CHAPTER
15
THERMODYNAMICS
CONCEPTUAL QUESTIONS
____________________________________________________________________________________________
1.
SSM
REASONING AND SOLUTION
The plunger of a bicycle tire pump is pushed
down rapidly with the end of the pump sealed so that no air escapes.
Since the compression
occurs rapidly, there is no time for heat to flow into or out of the system.
Therefore, to a
very good approximation, the process may be treated as an adiabatic compression that is
described by Equation 15.4:
W
=
(3 /2)
nR
(
T
i

T
f
)
The person who pushes the plunger down does work on the system, therefore
W
is negative.
It follows that the term
(
T
i

T
f
)
must also be negative.
Thus, the final temperature
T
f
must
be greater than the initial temperature
T
i
.
This increase in temperature is evidenced by the
fact that the pump becomes warm to the touch.
Alternate
Explanation:
Since the compression occurs rapidly, there is no time for heat to flow into or out of the
system.
Therefore, to a very good approximation, the process may be treated as an adiabatic
compression.
According to the first law of thermodynamics, the change in the internal
energy is
∆
U
=
Q

W
= 
W
, since
Q
= 0 for adiabatic processes.
Since work is done on
the system,
W
is negative; therefore the change in the internal energy,
∆
U
, is positive.
The
work done by the person pushing the plunger is manifested as an increase in the internal
energy of the air in the pump.
The internal energy of an ideal gas is proportional to the
Kelvin temperature.
Since the internal energy of the gas increases, the temperature of the air
in the pump must also increase.
This increase in temperature is evidenced by the fact that
the pump becomes warm to the touch.
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2
.
REASONING AND SOLUTION
The work done in an isobaric process is given by
Equation 15.2:
W
=
P
(
V
f

V
i
)
.
According to the first law of thermodynamics,
the change
in the internal energy is
∆
U
=
Q

W
=
Q

P
(
V
f

V
i
)
.
One hundred joules of heat is added to a gas, and the gas expands at constant pressure
(isobarically).
Since the gas expands, the final volume will be greater than the initial
volume.
Therefore, the term
P
(
V
f

V
i
)
will be positive.
Since
Q
= +100 J, and the term
P
(
V
f

V
i
)
is positive, the change in the internal energy must be less than 100 J.
It is not
possible that the internal energy increases by 200 J.
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3
.
REASONING AND SOLUTION
The internal energy of an ideal gas is proportional to its
Kelvin temperature (see Equation 14.7).
In an isothermal process the temperature remains
constant; therefore, the internal energy of an ideal gas remains constant throughout an
isothermal process. Thus, if a gas is compressed isothermally and its internal energy
increases, the gas is not an ideal gas.
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THERMODYNAMICS
4
.
REASONING AND SOLUTION
According to the first law of thermodynamics
(Equation 15.1), the change in the internal energy is
∆
U
=
Q
–
W
.
The process is isochoric,
which means that the volume is constant.
Consequently, no work is done, so
W
= 0.
The
process is also adiabatic, which means that no heat enters or leaves the system, so
Q
= 0.
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 Spring '07
 Russell
 Thermodynamics, Entropy, Heat

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