Gas Densities,
Partial Pressures, and
Kinetic-Molecular Theory
Sections 10.5-10.8
Objectives
Apply the ideal-gas equation to real gas
situations.
 Interpret the kinetic-molecular theory of
gases

Key Terms
Partial pressures
 Dalton’s Law of Partial Pressures
 Mole fraction
 Kinetic-molecular theory
 Root-mean-square speed
 Effusion
 Graham’s Law
 Diffusion
 Mean free path

Gas Densities and Molar Mass
Rearrange the ideal-gas equation :
n=P
V RT
 Multiply both sides by molar mass, M
nM = PM
V RT
 Product of n/V and M = density in g/L
Moles x grams = grams
Liter
mole
liter

Gas Densities and Molar Mass

Density is expressed:
d = PM
RT

Density depends on pressure, molar
mass, and temperature
Example

Calculate the average molar mass of dry
air if it has a density of 1.17 g/L at 21 ºC
and 740.0 torr.
Gas Mixtures and Partial Pressure

Dalton’s Law of Partial Pressures:
– Total pressure of a mixture equals sum
of the pressures that each would exert if
present alone.
Pt = P1 + P2 + P3 + ….
Gas Mixtures and Partial
Pressures
P1 = n1 (RT); P2 = n2 (RT); P3 = n3 (RT);…
V
V
V

And
Pt = (n1 + n2 + n3 + ….) RT = nt (RT)
V
V
Example 1

A gaseous mixture made from 6.00 g
oxygen and 9.00 g methane is placed in
a 15.0 L vessel at 0 C. What is the
partial pressure of each gas, and what is
the total pressure of the vessel?
Example 2

What is the total pressure exerted by a
mixture of 2.00g hydrogen and 8.00 g
nitrogen at 273 K in a 10.0 L vessel?
Mole Fraction, X
P1 = n1 RT/ V = n1
Pt = nt RT/ V = nt
Thus…
P1 = (n1/nt)Pt = X1Pt
Partial press = mole frac x total press
Example 3

Mole fraction of N2 in air is 0.78 (78%).
If the total pressure is 760 torr, what is
the partial pressure of N2?
Homework

44, 48, and 60-68 even only
Sections 10.7 & 10.8
Kinetic-Molecular Theory
And
Effusion/Diffusion
Objectives
Understand why gas behave as they do
 Apply the Kinetic-Molecular Theory to
the Gas Laws
 Define molecular effusion and diffusion
 Solve problems using Graham’s Law of
Effusion

Key Terms
Kinetic-Molecular Theory
 Root-Mean-Square Speed
 Effusion
 Diffusion
 Graham’s Law of Effusion
 Mean Free Path

Kinetic-Molecular Theory
•Explains why gases behave as they do
•Developed over 100 year period
•Published in 1857 by Rudolf Clausius
Kinetic Molecular Theory
* Theory
of moving molecules
You Must Know the 5 Postulates
(page 421).
Five Postulates
1)
2)
3)
4)
5)
Gases consist of large numbers of molecules that are in
continuous, random motion.
The combined volume of the molecules is negligible relative
to the total volume in which the gas is contained.
Attractive and repulsive forces between gas molecules are
negligible.
Energy can be transferred between molecules during
collisions, but the average kinetic energy of the molecules
does not change with time, as long as T is constant
The average kinetic energy of the molecules is proportional
to T. At any given T, all molecules have same avg. kinetic
energy
Root-mean-square speed, u

Speed of a molecule possessing average
kinetic energy
Є = ½ mu2
Є is average kinetic energy
m is mass of molecule

Both Є and u increase as temperature
increases
Application to Gas Laws
1.
Effect of a V increase at constant T:
- Є does not change when T is
constant. Thus u is unchanged. With
V increase, there are fewer collisions
with container walls, and pressure
decreases (Boyle’s Law).
Application to Gas Laws
2. Effect of a T increase at constant V:
- Increase T means increase of Є and u.
No change in V means there will be
more collisions with walls (P increase).
Learning Check
A sample of carbon dioxide initially at
STP is compressed into a smaller
volume at constant temperature. How
does this effect:
(a) Average kinetic energy
(b) rms speed
(c) Total number of collisions
(d) Pressure

Molecular Effusion & Diffusion
u = 3RT
M
*Derived equation from the k-m theory
**Less massive gas molecules have higher rms speed
***Use R in units of J/mol-K
Example

Calculate the rms speed of a nitrogen
molecule at 298K.
Effusion

Escape of gas molecules through a tiny
hole into an evacuated space
Diffusion

Spread of one substance throughout a
space or throughout a second substance
Graham’s Law of Effusion
Effusion rate of a gas is inversely
proportional to the square root of its
molar mass.
 Rates of effusion of two gases under
identical conditions*:

* At same T and P in containers with identical pinholes
Graham’s Law of Effusion

Rate directly proportional to u:
u1 = 3RT/M1
u2
3RT/M2
Graham’s Law of Effusion
Diffusion and Mean Free Path
Similar to Effusion (faster for lower mass
molecules)
 BUT diffusion is slower than molecular
speeds because of molecular collisions


Mean Free Path: average distance traveled by
a molecule between collisions
– For air molecules at seal level = 6 x 10-8 m
– At about 100 km in altitude = 10 cm
Homework

69, 70, 73, 76, 77, 79