression and
W<0 for expansion.
張宏宜
To go from the state (Vi, Pi) by the path (a) to the state (Vf, Pf)
requires a different amount of work then by path (b). To
return to the initial point (1) requires the work to be nonzero.
The work done on a system depends on the path taken in

the PV diagram. The work done on a system during a
closed cycle can be nonzero.
張宏宜
An isothermal
process implies
that both P and
V of the gas
change (PV∝T).
張宏宜
Thermodynamics for an Ideal Gas
No work is done on a system

when its volume remains
constant (isochoric process).
For an ideal gas (provided the
number of moles remains
constant), the change in internal
energy is
Q = ΔU = nCV ΔT .
張宏宜
Thermodynamics for an Ideal Gas
For a constant pressure (isobaric) process, the change in

internal energy is ΔU = Q + W
where W = − PΔV = − nRΔT and Q = nC P ΔT .
CP is the molar specific heat at
constant pressure. For an ideal gas
CP = CV+R.
For a constant temperature

(isothermal) process, ΔU = 0 and
the work done on an ideal gas is
⎛ Vi ⎞
W = nRT ln⎜⎜ ⎟⎟.
⎝ Vf ⎠
張宏宜
Summary of Thermodynamic Process
張宏宜
Example (text problem 15.7)
An ideal monatomic gas is taken through a cycle in the PV

diagram. (a) If there are 0.0200 mol of this gas, what are
the temperature and pressure at point C?
From the graph:

Pc = 98.0 kPa
Using the ideal gas law
PcVc
Tc = = 1180 K.
nR
張宏宜
(b) What is the change in internal energy of the gas as

it is taken from point A to B?
This is an isochoric process so W = 0 and ΔU = Q.
⎛ 3 ⎞⎛ PBVB PAVA ⎞
ΔU = Q = nCV ΔT = n⎜ R ⎟⎜ − ⎟
⎝ 2 ⎠⎝ nR nR ⎠
= (PBVB − PAVA )
3
2
= V (PB − PA ) = −200 J
3
2
張宏宜
(c) How much work is done by this gas per cycle?
The work done per cycle is the area between the

curves on the PV diagram. Here W=½ΔVΔP = 66 J.
(d) What is the total change in internal energy of this

gas in one cycle?
⎛ 3 ⎞⎛ Pf Vf PiVi ⎞
ΔU = nCV ΔT = n⎜ R ⎟⎜ − ⎟
⎝ 2 ⎠⎝ nR nR ⎠
= (Pf Vf − PiVi ) = 0
3
2
The cycle ends where
11
普物AB—海大輪機系 it began (ΔT = 0).
張宏宜
An ideal gas is in contact with a heat reservoir so that it

remains at constant temperature of 300.0 K. The gas is
compressed from a volume of 24.0 L to a volume of 14.0 L.
During the process, the mechanical device pushing the
piston to compress the gas is found to expend 5.00 kJ of
energy. How much heat flows between the heat reservoir
and the gas, and in what direction does the heat flow occur?
This is an isothermal process, so ΔU = Q + W = 0 (for

an ideal gas) and W = -Q = -5.00 kJ. Heat flows from
the gas to the reservoir.
張宏宜
Reversible and Irreversible Processes
A process is reversible if it does not violate any law of

physics when it is run backwards in time. For example an
ice cube placed on a countertop (廚具檯面) in a warm room
will melt. The reverse process cannot occur: an ice cube
will not form out of the puddle of water (一灘水) on the
countertop in a warm room.
A collision between two billiard balls is reversible.

Momentum is conserved if time is run forward; momentum
is still conserved if time runs backwards.
張宏宜
Reversible and Irreversible Processes
Any process that involves dissipation of energy is not

reversible.
Any process that involves heat transfer from a hotter object
to a colder object is not reversible.
The second law of thermodynamics (Clausius Statement):

Heat never flows spontaneously from a colder body to a
hotter body.
張宏宜
Heat Engines
A heat engine is a device designed to convert disordered

energy into ordered energy. The net work done by an engine
during one cycle is equal to the net heat flow into the engine
during the cycle (ΔU= 0).
W net = Qnet
The efficiency of an engine is defined as
net work done by the engine Wnet

e= = .
heat input Qin
Note: Qnet = Qin - Qout
張宏宜
Internal Combustion Engine
1. Intake stroke: The piston is pulled out,

drawing the fuel-air mixture into the cylinder
at atmospheric pressure.
2. Compression stroke: The piston is pushed
back in, compressing the fuel-air mixture and
work is done on the gas.
3. Ignition: A spark ignites the gases, quickly
and dramatically raising the temperature and
pressure.
4. Power stroke: The high pressure that results
from ignition pushes the piston out. The gases
do work on the piston and some heat flows
out of the cylinder.
5. Exhaust stroke: a valve opened and the
exhaust gases are pushed out of the cylinder.
張宏宜
Refrigerators and Heat Pumps
In a heat engine, heat flows

from hot to cold, with work
as the output. In a heat
pump, heat flows from cold
to hot, with work as the input.
The efficiency of a heat engine

can be rewritten as
net work output Wnet
e= =
heat input QH
QH − QC QC
= = 1− .
QH QH
張宏宜
(a) How much heat does an engine with efficiency of 33.3 %

absorb in order to deliver 1.00 kJ of work?
Wnet 1.00 kJ
QH = = = 3.00 kJ
e 0.333
(b) How much heat is exhausted by the engine?
QC
e = 1−
QH
QC = (1 − e ) QH = 2.00 kJ
張宏宜
Reversible Engines and Heat Pumps
A reversible engine can be used as an engine (heat input from
a hot reservoir and exhausted to a cold reservoir) or as a heat
pump (heat is taken from cold reservoir and exhausted to a
hot reservoir).
From the second law of thermodynamics, no engine can
have an efficiency greater than that of an ideal reversible
engine that uses the same two reservoirs. The efficiency of
this ideal reversible engine is TC
er = 1 − .
TH
張宏宜
Carnot Cycle
The ideal engine of the previous section is known as a

Carnot engine.
The Carnot cycle has four steps:
1. Isothermal expansion: takes in heat from hot reservoir;

keeping the gas temperature at TH.
2. Adiabatic expansion: the gas does work without heat flow
into the gas; gas temperature decreases to TC.
3. Isothermal compression: Heat QC is exhausted; gas
temperature remains at TC.
4. Adiabatic compression: raises the temperature back to TH.
張宏宜
Carnot Cycle
張宏宜
An engine operates between temperatures 650 K and

350 K at 65.0% of its maximum efficiency. (a) What is the
efficiency of this engine?
The maximum possible efficiency is
TC 350 K
er = 1 − = 1− = 0.462.
TH 650K
The engine operates at e = 0.65er = 0.30 or 30% efficiency.
張宏宜
(b) If 6.3×103 J is exhausted to the low temperature reservoir,

how much work does the engine do?
The heats exchanged at the reservoirs are related to each

other through
QC = (1 − e ) QH .
Wnet = QH − QC
QC ⎛ e ⎞
= − QC = ⎜ ⎟ QC = 2.7 kJ
(1 − e) ⎝1− e ⎠
張宏宜
Entropy
Heat flows from objects of high temperature to objects at

low temperature because this process increases the
disorder of the system. Entropy is a measure of a system’s
disorder. Entropy is a state variable.
If an amount of heat Q flows into a system at constant
temperature, then the change in entropy is Q
ΔS = .
T
Every irreversible process increases the total entropy of the
universe. Reversible processes do not increase the total
entropy of the universe.
The second law of thermodynamics (Entropy Statement):
The entropy of the universe never decreases.
張宏宜
An ice cube at 0.0 °C is slowly melting. What is the change

in the ice cube’s entropy for each 1.00 g of ice that melts?
To melt ice requires Q = mLf joules of heat. To melt one

gram of ice requires 333.7 J of energy.
The entropy change is
Q 333.7 J
ΔS = = = 1.22 J/K.
T 273 K
張宏宜
Statistical Interpretation of Entropy
A microstate specifies the state of each constituent particle
in a thermodynamic system. A macrostate is determined
by the values of the thermodynamic state variables.
number of microstates corresponding to the macrostate

probability of a macrostate =
total number of microstates for all possible macrostates
張宏宜
Statistical Interpretation of Entropy
The number of microstates for a given macrostate is related

to the entropy.
where Ω is the number
S = k ln Ω of microstates.
Number of microstate-n/N (n=0, 1, 2,--- N),

Where n=number of heads for N=4 coins,
10 coins, 100 coins, 6×1023 coins. 27
普物AB—海大輪機系
張宏宜
For a system composed of two identical dice, let the macrostate be
defined as the sum of the numbers showing on the top faces. What is the
maximum entropy of this system in units of Boltzmann’s constant?
Sum Possible microstates
2 (1,1)
The maximum entropy
corresponds to a sum of 3 (1,2); (2,1)
7 on the dice. For this 4 (1,3); (2,2); (3,1)
macrostate, Ω = 6 with an 5 (1,4); (2,3); (3,2); (4,1)
entropy of 6 (1,5); (2,4), (3,3); (4,2); (5,1)
S = k ln Ω = k ln 6 = 1.79k . 7 (1,6); (2,5); (3,4), (4,3); (5,2); (6,1)
8 (2,6); (3,5); (4,4) (5,3); (6,2)
9 (3,6); (4,5); (5,4) (6,3)
10 (4,6); (5,5); (6,4)
11 (5,6); (6,5)
普物AB—海大輪機系 12 (6,6) 28
張宏宜
The Third Law of Thermodynamics
It is impossible to cool a system to absolute zero.
張宏宜
Summary
•The Three Laws of Thermodynamics

•Thermodynamic Processes
•Reversible and Irreversible Processes
•Heat Engines and Heat Pumps
•Efficiency of an Engine
•Entropy
張宏宜

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Chapter 14

The first law of thermodynamics says

A state variable describes the state of a system at time t,

The area under a PV curve

The work done on a system depends on the path taken in

No work is done on a system

For a constant pressure (isobaric) process, the change in

For a constant temperature

An ideal monatomic gas is taken through a cycle in the PV

From the graph:

Using the ideal gas law

(b) What is the change in internal energy of the gas as

This is an isochoric process so W = 0 and ΔU = Q.

(c) How much work is done by this gas per cycle?

The work done per cycle is the area between the

(d) What is the total change in internal energy of this

An ideal gas is in contact with a heat reservoir so that it

This is an isothermal process, so ΔU = Q + W = 0 (for

A process is reversible if it does not violate any law of

A collision between two billiard balls is reversible.

Any process that involves dissipation of energy is not

The second law of thermodynamics (Clausius Statement):

A heat engine is a device designed to convert disordered

The efficiency of an engine is defined as

net work done by the engine Wnet

Note: Qnet = Qin - Qout

1. Intake stroke: The piston is pulled out,

In a heat engine, heat flows

The efficiency of a heat engine

(a) How much heat does an engine with efficiency of 33.3 %

(b) How much heat is exhausted by the engine?

The ideal engine of the previous section is known as a

1. Isothermal expansion: takes in heat from hot reservoir;

An engine operates between temperatures 650 K and

The maximum possible efficiency is

The engine operates at e = 0.65er = 0.30 or 30% efficiency.

(b) If 6.3×103 J is exhausted to the low temperature reservoir,

The heats exchanged at the reservoirs are related to each

Heat flows from objects of high temperature to objects at

An ice cube at 0.0 °C is slowly melting. What is the change

To melt ice requires Q = mLf joules of heat. To melt one

The entropy change is

number of microstates corresponding to the macrostate

The number of microstates for a given macrostate is related

Number of microstate-n/N (n=0, 1, 2,--- N),

It is impossible to cool a system to absolute zero.

•The Three Laws of Thermodynamics

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