PHYS 253 Assignment 4 Solutions
October 20 2010
Total: 12 marks
1) – 2 marks
Let the larger bulb be denoted as bulb 1 and the smaller one as bulb 2.
V
1
=3V
2
Because there is no heat transfer through the tube: T
2i
=T
2f
=T
0
Because of the tube pressure is equalized: P
i1
=P
i2
and P
f1
=P
f2
We approximate the air as an ideal gas: PV=nRT
Initially:
RT
0
P
0
=
V
1
n
1i
=
V
2
n
2i
Thus:
n
1i
=
n
2i
V
1
V
2
=
3n
2i
And:
n
1i
n
2i
=
n
1f
n
2f
=
4n
2i
When the pressure in both bulbs is doubled:
P
1f
=
P
2f
=
2P
0
=
n
1f
RT
1f
V
1
=
n
2f
RT
0
V
2
T
1f
=
V
1
V
2
n
2f
n
1f
T
0
n
2f
=
2
P
0
V
2
RT
0
n
1f
=
4n
2i
−
n
2f
n
1f
=
4
P
0
V
2
RT
0
−
2
P
0
V
2
RT
0
=
n
2f
T
1f
=
3T
0
2) 4– marks
a)
V
1
V
2
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P
1
LA
=
nRT
P
2
L
−
x
A
=
nRT
=
P
1
LA
F
2
L
−
x
=
F
1
L
k
=
F
x
x
=
P
x
=
F
1
L
L
−
x
−
1
x
=
F
1
L
−
x
=
P
0
A
L
−
x
For small x:
K
=
P
0
A
L
b) When adiabatic PV
γ
=Constant
P
0
V
0
=
P
f
V
f
P
=
P
f
−
P
i
=
P
0
[
V
0
V
f
−
1
]
P
=
P
0
[
AL
A
L
−
x
−
1
]
k
=
F
x
x
=
A
P
x
=
P
0
A
x
[
L
L
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 Fall '10
 PeterGrutter
 Thermodynamics, Heat, Trigraph, PV =Constant nRT

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