Transcript V. Two More Gas Laws

Gases
Two More Laws
Chapter 14
Dalton’s Law
 The
total pressure of a mixture
of gases equals the sum of the
partial pressures of the
individual gases.
Ptotal = P1 + P2 + ...
When a H2 gas is
collected by water
displacement, the gas in
the collection bottle is
actually a mixture of H2
and water vapor.
Dalton’s Law

Determine the total pressure of a gas mixture
that contains O2, N2, and He if the partial
pressures of the gases are as listed below.
GIVEN:
Ptotal = ?
PO2 = 20.0 kPa
PN2 = 46.7 kPa
PHe= 26.7 kPa
WORK:
Ptotal = Po2 + PN2 + P He
P total = 20.0 + 46.7 + 26.7
P total = 93.4 kPa
Dalton’s Law

A gas is collected over water at a temp of
35.0°C when the barometric pressure is 742.0
torr. What is the partial pressure of the dry gas?
The total pressure in the collection bottle is equal to barometric
pressure and is a mixture of the “gas” and water vapor.
GIVEN:
Pgas = ?
Ptotal = 742.0 torr
PH2O = 42.2 torr
Given: water-vapor pressure for
35.0°C.
WORK:
Ptotal = Pgas + PH2O
742.0 torr = PH2 + 42.2 torr
Pgas = 699.8 torr
Graham’s Law
 Diffusion
• Spreading of gas molecules
throughout a container until
evenly distributed.
 Effusion
• Passing of gas molecules
through a tiny opening in a
container
Graham’s Law
 Speed
of diffusion/effusion
• Kinetic energy is determined by
the temperature of the gas.
• At the same temp & KE, heavier
molecules move more slowly.
 Larger m  smaller v
KE =
2
½mv
Graham’s Law
 Graham’s
Law
• Rate of diffusion of a gas is
inversely related to the square root
of its molar mass.
• The equation shows the ratio of
Gas A’s speed to Gas B’s speed.
vA

vB
mB
mA
Graham’s Law
 Determine
the relative rate of diffusion
for krypton and bromine.
The first gas is “Gas A” and the second gas is “Gas B”.
Relative rate means find the ratio “vA/vB”.
vA

vB
v Kr

v Br2
m Br2
m Kr
mB
mA
159.80g/mol
 1.381

83.80g/mol
Kr diffuses 1.381 times faster than Br2.
Graham’s Law

vA

vB
A molecule of oxygen gas has an average
speed of 12.3 m/s at a given temp and pressure.
What is the average speed of hydrogen
molecules at the same conditions?
mB
mA
vH 2
12.3 m/s

32.00 g/mol
2.02 g/mol
vH 2
vH 2
vO2

mO2
mH 2
Put the gas with
the unknown
speed as
“Gas A”.
12.3 m/s
 3.980
vH2  49.0m/s
Graham’s Law

An unknown gas diffuses 4.0 times faster than
O2. Find its molar mass.
The first gas is “Gas A” and the second gas is “Gas B”.
The ratio “vA/vB” is 4.0.
vA

vB
vA

v O2
mB
mA
mO2
mA
Square both
sides to get rid
of the square
root sign.

32.00 g/mol
g/mol 
32.00
 4.0 



m
A
A


32.00 g/mol
16 
mA
32.00 g/mol
mA 
 2.0 g/mol
16
2
Questions!
1) How is the partial pressure of a
gas in a mixture calculated? How is
the rate of effusion of a gas
calculated?
Questions!
2) At the same temperature, the
rates of diffusion of carbon
monoxide and nitrogen gas are
virtually identical. Explain.
Calculations
3) Determine the relative rate of
diffusion for helium atoms to
fluorine molecules at the same
temperature.
Calculations!
4) Carbon dioxide molecules have
an average speed of 25.0 m/s at a
given temperature and pressure.
What is the average speed of
carbon monoxide molecules at the
same conditions?
Calculations!
5) Argon is collected over water at
30C. Find the pressure of the dry
gas if the barometric pressure is
0.975 atm. Water-vapor pressure
for 30C is 0.0418 atm.