Chemistry 111
KEY
Fall, 2005
Prof. Simpson
1
Chemistry 111
Handout #9
– KEY
Gas Laws, Kinetics, Thermodynamics & Thermochemistry
Gas Laws.
G1
. Oxygen in respiration therapy is stored in 43.0 L gas cylinders.
At 20.0
o
C, the pressure inside
the tank is 1.50 x 10
2
atm.
(a)
If all of the gas was let out of the tank, how much volume would it occupy at 1.00 atm
pressure and 20
o
C?
This corresponds to a change in pressure and volume at a constant temperature and
number of moles, so the appropriate equation to use is
P
1
V
1
= P
2
V
2
V
2
= (P
1
V
1
)/P
2
=
(150 atm x 43.0 L) / (1.0 atm) =
6,450 L
!!
(b)
What would the pressure inside the tank be if the temperature were doubled to 40.0
o
C?
Here, we are changing the pressure and temperature while keeping the volume and number
of moles constant.
Make sure you change to Kelvin, though!
Although the temperature is
doubled on the Celcius scale, it is not doubled on the Kelvin scale
!
P
1
/ T
1
=
P
2
/ T
2
P
2
= (P
1
T
2
) / T
1
=
(150 atm x 313 K) / (293 K) =
160.2 atm
(c)
A patient uses 30% of the oxygen in the tank.
What is the new pressure (in atm)?
Assume
the temperature is 20.0
o
C.
Ok, so now we're changing the number of moles and the pressure while keeping the
volume and temperature constant:
P
1
/ n
1
=
P
2
/ n
2
P
2
= P
1
x (n
2
/ n
1
) = 150 atm x (0.70) =
105 atm
Notice that we didn't actually have to calculate the number of moles.
We know that if the
patient used 30% of the amount of the oxygen, there must be 70% left behind, and this
corresponds to an n
2
/ n
1
ratio of 0.700.
(d)
Sketch the Boltzmann distribution of speeds for the oxygen molecules in the tank for
two different temperatures T
1
and T
2
.
Make sure you clearly label which curve goes
with which temperature, and which temperature is higher.
This is a pretty straightforward sketching exercise … we saw this several times in class.
#
O
T
>
w
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t
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g
v
e
n
s
p
d
speed
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KEY
Fall, 2005
Prof. Simpson
2
(e)
What is the average kinetic energy per mole of the oxygen atoms in the tank originally
(43.0 L, 20.0
o
C, 1.50 x 10
2
atm)?
The average kinetic energy per mole for a gas behaving ideally is given by the expression
KE
avg
= 3/2 RT
= (3/2) (8.314 J K
1
mol
1
) (293 K) =
3.65 kJ mol
1
Unless you're told otherwise, you can assume the gases on this test behave ideally.
(f)
Does the average kinetic energy per mole increase
with the process described in part (a)?
What about in part (b)?
What about in part (c)?
Explain.
In part (a), we changed the volume without changing the temperature or the number of
moles.
The pressure decreases, but not because the KE
avg
decreases.
KE
doesn't
change.
The pressure decreases because the surface area of the container increases, and
therefore the same number of particles encounter the walls of the container less frequently,
but with the same average force.
In part (b), we changed the temperature.
Clearly, from the expression in part (e), we can
see that
the KE
will increase
as the temperature increases.
In part (c), we changed the number of moles of gas in the container without changing the
temperature.
While the total kinetic energy in the container will decrease in this process,
the
per mole of gas
will not change.
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 Spring '08
 Kenney
 Thermochemistry, Kinetics, Prof. Simpson

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