Version PREVIEW – Homework 9 – Savrasov – (39825)
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Expanding Gas 03
001
10.0 points
A sample of ideal gas is expanded to twice
its original volume of 0
.
4 m
3
in a quasistatic
process for which
P
=
α V
2
with
α
= 4 atm
/
m
6
,
V
a
= 0
.
4 m
3
and
V
b
=
0
.
8 m
3
, as shown in the figure.
V
P
P
=
(
4
a
t
m
/
m
6
)
V
2
0
.
4 m
3
0
.
8 m
3
How much work was done by the expanding
gas?
Correct answer: 60509
.
9 J.
Explanation:
Given :
α
= 4 atm
/
m
6
,
V
a
= 0
.
4 m
3
,
and
V
b
= 0
.
8 m
3
.
The work done by the gas is the area under
the curve
P
=
α V
2
between
V
a
and
V
b
:
W
ab
=
integraldisplay
b
a
P d V
=
integraldisplay
b
a
α V
2
d V
=
α
(
V
3
b

V
3
a
)
3
=
(4 atm
/
m
6
)
bracketleftbig
(0
.
8 m
3
)
3

(0
.
4 m
3
)
3
bracketrightbig
3
=
60509
.
9 J
.
Gas Expands Isothermally
002
10.0 points
Given:
R
= 8
.
31 J
/
K
·
mol
.
1 mol of an ideal gas expands isothermally
at 373 K to 2
.
3 its initial volume.
Find the heat flow into the system.
Correct answer: 2581
.
71 J.
Explanation:
Given :
n
= 1 mol
,
R
= 8
.
31 J
/
K
·
mol
,
T
= 373 K
,
and
V
f
= 2
.
3
V
i
.
This is an isothermal process, so the inter
nal energy of the ideal gas is constant. Thus
the heat flowing into the system is equal to
the work done to the system:
W
=
n R T
ln
parenleftbigg
V
f
V
i
parenrightbigg
= (1 mol) (8
.
31 J
/
K
·
mol) (373 K)
×
ln (2
.
3)
= 2581
.
71 J
.
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 Fall '09
 RichardScalettar
 Energy, Work, Entropy

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