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•
process 1
→
2
(d) The work is given by Eq. 1914:
W = nRT
1
ln (
V
2
/
V
1
) =
RT
1
ln3 =1.10
RT
1
.
Thus,
W
/
nRT
1
= ln3 = 1.10.
(e) The internal energy change is
Δ
E
int
= 0 since this is an ideal gas process without a
temperature change (see Eq. 1945). Thus, the energy absorbed as heat is given by the
first law of thermodynamics:
Q =
Δ
E
int
+
W
≈
1.10
RT
1
, or
Q
/
nRT
1
= ln3 = 1.10.
(f)
Δ
E
int
= 0 or
Δ
E
int
/
nRT
1
=0
(g) The entropy change is
Δ
S = Q
/
T
1
= 1.10
R
, or
Δ
S
/
R
= 1.10.
•
process 2
→
3
(h) The work is zero since there is no volume change. Therefore,
W
/
nRT
1
= 0
13. The connection between molar heat capacity and the degrees of freedom of a
diatomic gas is given by setting
f
= 5 in Eq. 1951. Thus,
5/
2
,
7/
2
Vp
CR
==
, and
7/5
γ
=
. In addition to various equations from Chapter 19, we also make use of Eq. 204
of this chapter. We note that we are asked to use the ideal gas constant as
R
and not plug
in its numerical value. We also recall that isothermal means constanttemperature, so
T
2
=
T
1
for the 1
→
2 process. The statement (at the end of the problem) regarding “per mole”
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 Spring '08
 Any
 Physics, Heat

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