LMALaput
Chemistry 16
Lecture 19
Chapter Outline
1. Mole Fraction
2. Dalton’s Law of Partial Pressures
3. Stoichiometry involving gases
4. Kinetic Molecular Theory
5. Diffusion and Effusion
6. Graham’s Law of effusion
7. Deviations from the Ideal Gas Law
DALTON’S LAW OF PARTIAL PRESSURES
•Dalton’s law states that the pressure exerted by a mixture of
gases is the sum of the partial pressures of the individual
gases.
P
total
= P
A
+ P
B
+ P
C
+ .
....
The partial pressure of each gas is equal to its mole fraction
in the gaseous mixture times the total pressure of the mixture.
•A part of any mixture can be described as a mole fraction,
X
A
:
X
A
= No. mol A / total no. mol of all components
For gaseous mixture we can relate the mole fraction of each
component to its partial pressure as follows. From the ideal
gas equation, the number of moles of each component can
be written as
°
A
= P
A
V/ RT,
°
B
= P
b
V / RT and so on
And the total number of moles is
°
TOTAL
= P
TOTAL
V/RT
Substituting into the definition of X
A
,
°
A
/
°
A
+
°
B
+ …
= P
A
V/RT ÷ P
TOTAL
V/RT
X
A
= P
A
/P
TOTAL
Similarly, X
B
=P
B
/P
TOTAL
Example:
If 100 mL of hydrogen, measured at 25.0° C and
3.00 atm pressure, and 100 mL of oxygen, measured at
25.0°C and 2.00 atm pressure, were forced into one of the
containers at 25.0°C, what would be the pressure of the
mixture of gases?
P
TOTAL
= P
H2
+ P
O2
3.00 atm + 2.00 atm
= 5.00 atm
The mole fraction of oxygen in the atmosphere is 0.2094.
Calculate the partial pressure of O
2
in the air when the
atmospheric pressure is 760 torr.
(The partial pressure of each gas in a mixture is equal to its
mole fraction in the mixture times the total pressure of the
mixture)
P
O2
= X
02
× P
TOTAL
= 0.2094 × 760 torr = 159 torr
Two tanks are connected by a closed valve. Each tank is
filled with gas as shown and both tanks are held at the same
temperature. We open the valve and allow the gases to mix.
(a) After the gases mix, what partial pressure of each gas
and what is the total pressure? (b) What is the mole fraction
of each gas in the mixture?
(a) For O
2
: P
2,O2
= P
1
V
1
/V
2
= 24.0 atm × 5.0 L / 8.0 L = 15.0 atm
For N
2
: P
2,n2
= P
1
V
1
/V
2
= 32.0 atm × 3.0 L / 8.0 L = 12.0 atm
P
TOTAL
= P
2,O2
+ P
2,N2
= 15.0 atm + 12.0 atm = 27.0 atm
(b) X
O2
= P
2,O2
/P
TOTAL
= 15.0atm / 27.0 atm = 0.556
X
N2
= P
2,N2
/P
TOTAL
= 12.0atm/27.0atm = 0.444
X
O2
+ X
N2
should be equal to 1.
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 Spring '11
 UPERG
 Mole, Partial Pressure, Stoichiometry

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