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aljabr (faa335) – Hw28 – Ross – (89251)
1
This printout should have 12 questions.
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001
(part 1 of 2) 10.0 points
One mole of an ideal gas does 1296 J of work
on the surroundings as it expands isother
mally to a Fnal pressure of 1 atm and volume
of 30 L.
Determine
the
initial
volume.(
R
=
8
.
31451 J
/
K
·
mol.)
Correct answer: 19
.
5846 L.
Explanation:
Given :
n
= 1 mol
,
W
= 1296 J
,
R
= 8
.
31451 J
/
K
·
mol
,
P
f
= 1 atm
,
and
V
f
= 30 L
.
By the ideal gas law,
P V
=
nRT .
The work done in an isothermal process is
W
=
nRT
ln
p
V
f
V
i
P
=
P V
ln
p
V
f
V
i
P
ln
p
V
f
V
i
P
=
W
P V
V
f
V
i
=
e
W/
(
P V
)
V
i
=
V
f
e

W/
(
P V
)
Since
W
PV
=
1296 J
(1 atm) (1
.
013
×
10
5
) (30 L) (1
×
10

3
)
= 0
.
426456
V
i
= (30 L)
e
(

0
.
426456)
= 19
.
5846 L
.
002
(part 2 of 2) 10.0 points
Determine the temperature of the gas.
Correct answer: 365
.
506 K.
Explanation:
According to the equation of state for an ideal
gas, the temperature is
T
f
=
P
f
V
f
nR
=
(1 atm)(1
.
013
×
10
5
)(30 L)(1
e

3)
(1 mol)(8
.
31451 J
/
K
·
mol)
= 365
.
506 K
.
003
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 Spring '10
 Gavrine
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