General Physics III Practice Exam I Solutions Fall 2007

General Physics III Practice Exam I Solutions Fall 2007
PROBLEM SCORE
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2. _______/ 38
3. _______/ 22 TOTAL SCORE ______________ / 100
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PROBLEMS
1. Determine the latent heat of fusion of mercury (Hg) using the following calorimetric data: 1.00 kg of solid Hg
at its melting point of -39°C is placed in a 0.62 kg aluminum calorimeter with 0.43 kg of water at 12.80°C; the resulting
equilibrium temperature is 5.06°C.
The specific heat of aluminum is c Al = 900 J/kg°C, of water is c water = 4186 J/kg°C and of mercury is c Hg = 138 J/kg°C.
Solution:
(20 points)
Heat lost = Heat gained:
(m Al c Al + m water c water ) ΔT water = m Hg (L Hg + c Hg ΔT Hg )
[(0.620 kg)(900 J/kg · °C) + (0.430 kg)(4186 J/kg · °C)](12.80°C – 5.06°C) =
(1.00 kg){L Hg + (138 J/kg · °C)[5.06°C – (– 39.0°C)]},
which gives L Hg = 1.22x10 4 J/kg

2. A heat engine uses 1.00 mol of a monatomic gas and follows the cycle shown in the diagram below. The
temperature of state 1 is 300 K. The energy transferred by heat to the gas during the process 1 to 2 is 3750
J.
(a) Calculate the volume of the gas for state 1, and the pressures, volumes and temperatures for the gas
at states 2, 3 and 4. Show your work and list the values in the table below.
State P (kPa) V (m 3 ) T (K)
1 300 8.31 x10 -3 300
2 601 8.31 x10 -3 601
3 601 1.66 x10 -2 1202
4 300 1.66 x10 -2 600
(a) 12 points
(1pt for V 1 ,V 2 , V 3 ,V 4 , P 2 , P 3 )
(2 pts. for T 2 , T 3 , T 4 )
(a) The first law of thermodynamics is Q =Δ Eth + WS
. For the isochoric process 1 → 2, W S 1 → 2 = 0 J. Thus,
Q = 1 2
3750 J = Δ E = th
nC Δ
→ V
T
3750 J 3750 J 3750 J
⇒Δ T = = = = 301 K
3 3
nCV ( 1.0 mol)( 2
R) ( 1.0 mol)( 2)( 8.31 J/mol K)
⇒T − T = 300.8 K ⇒ T = 300.8 K + 300 K = 601 K
To find volumes V 2 and V 1 ,
V
2 1 2
( 1.0 mol )( 8.31 J/mol K)( 300 K) −3 3
nRT
= V = = = 8.31×
10 m
3.0 10 Pa
1
2 1 5
p1
×
The pressure p 2 can be obtained from the isochoric condition as follows:
p p T ⎛
= ⇒ p = p = ( )
T T T
⎜
× = ×
300 K
⎟
⎝ ⎠
2 1 2
2 1
2 1 1
601 K ⎞ 3.00 10
5 Pa 6.01 10
5 Pa
With the above values of p 2 , V 2 and T 2 , we can now obtain p 3 , V 3 and T 3 . We have
The value of p 4 = p 3 , and V 4 = V 1 ; and
V = 2V = 1.662 × 10 m p = p = 6.01×
10 Pa
−2 3 5
3 2 3 2
T T V
= ⇒ T = T = 2T
= 1202 K
V V V
3 2
3
3 2 2
3 2 2
T T V
= ⇒ T = T = 2T
= 600 K
V V V
4 1 4
4 1 1
4 1 1

(b)
Determine the work done by the gas system W S , the heat Q, and the change in
internal energy for each of the four processes in the cycle. Show your work and
list your answers in the table below.
Process W S (J) Q (J) ΔE int (J)
1 → 2 0 3750 3750
2 → 3 4990 12,480 7490
3 → 4 0 –7500 –7500
4 → 1 –2490 –6230 –3740
(b) (17 Points) (1 pt. for W 12 , W 34 , Q 34, ΔE int12. ΔE int34 ), (2 pts. for W 23 , W 41 , Q 13 , Q 41 , ΔE int23. ΔE int41 )
For the isochoric process 1 → 2,
W S 1 → 2 = 0 J. Thus, ΔE th = Q 12 = 3750 J
For the isobaric process 2 → 3,
5 5
( )( )( ) ( )( )( )( )
Q = nC Δ T = 1.0 mol R T − T = 1.0 mol 8.31 J/mol K 601 K = 12,480 J
2→3 P 2 3 2
2
S 2→3 3 3 2
We are now able to obtain p 4 , V 4 and T 4 . We have
5 −3 3
( ) ( )( )
W = p V − V = 6.01× 10 Pa 8.31× 10 m = 4990 J
Δ E = Q − W = 12,480 J − 4990 J = 7490 J
th 2→3 S 2→3
V = V = 1.662 × 10 m p = p = 3.00 × 10 Pa
−2 3 5
4 3 4 1
5
T4 T3
p ⎛
4
3.00 × 10 Pa ⎞
= ⇒ T4 = T3 = ⎜
5 ⎟( 1202 K)
= 600 K
p4 p3 p3
⎝ 6.01×
10 Pa ⎠
For isochoric process 3 → 4,
3 3
( )( )( ) ( )( )( )( )
Q = nC Δ T = 1.0 mol R T − T = 1.0 mol 8.31 J/mol K − 602 = −7500 J
3→4 V 2 4 3
2
W = 0 J ⇒ Δ E = Q − W = −7500 J
S 3→4 th 3→4 S 3→4
For isobaric process 4 → 1,
Q = nC Δ T = 1.0 mol
5
8.31 J/mol K 300 K − 600 K = −6230 J
4→1 P 2
S 4→1 4 1 4
( ) ( )( )
5 −3 3 −2 3
( ) ( ) ( )
W = p V − V = 3.00 × 10 Pa × 8.31× 10 m − 1.662 × 10 m = −2490 J
th 4→1 S 4→1
( )
Δ E = Q − W = −6230 J − − 2490 J = −3740 J
(c) Calculate the efficiency of this heat engine and compare it to the efficiency of a Carnot heat engine
operating between the same maximum and minimum temperatures as this cycle.
(5 points)
The thermal efficiency of this heat engine is
η
W W
2500 J
Q Q + Q 3750 J + 12,480 J
out
out
= = = = =
H 1→2 2→3
0.154 15.4%
The efficiency of a Carnot heat engine operating between the same maximum and minimum temperatures
would be (4 points)
TC
300K
e = 1 − C
1 1 0.25 0.75 75%
T
= − 1200K
= − = =
H

3. Two samples of an ideal gas are initially at the same temperature and pressure. They are each compressed
reversibly from volume V to volume V/2, one isothermally, the other adiabatically.
(a) In which sample is the final pressure greater?
(b) Determine the change in entropy of the gas for each process.
(c) What is the entropy change of the environment for each process?
(22 points)
Solution:
(a) For the isothermal process we find the ratio of pressures from the ideal gas equation:
(P 2 V 2 /P 1 V 1 ) = T 2 /T 1 = 1, so P 2 /P 1 = V 1 /V 2 = 2.
(4 points)
For the adiabatic process,
P 1 V γ 1 = P 2 V γ 2 , so P 2 /P 1 = (V 1 /V 2 ) γ = 2 γ .
(4 points)
Because γ > 1, P2/P1 > 2 in the adiabatic case, so the final pressure is greater for the adiabatic process.
(b)
For the isothermal process we have
ΔE int = 0; Q = -W.
Thus ΔS = -W/T = [nRT ln(V 2 /V 1 )]/T = nR ln(1/2) = – nR ln 2.
(4 points)
For the adiabatic process we have
Q = 0.
Thus ΔS a = 0. (3 points)
(c)
Because each process is reversible, the energy change of the universe is zero.
For the isothermal process we have
ΔS surr = – ΔSi = nR ln 2. (4 points)
For the adiabatic process we have
ΔS surr = –ΔS = 0. (3 points)

4A. This P-V diagram represents a system consisting of a fixed amount of ideal gas that undergoes two
different processes in going from state A to state B:
A. Process #1
A. State B
Pressure
II. State A
B. Process #2
III. Volume
[In these questions, W represents the work done by the system during a process; Q represents the heat transfer to
the system during a process.]
1. Is W for Process #1 greater than, less than, or equal to that for Process #2? Explain.
Answer: (5 points)
The work done by the system in Process #1 is greater than the work done by the system in Process #2 because this work is
V
given by the integral ∫ B PdV , equivalent to the area under the curve, and this is greater for Process #1.
VA
2. Is Q for Process #1 greater than, less than, or equal to that for Process #2? Please explain your answer.
Answer: (5 points)
Q = ΔU + W; since ΔU is defined as U final state – U initial state , ΔU will be the same for both processes. The work done by the
system in Process #1 is greater than that in Process #2, therefore the heat transfer in Process #1 is greater than the heat
transfer in Process #2.
3. Which would produce the largest change in the total energy (kinetic plus potential) of all the atoms in the
system: Process #1, Process #2, or both processes produce the same change?
Answer: (5 points)
Both processes would have the same change in total energy since they have the same initial and final states.
4B. A subsystem A is in thermal contact with its environment B and they together comprise an isolated
system that is undergoing an irreversible process. Consider the following situations:
I. Entropy of subsystem A increases by 5 J/K; entropy of the environment B decreases by 5 J/K.
II. Entropy of subsystem A increases by 5 J/K; entropy of the environment B decreases by 3 J/K.
III. Entropy of subsystem A increases by 3 J/K; entropy of the environment B decreases by 5 J/K.
IV. Entropy of subsystem A decreases by 3 J/K; entropy of the environment B increases by 5 J/K.
Which of the above four situations can actually occur?
(Circle the correct answer listed below)
a. I only
b. II only
c. III only
d. II and IV only
e. I, II, and IV only
Answer: d. II and IV only. (5 points)
The total entropy of an isolated system (i.e. system + environment) must increase during any irreversible
process. There is no separate constraint on the entropy of a subsystem or of its environment.