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Q18. First Law of Thermodynamics. 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. - PowerPoint PPT Presentation
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Q18. First Law of Thermodynamics
1. 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.
1. adiabatic, isothermal,
isobaric
2. adiabatic, isobaric,
isothermal
3. isothermal, adiabatic,
isobaric
4. isobaric, adiabatic,
isothermal
5. isobaric, isothermal,
adiabatic
AdiabaticIsothermalIsobaric
P
2P
2 P
2. When an ideal gas undergoes a slow isothermal expansion:
1. (work done by the gas) = (energy absorbed as heat )
2. (work done by the environment ) = (energy absorbed as heat )
3. (increase in internal energy ) = ( heat absorbed )
4. (increase in internal energy ) = ( work done by the gas )
5. ( increase in internal energy ) = ( work done by the
environment )
Isothermal dU = 0 d Q = d W
Expansion d W = P d V > 0 (Work done by gas)
d Q > 0 Heat absorbed
dU dQ dW 3
2U n R T
1. (work done by the gas) = (energy absorbed as heat ) d W =
d Q
2. (work done by the environment ) = (energy absorbed as heat ) d W =
d Q
3. (increase in internal energy ) = ( heat absorbed ) d U = d Q
4. (increase in internal energy ) = ( work done by the gas ) d U = d W
5. ( increase in internal energy ) = ( work done by the environment ) d U = d W
3. An ideal gas of N monatomic molecules is in thermal
equilibrium with an ideal gas of the same number of diatomic
molecules and equilibrium is maintained as temperature is
increased. The ratio of the changes in the internal energies
ΔEdia / Emon is :
1. 1/2
2. 3/5
3. 1
4. 5/3
5. 2
2
fU N k T
2
fU N k T
5
3dia dia
mon mon
U f
U f
4. The pressure of an ideal gas of diatomic molecules is doubled
by halving the volume. The ratio of the new internal energy
to the old, both measured relative to the internal energy at 0
K, is:
1. 1/4
2. 1/2
3. 1
4. 2
5. 4
P V n R T2
fU N k T
22
VP P V
T remains the same.
So does U.
5. When work W is done on an ideal gas of N diatomic
molecules in thermal isolation the temperature increases by
1. W / 2 N k
2. W / 3 N k
3. 2 W / 3 N k
4. 2 W / 5 N k
5. W / N k
dU dQ dW
5
2U N k T
2
5T W
N k
Diatomic molecules :
Thermal isolation : dQ = 0 U W Work W done on gas:
6. When work W is done on an ideal gas of diatomic molecules
in thermal isolation the increase in the total rotational energy
of the molecules is:
1. 0
2. W / 3
3. 2 W / 3
4. 2 W / 5
5. W
dU dQ dW
2
5rotU U
2
5rotU W
Diatomic molecules :
Thermal isolation : dQ = 0 U W
3
5tranU U
Work W done on gas:
7. The pressure of an ideal gas is doubled during a process in
which the energy given up as heat by the gas equals the work
done on the gas. As a result, the volume is:
1. doubled
2. halved
3. unchanged
4. need more information to
answer
5. nonsense, the process is
impossible
dU dQ dW
T const P V n R T const
0U Q W 2
fU n R T
2P P 2
VV
Heat Q given up by the gas equals work W done on the gas
8. 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:
1. Q = (5/2)nRT, Eint = 0, W = –nRT
2. Q = (3/2)nRT, Eint = (5/2)nRT, W = –
(3/2)nRT
3. Q = (5/2)nRT, Eint = (3/2)nRT, W = nRT
4. Q = (3/2)nRT, Eint = 0, W = nRT
5. Q = (5/2)nRT, Eint = (3/2)nRT, W = –nRT
51
2 2P
fC nR nR
3
2U n R T
n RW P V P T n R T
P
5
2PQ C T n R T