Version 078/ABADC – TEST4 – Tsoi – (58020)
1
This printout should have 16 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
10.0 points
A heliumflled balloon has a volume oF 1 m
3
.
As it rises in the Earth’s atmosphere, its vol
ume changes.
What is its new volume iF its original tem
perature and pressure are 32
◦
C and 1
.
6 atm
and its fnal temperature and pressure are

26
◦
C and 0
.
4 atm?
1. 0.359202
2. 3.31507
3. 0.753125
4. 0.837191
5. 0.367759
6. 1.25638
7. 0.335075
8. 0.977586
9. 0.911141
10. 3.23934
Correct answer: 3
.
23934 m
3
.
Explanation:
Given :
V
i
= 1 m
3
,
T
i
= 32
◦
C = 305 K
,
P
i
= 1
.
6 atm
,
T
f
=

26
◦
C = 247 K
,
and
P
f
= 0
.
4 atm
.
Using ideal gas law
P V
=
nRT
we obtain
P
i
V
i
T
i
=
P
f
V
f
T
f
V
f
=
P
i
V
i
T
f
T
i
P
f
=
(1
.
6 atm)
(
1 m
3
)
(247 K)
(305 K)(0
.
4 atm)
=
3
.
23934 m
3
.
002
10.0 points
A gold ring has an inner diameter oF 1
.
6214 cm
at a temperature oF 21
◦
C.
Determine its inner diameter at 129
◦
C.
(The coe±cient oF linear expansion is 1
.
42
×
10
−
5
(
◦
C)
−
1
.)
1. 2.05116
2. 1.41082
3. 1.6478
4. 2.63208
5. 1.76277
6. 1.2111
7. 1.62389
8. 2.76597
9. 2.67144
10. 2.14839
Correct answer: 1
.
62389 cm.
Explanation:
Given :
L
1
= 1
.
6214 cm
,
T
1
= 21
◦
C
,
T
2
= 129
◦
C
,
and
α
= 1
.
42
×
10
−
5
(
◦
C)
−
1
.
The diameter expands by
Δ
L
=
αL
1
Δ
T
=
αL
1
(
T
2

T
1
)
=
b
1
.
42
×
10
−
5
(
◦
C)
−
1
B
×
(1
.
6214 cm) (129
◦
C

21
◦
C)
= 0
.
00248658 cm
.
and
L
2
=
L
1
+ Δ
L
= 1
.
6214 cm + 0
.
00248658 cm
=
1
.
62389 cm
.
003
(part 1 oF 2) 10.0 points
A vertical cylinder oF crosssectional area
0
.
2 m
2
is ftted with a tightftting, Frictionless
piston oF mass 35 kg.
The acceleration oF gravity is 9
.
8 m
/
s
2
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentVersion 078/ABADC – TEST4 – Tsoi – (58020)
2
h
gas
m
If there are 2 mol of an ideal gas in the
cylinder at 363 K, Fnd the pressure inside
the cylinder. (Assume that the system is in
equilibrium.)
1. 101572.0
2. 106880.0
3. 124520.0
4. 108644.0
5. 137260.0
6. 111045.0
7. 102715.0
8. 121580.0
9. 114720.0
10. 104296.0
Correct answer: 1
.
02715
×
10
5
Pa.
Explanation:
Given :
A
= 0
.
2 m
2
,
m
= 35 kg
,
R
= 8
.
314 J
/
Kmol
,
and
P
atm
= 1
.
01
×
10
5
Pa
.
In equilibrium,
P
gas
A
=
mg
+
P
atm
A
P
gas
=
mg
A
+
P
atm
=
(35 kg)
(
9
.
8 m
/
s
2
)
0
.
2 m
2
+ 1
.
01
×
10
5
Pa
=
1
.
02715
×
10
5
Pa
.
004
(part 2 of 2) 10.0 points
At what height will the piston be in equilib
rium under its own weight?
1. 0.334885
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '09
 Staff
 Physics, Thermodynamics, Energy, Correct Answer

Click to edit the document details