Solutions to Quiz #1 for GP II, MATH,
Spring 2012.
1. (B)
When a certain constant volume gas thermometer is in thermal contact with water at
its triple point (273.16 K) the pressure is 6.30  104 Pa.
kelvin corresponds to a change in pressure of about:
For this thermometer a
A) 4.34  102 Pa
B) 2.31  102 Pa
C) 1.72  103 Pa
D) 2.31  103 Pa
E) 1.72  107 Pa
Solution:
pV  nRT
Ideal gas:
Constant volume:
From given:

p 
nR
T
V
nR p 6.30  104
 
 2.31  102
V
T
273.16
for T  1K
p  2.31  102 Pa
2. (B)
The coefficient of linear expansion of iron is 10–5 per C. The volume of an iron
cube, 5 cm on edge, will increase by what amount if it is heated from 10C to 60C?
A) 0.00375 cm3
B) 0.1875 cm3
C) 0.0225 cm3
D) 0.00125 cm3
E) 0.0625 cm3
Solution:
V  3 V T  3 105  53   60  10  3 105  53   60  10  0.1875 cm3
3. (D)
Take the mechanical equivalent of heat as 4 J/cal.
A 10-gram bullet moving at 2000
m/s plunges into 1 kg of paraffin wax (specific heat 0.7 cal/g C). he wax was
initially at 20C. Assuming that all the bullet's energy heats the wax, its final
temperature (C) is:
A) 20.14
B) 23.5
C) 20.006
D) 27.1
E) 30.23
Solution:
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4. (E)
A quantity of an ideal gas is compressed to half its initial volume.
The process may
be adiabatic, isothermal or isobaric. Rank those three processes in order of the work
required of an external agent, least to greatest.
A) adiabatic, isothermal, isobaric
B) adiabatic, isobaric, isothermal
C) isothermal, adiabatic, isobaric
D) isobaric, adiabatic, isothermal
E) isobaric, isothermal, adiabatic
Solution:
Adiabatic:


p1V1  p2V2
Wad  p1V1
V2
V1
V1
V1
V2
V1 /2
W    pdV   pdV  
External work required:

pdV
p1V1
p 
V
  1
dV
p V  1
p1V1
 1  1

V1 /2 V    1 V1  2 V1   1 11  2  1
V1


Isothermal:
p
nRT1
V
WT  nRT1

V1
V1 /2
dV
 nRT1 ln 2  p1V1 ln 2  0.69 p1V1
V
Isobaric:
WP  p1

V1
V1 /2
dV 
2  e ln 2  1   ln 2 
1
p1V1  0.5 p1V1
2
1
2
 ln 2  
2!

2  1


ln 2  ln 2
 1  1

WP  WT  Wad
Note:
 1   ln 2
where   1
Easiest way to get answer is graphically.
5. (E)
As the pressure in an ideal gas is increased isothermally the average molecular speed:
A) increases
B) decreases
C) increases at high temperature, decreases at low
D) decreases at high temperature, increases at low
E) stays the same
Solution:
Ideal gas:
T  const
U
3
1
1
NkT  3N  m v 2  3N  m v
2
2
2

2
v  const
6. (E)
The temperature of n moles of an ideal monatomic gas is increased by T at constant
pressure. The energy Q absorbed as heat, change Eint in internal energy, and work
W done by the environment are given by:
A) Q = (5/2)nRT, Eint = 0, W = –nRT
B) Q = (3/2)nRT, Eint = (5/2)nRT, W = –(3/2)nRT
C) Q = (5/2)nRT, Eint = (5/2)nRT, W = 0
D) Q = (3/2)nRT, Eint = 0, W = –nRT
E) Q = (5/2)nRT, Eint = (3/2)nRT, W = –nRT
Solution:
Ideal gas:
pV  nRT

Eint 
Constant
p:
dV 
Eint 
3
nRT
2
3
nR T
2
nR
dT
p
Wgas  p  dV  nRT
Wenv  Wgas   nRT
Q  Eint  Wgas 
5
nRT
2
7. (C)
An ideal gas, consisting of n moles, undergoes a reversible isothermal process during
which the volume changes from Vi to Vf. The change in entropy of the thermal
reservoir in contact with the gas is given by:
A) nR(Vf – Vi)
B) nR ln(Vf – Vi)
C) nR ln(Vi /Vf)
D) nR ln(Vf /Vi)
E) none of the above (entropy can't be calculated for an irreversible process)
Solution:
Ideal gas:
U  U T  .
Isothermal 
S gas 
U  0
V
Q
 n R ln f
T
Vi

Q W  
Vf
Vi
p d V 
Vf
Vi
V
n RT
dV  n R T ln f
V
Vi
Sres  S gas  n R ln
Vi
Vf
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8. (D)
A Carnot heat engine runs between a cold reservoir at temperature TC and a hot
reservoir at temperature TH. You want to increase its efficiency. Of the following,
which change results in the greatest increase in efficiency?
same for all changes.
The value of T is the
A) Raise the temperature of the hot reservoir by T
B) Raise the temperature of the cold reservoir by T
C) Lower the temperature of the hot reservoir by T
D) Lower the temperature of the cold reservoir by T
E) Lower the temperature of the hot reservoir by (1/2)T and raise the temperature
of the cold reservoir by (1/2)T
Solution:
Carnot  1 
TC
TH
TH  TH  T :
T
 1 C
TH
TC  TC  T :
 1
TH  TH  T :
T
 1 C
TH
TC  TC  T :
 1
1
TH  TH  T ;
2
1
 T 
 TC  T
1 
  0   
TH 

 TH  TH
TC T
T

 0 
TH TH
TH
1
 T 
 TC  T
1 
  0   
TH 

 TH  TH
TC T
T

 0 
TH TH
TH
1
TC  TC  T :
2
1
T
T  
T 
T  TC  T
 1  C 
 
 1 
  0 
2TH  TH  2TH
 TH 2TH   2TH 
9. (B)
A Carnot heat engine and an irreversible heat engine both operate between the same
high temperature and low temperature reservoirs. They absorb the same heat from the
high temperature reservoir as heat. The irreversible engine:
A) does more work
B) rejects more energy to the low temperature reservoir as heat
C) has the greater efficiency
D) has the same efficiency as the reversible engine
E) cannot absorb the same energy from the high temperature reservoir as heat
without violating the second law of thermodynamics
Solution:
  1
QC
QH
Carnot  irrev 
QCirrev  QCCarnot
10. (D)
Let k be the Boltzmann constant. If the configuration of the molecules in a gas
changes so that the multiplicity is reduced to one-third its previous value, the entropy
of the gas changes by:
A)
B)
C)
D)
E)
S = 0
S = 3k ln 2
S = –3k ln 2
S = –k ln 3
S = k ln 3
Solution:
S  k ln 
S  k ln
f
i
 k ln 3
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