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Chemistry 456
Winter 2010
Bruce H. Robinson
Problems:
Z5.22, P1.2, PX1, P1.24, PX2,P1.37
(Formerly 1
st
Ed: P1.4, P1.5, P1.14, P1.15)
Homework
1A key (
due Friday
).
Problems denoted by Z mean they come from Zumdahl’s text (Chemical
Principles).
You may work all problems to a single significant figure, taking
care to notice differences (such as in P1.15) and explain them.
The PX
problems are those that were in the 1
st
Ed but are not in the second edtion.
Z5.22)
Two examples for an opentube manometer are shown.
If the flask
(attached on the right but not shown)
is open to the atmosphere, the mercury
levels will be equal in height (The height difference will be zero.)
For each
of the following situations, in which a gas is contained in the flask, calculate
the pressure in the flask in torr, atmospheres, and Pascals.
The pressure of the mercury is :
Fm
gm
g
P
h
dgh
AAA
h
==
=
=
, where d is the density of
mercury and h is the height of the column, g is the gravitational constant (9.8S.I.).
The
density of mercury is about 13.6 g/cc.
Because 1 torr = 1mm Hg, we can just read off the
pressure in torr.
For A:
P=760118 = 660 torr, and B:
P=760+215 = 960 torr .
Noticed
how I rounded the difference and just added and subtracted simple numbers from the
reference of 760.
So now we can look up conversion tables
that tell us 1Atm ~ 1Bar =
5
11
0
Bar
Pa
=
.
So the pressure is also
660
960
:
~ .85
1.25
760
760
A
BA
∼
t
m
C If the pressure of the atmosphere is 635 torr instead of 760 (but the height differences
are the same) calculate the pressures in the flasks in parts a and b (in torr only).
For A:
P=635118 = 520 torr, and B:
P=635+215 = 850 torr .
1
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View Full DocumentChemistry 456
Winter 2010
Bruce H. Robinson
P1.2
)
A compressed cylinder of gas contains 1.50
×
10
3
g of N
2
gas at a pressure of
2.00
×
10
7
Pa and a temperature of 17.1ºC.
What volume of gas has been released into the
atmosphere if the final pressure in the cylinder is 1.80
×
10
5
Pa?
Assume ideal behavior
and that the gas temperature is unchanged.
For I.G.
assume the volume of the gas cylinder does not change and there is no
temperature change, so from the pressure change we compute the change in the number
of moles, which is the number of moles released from the cylinder.
The volume of the cylinder is computed from the initial conditions of the cylinder:
3
3
33
7
1.5
10
50
28
2.4 10
50
0.05
210
ii
PV
n RT
nm
o
Vm
l
e
s
m
=
=⋅=
⋅
==
⋅
The number of moles released then is found from the pressure difference.
ff
i
i
f
if
PV
nRT
nRT
PP P
nn n
RT
Pn
V
=
Δ≡ −
Δ=
Δ
i
Notice that almost all of the moles are released.
Those moles occupy a volume at 1Atm and the same temperature,
and that is:
air
air
air
air
air
nR
T
n
R
T
P
VV
PP
P
Δ
Δ
===
This could all be done more simply, by realizing that nearly all the gas is released, so 50
moles of gas at STP occupies
.
3
50 22.4
1
1
km
⋅=
=
AA
PX1)
A balloon filled with 10.50 L of Ar at 18.0ºC and 1 atm rises to a height in the
atmosphere where the pressure is 248 Torr and the temperature is –30.5ºC. What is the
final volume of the balloon?
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 Chemistry, Physical chemistry, pH

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