Mid Term 105A
10/31/06
1/5
Mid Term #1
105A Fall 2006
Question 1
One mole of gas in a chamber is compressed and decompressed using a piston which
slides without friction. The initial state is at p
1
= 2x10
5
Pa and V
1
=14x10
3
m
3
. This gas
goes successively through:
1.
An isobaric decompression, which double its volume;
2.
An isothermal compression, which bring the volume back to V
1
;
3.
An isochoric (constant volume) cooling, which brings it back to p
1
.
We suppose this gas ideal.
a
What is the temperature at which the compression happens?
b
Draw the different processes on a (p,V) diagram. How is this type of
transformation called?
c
What is the work and heat exchanged by this system during one of these
transformations?
a (5 pts) We start with one mole of gas at 2x10
5
Pa and 14x10
3
m
3
. Using
the ideal gas law the temperature is found to
==
=
22
1
1
21
2
2
p
Vp
V
T
RR
or
2
= 673.56 K.
b (10 pts)
Using the ideal gas law at constant
temperature for closed system we
also find that
=
33
. Since the
isothermal process brings us back
to
V
1
w
e
h
a
v
e
t
h
a
t
V
3
= V
1
. So
=
31
2
1
2
and because
p
2
= p
1
we finally find
=
2
Since we come back to the original
state this transformation is called
a cycle.
c (15 pts) Using the first law of thermodynamics we have for the whole
cycle
1
2
3
p
V
V
1
2V
1
p
1
2p
1
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∆=
−
cycle
UQ
W
.Since
0
U
,
Q
=
. The elementary work for a simple
compressional system such as this one is pdV and the total work is
→→→
=++
12
23
31
WWW
. So
=+
+
∫∫∫
231
123
pdV
. Since the process
3
→
1 is isochoric
→
=
0
, since the process 1
→
2 is isobaric
→
=−
=
1 2
1
11
()
pV V
pV
and
since
2
→
3
is
isothermal
→
==
∫∫
33
22
1 1
2
dV
p V
VV
or
→
−
3
2
2l
n
2
l
n
2
V
. The total work
is
=−<
l
n
2 0
or
= 1081.6 J = 
Question 2
A closed chamber has adiabatic wall (no
exchange of heat with the surroundings).
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 Fall '06
 PierreGourdain
 Thermodynamics

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