Problems du Jour
Consider a 30.0 L cylinder of fluorine gas. The gas is at a
pressure of 160 lb/in2 at 26oC: (1 atm = 14.7 lb/in2)
What mass of fluorine gas is contained in the cylinder?
What volume would the gas occupy at STP?
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Applications of the ideal gas law
Molar mass & gas density
How to relate PV = nRT to density?
Rearrange:
P
n

RT V
What are the units of the right-hand side?
Multiply both sides by molar mass, M
PM n
 M
RT V
What are the units of the right hand side now?
Notice that gas densities are usually given in units of g/L
From the above, we obtain
d
PM
RT
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E.g., calculate the density of sulfur trioxide gas at 2.50 atm
and 25oC.
Gas mixtures and partial pressures
Thus far we have considered only the behavior of pure
gases (those consisting of only one substance)
What if a gas is a mixture of two or more different
substances? How do we think about the relationship
between P, V, and T?
134
Dalton's Law of Partial Pressures states that the total
pressure of a mixture of gases equals the sum of the
pressures that each gas would exert if it were present
alone
If the total pressure is written as Ptot,, and if P1, P2, etc.
represent the partial pressures of substances 1,2, etc.,
then
Ptot = P1 + P2 + P3 + . . .
is the mathematical statement of Dalton's Law.
If each gas in the mixture is ideal,
P 
1
n RT
n RT
; P 
etc.;
V
V
1
2
2
Substitute into our statement of Dalton's Law to obtain . . .
P 
tot
RT
n  n  ...
V
1
2
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E.g., calculate Ptot for a mixture containing 0.125 mol CH4,
0.250 mol C2H6, and 0.075 mol O2 at 22oC in a 2.00 L flask
Hmmm... what about this mixture of ideal gases?
A mixture of gases contains 3.50 g of N2 and 1.30 g of H2.
If Ptot = 2.50 atm, find the partial pressure of each
component.
Based on the above form of Dalton’s law, do we have
enough info to solve the problem?
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How to specify the composition of a gas mixture?
Use the mol fraction of each gas in a mixture
recall the mol fraction X1 of component 1 is :
X 
1
n
n

n  n  .... n
1
1
1
2
tot
So, for component 1, P 
1
n RT
V
1
For all components in the mixture, P 
tot
Divide P1 by Ptot: see that
Or
P
n

X,
P
n
1
1
tot
tot
1
P  XP
1
1
tot
Now we can solve the problem above!!
137
n RT
V
tot
Gas volumes in chemical reactions
Ch 3: given gram or mol amounts of a species in a
reaction- convert to moles of something else and ...
Ch 4: given molarity and volumes for solution reactions –
find moles and go!
Now: ideal gases as products/reactants in reactions
What information must be given in order for us to
calculate moles of a species in a reaction?
RECALL: what do the stoichiometric coefficients in a
balanced chemical equation tell us?
e.g., Consider the reaction
CaH2(s) + 2H2O(l)  Ca(OH)2(aq) + 2H2(g)
How many grams of CaH2 are needed to generate 10.0 L of
hydrogen gas if the partial pressure of H2 is 740 torr at
23oC?
138
Kinetic-molecular theory of gases
The ideal gas law describes how gases behave, i.e., P-V-T
behavior
The kinetic-molecular theory of gases helps explain why
gases behave as they do
E.g. How does K-M theory explain the pressure of a
gas?
The pressure of a gas is caused by the force exerted on
the walls of the container due to collisions between the
particles that make up the gas and the wall
Any factor which increases the force with which the
particles collide with the container wall will increase
the pressure
The absolute temperature of a gas is a measure of the
average kinetic energy of its molecules
E.g., why does the volume of a gas increase with a
temperature increase at constant pressure?
139
Summary of the main points of K-M theory
Gases consist of large numbers of particles in
continuous, random motion
Volume of gas particles is negligible relative to the
total volume of the container
Attractive/repulsive interactions between gas particles
are negligible
Collisions between gas particles are perfectly elastic
Average kinetic energy KE of the particles in a gas is
proportional to absolute temperature
Recall that KE 
1
mv , in general.........
2
2
1
Mu , where M=molar mass and u is
2
the average speed of particles in the gas
For a gas, KE 
2
140
3
RT for 1 mol of a gas.......this is a result
2
from K-M theory (we’ll use it in a second)
We have KE 
If two different gases are at the same temperature, then
their particles have the same average kinetic energy
What are the implications of this?
Molecular motion increases with increasing temperature:
if the temperature is doubled, average kinetic energy
doubles
Although particles in a gas sample have an average
kinetic energy KE (and hence an average speed u ),
individual particles move at varying speeds
Why is this so?
In a sample on the order of 1 mol of particles, how can we
calculate the average speed of a single particle?
We can't - we must look at distribution of speeds in the
sample and find the average from this
141
Demo # 1: Maxwell distribution of speeds in a gas
Demo #2: Effect of molar mass and temperature on speeds
in a gas
142
Notice:
3
1
RT  Mu
2
2
2
We define the root-mean-square (rms) speed urms as
u

u
u

3RT
M
rms
2
So, from above,
rms
Where M is the molar mass.
Since the molar mass M doesn't change with temperature,
as T increases, the rms speed of the particles must
increase
How does the rms speed depend on molar mass at
constant temperature?
143
E.g., Place the following gases in order of increasing rms
speed at 25oC: Ne, HBr, SO2, NF3, CO.
E.g., Calculate and compare the rms speeds of H2 and CO2
molecules at 300K.
144
Other applications of K-M Theory
Effusion: escape of gas particles through a pinhole in to
an evacuated space
E.g., He leaking (effusing) from a hole in a balloon – a
He atom can exit only when it happens to hit the hole
Diffusion: spread of one substance throughout a space or
throughout a second substance
E.g., a perfume moves by diffusion throughout a room
How do these phenomena depend on molar mass?
Suppose we have two gases of molar masses M1 and M2 at
the same temperature
Then,
u
rms1

3RT
, and
M
1
u
rms2

3RT
M
2
Form the ratio of the rms speeds.....
u
u
rms1
rms2

M
M
2
1
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This tells us that the faster a particle is moving, the more
likely it is to hit the ‘hole’ and effuse!
Deviations from ideal gas behavior
When do gases not obey the ideal gas law?
Real gases deviate from ideal behavior most notably
at high pressure and low temperature
Why? what does the ideal gas law neglect?
How to account for these factors?
One way: Van der Waals equation
2
nRT n a
P

V  nb V
2
What are a & b?
Next: Chapter 14
146
Problems du Jour
The planet Jupiter has a mass 318 times that of Earth, and
its surface temperature is 140K. Mercury has a mass 0.05
times that of Earth, and its surface temperature is 600700K. On which planet is the atmosphere more likely to
obey the ideal gas law?
Calculate the pressure that CCl4 will exert at 40oC if 1.00
mol occupies 28.0 L, assuming:
(a) CCl4 obeys the ideal gas law
(b) CCl4 obeys the van der Waals equation
147