shah (rps587) – HW 14 – Kleinman – (58225)
1
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23
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before answering.
This HW covers Chs. 20 and 21.
001
10.0 points
During a certain thermodynamic process a
sample of gas expands and cools, reducing its
internal energy by 2380 J, while no heat is
added or taken away.
How much work is done during this process?
Correct answer: 2380 J.
Explanation:
Let :
Δ
E
=

2380 J
and
Q
= 0 J
.
Q
= Δ
E
+
W
W
=
Q

Δ
E
= 0 J

(

2380 J)
=
2380 J
.
002
5.0 points
Suppose you squeeze an airfilled hollow rub
ber ball in your hand.
Assuming no heat escapes, what happens
to the internal energy of the air inside?
1.
It increases.
correct
2.
Unable to determine
3.
It decreases.
4.
It doesn’t change.
Explanation:
You do work on the air by decreasing its
volume while no heat escapes, so by the first
law of thermodynamics, the internal energy
increases.
003
(part 1 of 2) 10.0 points
A gas is compressed at a constant pressure
of 0
.
586 atm from 5
.
26 L to 2
.
75 L.
In the
process, 344 J of energy leaves the gas by heat.
What is the work done on the gas?
Correct answer: 148
.
998 J.
Explanation:
Given :
P
= 0
.
586 atm = 59361
.
8 Pa
,
V
1
= 5
.
26 L = 0
.
00526 m
3
,
and
V
2
= 2
.
75 L = 0
.
00275 m
3
.
The work done on the gas is
W
=

P
Δ
V
=

(59361
.
8 Pa)
(
0
.
00275 m
3

0
.
00526 m
3
)
=
148
.
998 J
.
004
(part 2 of 2) 10.0 points
What is the change in its internal energy?
Correct answer:

195
.
002 J.
Explanation:
Given :
Q
=

344 J
.
The change in the internal energy is
Δ
U
=
Q
+
W
=

344 J + 148
.
998 J
=

195
.
002 J
.
005
(part 1 of 3) 10.0 points
Hint: look at equation (20.23) for part 2.
Given:
R
= 8
.
31451 J
/
K
·
mol
.
Two moles of helium gas initially at 366 K
and 0
.
21 atm are compressed isothermally to
0
.
74 atm.
Find the final volume of the gas.
Assume
the helium to behave as an ideal gas.
Correct answer: 0
.
0811908 m
3
.
Explanation:
Given :
n
= 2 mol
,
R
= 8
.
31451 J
/
K
·
mol
,
T
f
= 366 K
,
and
P
f
= 0
.
74 atm
.
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shah (rps587) – HW 14 – Kleinman – (58225)
2
From the ideal gas law,
P
f
V
f
=
n R T
V
f
=
n R T
P
f
=
(2 mol)(8
.
31451 J
/
K
·
mol)(366 K)
(0
.
74 atm)(1
.
013
×
10
5
Pa
/
atm)
= 0
.
0811908 m
3
.
006
(part 2 of 3) 10.0 points
Find the work done by the gas.
Correct answer:

7
.
66586 kJ.
Explanation:
Given :
P
i
= 0
.
21 atm
and
T
i
= 366 K
.
The initial volume was
V
i
=
n R T
P
i
=
(2 mol)(8
.
31451 J
/
K
·
mol)(366 K)
(0
.
21 atm)(1
.
013
×
10
5
Pa
/
atm)
= 0
.
286101 m
3
and the work done was
W
=
integraldisplay
P dV
=
n R T
ln
parenleftbigg
V
f
V
i
parenrightbigg
= (2 mol)(8
.
31451 J
/
K
·
mol)(366 K)
ln
parenleftbigg
0
.
0811908 m
3
0
.
286101 m
3
parenrightbigg
1 kJ
1000 J
=

7
.
66586 kJ
.
007
(part 3 of 3) 10.0 points
Find the thermal energy transferred.
Correct answer:

7
.
66586 kJ.
Explanation:
From the first law of thermodynamics, since
Δ
U
= 0,
Q
=
W
=

7
.
66586 kJ
.
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 Spring '09
 KLIENMAN
 Thermodynamics, Energy, Entropy, Heat

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