Phy 212: General Physics II
Chapter 20 Worksheet (2nd Law of Thermodynamics & Heat Engines)
1
Entropy:
1. A sample of 10.0 moles of a monatomic ideal gas, held at constant temperature
(1000K), is expanded from 0.10 m3 to 0.20 m3. Calculate the entropy change during this
process.
V 
nRT  ln  f 
Q
 Vi  = 10.0 mol 8.314 J ln 2 = 57.6 J
=
Ans. S =


 
molK 
K
T
T
2. A sample of 10.0 moles of a monatomic ideal gas, held at constant pressure (1.5 atm or
1.52x105 Pa), is compressed from 0.10 m3 to 0.05 m3. Calculate the entropy change during
this process.
Ans.
S =
Sf
Qf
 dS = 
Si
Qi
dQ
=
T
Tf 3
2

Ti
V
nRdT f pdV

=
T
T
Vi
Tf 3
2

Ti
S = 52 10.0 mol 8.314 molJ K  ln  21 = -144. KJ
V
T
f 5
nRdT f pdV
nRdT

= 2
=
T
T
T
Vi
Ti
5
2
T 
nRln  f 
 Ti 
3. A sample of 10.0 moles of a monatomic ideal gas, held at constant volume (1.0 m3), is
heated from 300 K to 400 K. Calculate the entropy change during this process.
Ans.
S =
Qf

Qi
dQ
=
T
Tf 3
2

Ti
T 
nRdT 3
400K
= 2 nRln  f = 23 10.0 mol 8.314 molJ K  ln  300K
= 35.9 KJ
T
T
 i
4. A sample of 1.5 moles of a thermally insulated (adiabatic) monatomic ideal gas is
expanded from 0.10 m3 to 0.20 m3. During the expansion, the pressure of the gas
decreases from 3.039x105 Pa to 2.026x105 Pa. Calculate the entropy change during this
process.
Ans. S =
Sf
Qf
dQ
= 0 KJ
T
Qi
 dS = 
Si
Phy 212: General Physics II
Chapter 20 Worksheet (2nd Law of Thermodynamics & Heat Engines)
2
Heat Engines:
5. A Carnot heat engine consists of 2 isothermal and 2 adiabatic processes. Steps AB and
CD are isothermic and steps BC and DA are adiabatic. The temperature at CD (TH)
is 1500 K and the temperature at AB (TC) 300 K. For state A, P=2490 Pa & V = 3.0 m3,
and Carnot Engine performs 4.5x104 J of work. Note: the working fluid is 3 moles of ideal
diatomic gas.
A) Sketch the P-V diagram for this heat engine.
B) What is the efficiency of this heat engine?
Ans.
e=1-
TC
= 0.80
TH
C) How much heat energy (QH) is absorbed during the isothermic phase BC?
Ans.
W
4.5 104 J
QH =
=
= 5.63 104J
e
0.80
D) How much heat energy (QC) is released during the isothermic phase DA?
Ans.
QC = -QH + W= -5.63 104 J + 4.5 104J = -1.13 104J
E) What is the pressure at each point A-D? Calculate using the ideal gas law.
Ans.
V 
Get VB from QC: QC = WA B= nRTln  B  = -1.13 104J
 VA 
VB = VA e
 PB=
QC
nRT

3

= 3.0m e
-1.13104 J
J )(300K)
(3mol)(8.314 molK

= 0.663m3
(3mol)(8.314 molJ K )(300K)
= 1.13 105Pa
3
0.663m

 nRTC 
Get PC from PC VC = PB VB: PC 

 PC 

 nRTB 
= PB 

 PB 


5
300K
4
PC = 1.13 10 Pa 

4

 (1500K)(1.13 10 Pa) 

7




  T 
 PC =  PB  B 
  TCPB 

1
1



 52
= 3.16 106Pa
5.63104 J
QH
J )(1500K)
 nRTH  nRT
 nRTH  (3mol)(8.314 molK

Get VD from QH: VD = 
= 
= 0.0531m3
e
e
 PC 
 PC 
(3mol)(8.314 molJ K )(1500K)
 PB=
= 7.05 105Pa
0.0531m3
F) How much work is performed by the gas during each step?
Ans.
WA B = QC = -1.13 104 J and WBC =
WCD = QH = 5.63 104 J and WDA =
3
2
3
2
nRT = -4.49 104 J
nRT = 4.49 104J
Phy 212: General Physics II
Chapter 20 Worksheet (2nd Law of Thermodynamics & Heat Engines)
3
H) What is the change in entropy (SA-B) during AB?
Ans. SAB =
QC
-1.13 104 J
=
= -37.7 KJ
TC
300K
I) What is the change in entropy (SC-D) during CD?
Ans. SCD =
QH
5.63 104 J
=
= 37.7 KJ
TH
1500K
J) What is the net change in entropy (Suniverse) between the steps in H and I?
Ans. Suniverse = SAB+ SAB = 0 KJ
6. Consider the following heat engine, where steps AB and CD are isobaric and steps
BC and DA are adiabatic (yes, the adiabatic steps should have a slight curvature).
Note: the working fluid in this engine is a monatomic ideal gas (10 moles).
3.50E+05
C
3.00E+05
D
PA=PB=1.0x105 Pa
PC=PD=3.0x105 Pa
Pressure (Pa)
2.50E+05
2.00E+05
VA=0.40m3
VB=0.20m3
1.50E+05
VC=0.103m3
1.00E+05
VD=0.207m3
A
B
5.00E+04
0.00E+00
0
0.1
0.2
0.3
0.4
0.5
Volume (m^3)
A) What is the temperature at each point A-D?
Ans. T=
PV
 TA =481 K, TB=241 K, TC=372 K, and TD=747 K
nR
B) How much work is performed by the gas during each step?
Ans.
WA B = pV = -2.00 104 J and WBC =
WCD = pV = 3.12  104 J and WDA =
3
2
3
2
nRT = -1.63 104 J
nRT = 3.3 104 J
C) What is the net work during the complete cycle?
Ans.
WNet=WAB+WBC+WCD+WDA=-2.00 104J-1.63 104J+3.12 104J+3.3 104 J
WNet=2.79 104J
Phy 212: General Physics II
Chapter 20 Worksheet (2nd Law of Thermodynamics & Heat Engines)
4
D) How much heat (QH) is absorbed by the gas?
Ans.
Qin = QCD = ncpT= 7.80 104J
E) How much heat (QC) is discarded by the gas?
Ans.
Qout = QCD = ncpT= -5.00 104J
F) What is the efficiency of the heat engine?
Ans.
W
2.79 104 J
e=
=
= 0.35
QH
7.80 104 J
G) What is the Carnot efficiency for the engine?
Ans.
e=1-
TC
241K
=1=0.68
TH
747K
7. Consider the following heat engine (the Stirling Engine). Steps AB and CD are
isothermic and steps BC and DA are isochoric. The temperature at CD (TH) is 2500 K
and the temperature at AB (TC) 500
K. Note: the working fluid in this
250000
engine is an ideal monatomic gas (10
moles).
A) What is the pressure at each point
A-D? Calculate using the ideal gas
law.
nRT
V
PA =8.31 104Pa
P=
PB=8.48 104Pa
B) How much work is performed by
the gas during each step?
B
A
150000
100000
50000
PC=4.24 105Pa
PD=4.16 105Pa
D
200000
Pressure (Pa)
Ans.
C
0
0.488
0.49
0.492
0.494
0.496
0.498
0.5
Volum e (m ^3)
V 
 0.49 
2
WAB = nRT  ln  B  = 10.0 mol 8.314 molJ K  500K  ln 
= -8.40 10 J
 0.50 
 VA 
 VB 
 0.50 
3
J
 = 10.0 mol  8.314 molK  2500K  ln 
= 4.20 10 J
V
0.49


 A
=0J
Ans. WCD = nRT  ln 
WBC = WDA
C) What is the net work during the complete cycle?
Ans. Wnet = WAB + WCD= -8.40 102 J + 4.20 103J = 3.36 103J
D) How much heat (Qin = QBC + QH) is absorbed by the gas?
Ans.
Qin = QCD + QDA= 4.20 103J + 2.49 105J= 2.53 105J
0.502
Phy 212: General Physics II
Chapter 20 Worksheet (2nd Law of Thermodynamics & Heat Engines)
5
E) How much heat (Qout = QDA + QC) is discarded by the gas?
Ans.
Qout = QAB + QBC= -8.40 102J - 2.49 105J= -2.50 105J
F) What is the efficiency of this Stirling engine?
Ans.
e=
W
3.36 103J
=
= 0.013
Qin
2.53 105J
G) What is the Carnot efficiency for this engine?
Ans.
e=1-
TC
500K
=1=0.80
TH
2500K
H) What is the change in entropy (SB-D) during BD?
Ans.
T  Q
S =SBC +SCD= 23 nRln  H + CD
 TC  TH
S = 23 10.0 mol  8.314 molJ K  ln  2500K
500K  +
4.20 103 J
= 202.4 KJ
2500K
I) What is the change in entropy (SD-B) during DB?
S =SAB+SBC=
Ans.
S =
T 
QAB 3
+ 2 nRln  H 
TC
 TC 
-8.40 102 J 3
500K
+ 2 10.0 mol 8.314 molJ K  ln  2500K
= -202.4 KJ

500K
J) What is the net change in entropy (Suniverse) between the steps in H and I?
Ans.
Ssystem =Suniverse=0 KJ
Statistical Interpretation of Entropy:
8. Consider a system of 2 identical coins, with equal probability of “heads” or “tails”,
respectively.
A) How many possible unique configurations are possible in this system?
Ans. Wtot = XN - 1 = 22 – 1 = 3
B) What is the multiplicity of configurations (W) that both coins will and on “tails”?
Ans.
W=
N!
2!
=
= 1, n1="heads" & n2="tails"
n1!n2!
0!2!
C) What is the entropy of the state when they are tossed and both land on “tails”?
Ans.


S = k  ln W = 1.38 10-23 KJ ln 1 = 0
D) What does your answer in (C) imply about that state of the 2 dice system? Be as
specific as possible.
Ans. All “tails” represents a perfectly ordered state and this corresponds to S=0, the
lowest possible entropy state.
Phy 212: General Physics II
Chapter 20 Worksheet (2nd Law of Thermodynamics & Heat Engines)
6
9. Consider a system of 5 identical dice, with equal probability of landing on any side,
respectively.
A) How many possible unique configurations are possible in this system?
Ans. Wtot = 6 (yahtzee) + 30 (full house) + 30 (4-of-a-kind) + 120 (3-of-a-kind) + 120 (2
pairs) and 360 (1 pair) = 667 unique combinations (states)
B) What is the multiplicity of configurations (W) that all dice will land on the same value
(i.e. a Yahtzee)?
Ans. for any particular Yahtzee value: W =
N!
5!
=
=1
n1!n2!n3!n4!n5!n6!
5!0!0!0!0!0!
C) What is the entropy of a “Yahtzee” state (all the same value) for the 5 dice system?
Ans.


S = k  ln W = 1.38 10-23 KJ ln 1 = 0
10. In information theory, entropy (measured in bits) is used to define a lower boundary
on the necessary number of numerical bit digits required to uniquely specify all possible
states of an information system (e.g. unique letters of an alphabet).
A) What is the entropy of a 36 character alphabet letter (incl. 10 numbers) system?
Assume that all letters and numbers are equally probable.
Ans: Sin bits=- 1.443
N
36
 1 
 p ln p =- 1.443   36 ln
i=1
i
i
i=1
 1 

=1.443  ln 36 =5.2 bits
 36 
A minimum of 6 bits is necessary to uniquely express this character system.
B) What is the entropy of this 36 character alphabet letter system, when numbers are 5
times more probable that letters?
Ans: Note, the probability of each character is: 26pL + 10 P# = 26pL + 10.5pL = 1
Therefore: pL = 1/76 & p# = 5/76.
 10
Sin bits=- 1.443   p# ln p#  +
 i=1

p ln p  
26
i=1
L
L
  50   5 
 26   1  
Sin bits=- 1.443  
 ln 
 + 
 ln 
 = 4.7 bits
 76   76  
  76   76 
A minimum of 5 bits is necessary to uniquely express this character system.