Dr. Zellmer
Chemistry 125
Thursday
Time: 30 mins
Winter Quarter 2012
January 26, 2012
Quiz I
Name
KEY
Rec. TA/time
1.
(6 pts) Consider a steel tank with a volume of 10.0 L that can withstand a pressure of 30.0 lb/in
2
. A
sample of He gas has a volume of 50.0 L at STP.
If the He gas is put in the steel tank, can liquid
nitrogen (boiling point of 196°C) be used to keep the tank from exploding?
(1 atm = 760 mmHg = 760 torr = 101.325 kPa = 14.7 lb/in
2
)
STP
=> Standard Temp. & Pressure:
T = 0°C (273.15K)
&
P = 1 atm
Since P, V & T all change use the Combined Gas Law:
P
2
V
2
P
1
V
1

=

T
MUST
be in
kelvin
T
2
T
1
Need to find T at which the pressure will be below or equal to 45.0 lb/in
2
.
P
1
= 1.00 atm
V
1
= 50.0 L T
1
= 0.00°C + 273.15 = 273.15 K
1 atm
P
2
= 30.0 lb/in
2
×  = 2.0
4
0 atm
V
2
= 10.0 L T
2
= ?
14.7 lb/in
2
P
2
V
2
T
1
P
2
V
2
2.0
4
0 atm
10.0 L
T
2
=

=
()()T
1
=
()() 273.15 K
=
11
1
.489 K
P
1
V
1
P
1
V
1
1.00 atm
50.0 L
T
2
=
11
1
.489 K  273.15
=
&
161
.66°C
=
&
162°C
Since N
2
boils at
&
196
°C it can be used to cool the tank to
&
162°C.
The tank will
NOT
explode.
Could also solve for P
2
using T
2
= 196°C and then compare P
2
to 30.0 lb/in
2
.
(P
2
= 1.41 atm = 20.8 lb/in
2
)
Copyright R. J. Zellmer, January 26, 2012
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(5 pts) If the temperature of a gas increases from 200.0 K to 800.0 K what happens to the average kinetic
energy of the gas particles?
Be
specific
and
quantitative
using Kinetic Molecular Theory to explain.
Also
, sketch a qualitative picture of the distribution of speeds at the two temperatures (i.e. sketch what
the two curves look like relative to each other).
KE
avg
= ½ m
:
2
The average KE of a single “average” particle moving with the rms speed
m = mass of the particle
:
= rms speed (type of average speed)
KE
total
= 3/2 RT
Total average KE per mole.
Note it depends only on T and is directly proportional
to T (in Kelvin).
The
average kinetic energy
is
directly
proportional
to the
absolute
temperature
(
Kelvin
temp.).
KE
avg
%
T
(T in Kelvin)
This means if the temperature increases then the average KE will increase.
More specifically, if the
temperature
quadruples
(200.0 K to 800.0 K) the
KE
avg
quadruples
(and vice versa), they are directly
proportional.
For 1 mole of a gas the average total KE per mole is equal to (3/2)RT.
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 Winter '08
 Zellmer
 Chemistry

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