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3. (a) Since the gas is ideal, its pressure
p
is given in terms of the number of moles
n
,thevo
lume
V
,and
the temperature
T
by
p
=
nRT/V
. The work done by the gas during the isothermal expansion is
W
=
Z
V
2
V
1
pdV
=
nRT
Z
V
2
V
1
dV
V
=
nRT
ln
V
2
V
1
.
We substitute
V
2
=2
V
1
to obtain
W
=
nRT
ln 2 = (4
.
00 mol)
µ
8
.
31
J
mol
·
K
¶
(400 K) ln2 = 9
.
22
×
10
3
J
.
(b) Since the expansion is isothermal, the change in entropy is given by ∆
S
=
R
(1
/T
)
dQ
=
Q/T
,
where
Q
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Unformatted text preview: E int = Q − W . Now the internal energy of an ideal gas depends only on the temperature and not on the pressure and volume. Since the expansion is isothermal, ∆ E int = 0 and Q = W . Thus, ∆ S = W T = 9 . 22 × 10 3 J 400 K = 23 . 1 J / K . (c) ∆ S = 0 for all reversible adiabatic processes....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Work

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