Solutions toQuiz #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. For this thermometer a kelvin corresponds to a change in pressure of about:
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:
Ideal gas:
Constant volume:
From given:
for
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 10C to 60C?
A) 0.00375 cm3
B) 0.1875 cm3
C) 0.0225 cm3
D) 0.00125 cm3
E) 0.0625 cm3
Solution:
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 20C. 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:
External work required:
Adiabatic:
Isothermal:
Isobaric:
where
Note: 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:
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)nRT,Eint = 0, W = –nRT
B) Q = (3/2)nRT,Eint = (5/2)nRT,W = –(3/2)nRT
C) Q = (5/2)nRT,Eint = (5/2)nRT,W = 0
D) Q = (3/2)nRT,Eint = 0, W = –nRT
E) Q = (5/2)nRT,Eint = (3/2)nRT,W = –nRT
Solution:
Ideal gas:
Constant p:
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) nRln(Vf – Vi)
C) nRln(Vi/Vf)
D) nRln(Vf/Vi)
E) none of the above (entropy can't be calculated for an irreversible process)
Solution:
Ideal gas: .
Isothermal
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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? The value of T is the same for all changes.
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:
:
:
:
:
; :
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:
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) S = 0
B) S = 3kln2
C) S = –3kln2
D) S = –kln3
E) S = kln3
Solution:
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