Chemistry 456
Winter 2010
Bruce H. Robinson
Problem Set 2A
Problems:
Z9.36 (6
th
Ed) {or Z9.28 (4
th
Ed) or Z9.30 (5
th
Ed)}, Q2.2,
P2.14, P2.19
(Q2.5 P2.11 P2.12 from 1
st
Ed.)
Work the problems to a single significant figure.
Problem set is due Friday
Problem Set 2A
Key
X1 A)
A cylinder has a diameter of 1/3
meter and contains 1 mole of an ideal gas at
atmospheric pressure and 25C (room temperature).
There is a piston on top of the gas.
How high is the cylinder?
3
22
1
1 0.082 298
24.43
0.024
36
36
0.024
0.275
PV
nRT
V
Vm
Ar
m
V
hm
A
π
A
h
m
=
=⋅
⋅
=
==
=
i
A
⋅
X1 B)
A weight is placed on the cylinder and the final height is ¾ of the initial height?
What is the mass of the weight? [Assume the temperature is constant.]
The new pressure is:
5
5
3
4
3
1
3
1.01
1
33
1.01
3
1
10
9.8
10
36
0.3 10
0.3
ff
i
i
i
fi
f
PV
V
PP
V
PP P A
t
m
Fm
g
P
AA
Atm
Pa
m
m
kg
kTons
=
⋅
Δ= − =
Δ= =
⋅=
⋅
=
This should make you stop to think, that to compress a gas will require an additional 1/3
of an Atmosphere which over this small area is 1/3 of a kiloTon, which is about the same
as an English ton.
The height of the cylinder went down only by 7 cm,
but it would take
a column of water 10 feet high to compress the gas this much.
Z9.36)
Consider a sample containing 2 moles of a monatomic ideal gas (I.G.) that
undergoes the following changes:
1
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Winter 2010
Bruce H. Robinson
L
12
3
20
10
10
20
5
10
5
25
C
AB
D
C
D
Pa
t
m
P
atm
P
atm
P
atm
VL
V
L
V
=
==
=
⎯⎯
→⎯
⎯
⎯
→
=
=
For each step, assume that the external pressure is constant and equals the final pressure
of the gas for that step.
Compute , ,
, and
qw E
H
Δ
Δ
for each step and for the overall
change from state A to state D.
[Hint:
Analyze each step, and determine the physical change that needs to happen;
isothermal?, etc. and the type of change,
system doing work?, etc.]
Before the transformations, using the I.G. E.o.S. we can determine the temperature (if we
need it later).
Because PV is in units of energy it is probably preferable to determine RT
rather than T so that all quantities will be in energy units.
Path 1:
(A to B)
volume decrease, pressure held constant (isobaric)
1
0
P
Δ
=
.
1
5
BA
VVV
Δ
=−=
−
A
Because it is isobaric we can use the E.o.S. to describe the change in T and V
simultaneously:
1
Aq
PV n
RT
Δ
=Δ
From here we can compute the work, heat and energy:
11 1
11
1
1
1
1
1
Work is determined by the environment
3
Internal Energy Change from I.G.
2
3
First Law Gives q
2
Constant pressure change is Ent
x
B
v
p
wP
V
Vn
R
T
Un
R
T
C
T
q
U
w
nR T
nR T
C
T
Hq
=− Δ
=− Δ =−
Δ
Δ=
Δ
−
=
Δ +
Δ =
Δ
halpy which is
p
q
To get everything in S.I. (or m,K,s) units:
1
10 5
50
5
0
B
R
T
A
t
m
k
J
⋅ =
=
>
A
The work is positive because work is transferred into the system, but the Heat is negative,
because heat was withdrawn from the system:
1
5
2
q
=−
1
w
.
So the cooling reduces the
temperature so the internal energy overall drops.
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 Spring '08
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 Chemistry, Physical chemistry, Thermodynamics, pH, Adiabatic process, Bruce H. Robinson

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