Transcript Graham`s Law

Additional Gas
Laws
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
C. Johannesson
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
C. Johannesson
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 mean 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
C. Johannesson
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
12.3 m/s
Put the gas with
the unknown
speed as
“Gas A”.C. Johannesson
 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
m A  C. Johannesson
 2.0 g/mol
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2
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 of partial pressures and a
table of known vapor pressures of water
can be used to determine the pressure
of dry gas that has been collected.
Dalton’s Law
 Hydrogen
gas is collected over water at
22.5°C. Find the pressure of the dry gas
if the atmospheric pressure is 94.4 kPa.
The total pressure in the collection bottle is equal to atmospheric
pressure and is a mixture of H2 and water vapor.
GIVEN:
PH2 = ?
Ptotal = 94.4 kPa
PH2O = 2.72 kPa
Look up water-vapor pressure
on chart for 22.5°C.
WORK:
Ptotal = PH2 + PH2O
94.4 kPa = PH2 + 2.72 kPa
PH2 = 91.7 kPa
Sig Figs: Round to least number
C. Johannesson of decimal places.
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
Look up water-vapor pressure
on table for 35.0°C and convert
to torr.
WORK:
Ptotal = Pgas + PH2O
742.0 torr = PH2 + 42.2 torr
Pgas = 699.8 torr
Sig Figs: Round to least number
C. Johannesson of decimal places.
Sample Problem #1
Mixtures of helium and oxygen are used in the air tanks of
underwater divers for deep dives. For a particular dive, 12 L of O2
at 25oC and 1.0 atm and 46 L of He at 25oC and 1.0 atm were both
pumped into a 5.0-L tank.
A)
B)
Calculate the partial pressure of each gas
Calculate the total pressure in the tank at 25oC (298 K).
PV = nRT *Because the partial pressure of each gas depends on the total moles
present, we must first calculate n for each gas using the ideal gas law.
n = PV ÷ RT
nO2 = (1.0 x 12) ÷ (0.08206 x 298) = 0.49 mol
nHe = (1.0 x 46) ÷ (0.08206 x 298) = 1.9 mol
*The tank has a volume of 5.0-L and temperature of 298 K, so we can
V figure out the partial pressure of each gas
P = nRT ÷
PO2 = (0.49 x 0.08206 x 298) ÷ 5.0 = 2.4 atm
PHe = (1.9 x 0.08206 x 298) ÷ 5.0 = 9.3 atm
Determine the total pressure:
Ptotal = PO2 + PHe = 2.4 atm + 9.3 atm = 11.7 atm
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Sample Problem #2
A sample of solid potassium chlorate, KClO3, was heated in
a test tube and decomposed according to the reaction:
2KClO3 (s)  2KCl (s) + 3O2 (g)
The oxygen produced was collected by displacement of
water at 22oC. The resulting mixture of O2 and H2O vapor
had a pressure of 754 torr and a volume of 0.650-L.
Calculate the partial pressure of O2 in the gas collected
and the number of moles of O2 present. The vapor
pressure of water (PH2O) at 22oC is 21 torr.
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Sample Problem #2 Solution
We know:
• Ptotal = 754 torr
• PH2O = 21 torr
So we can determine:
Ptotal = PH2O + PO2 OR Ptotal - PH2O = PO2
SO… 754 torr – 21 torr = 733 torr
PO2 = 733 torr
Next, we solve the ideal gas law for nO2
Convert pressure from torr to atm:
nO2 = (PO2 V) ÷ (RT)
733 torr x (1 atm/760 torr) = 0.964 atm
Then solve:
P = 0.964 atm
V = 0.650 L
T = 22oC = 22 + 273 = 295 K
R = 0.08206 L atm/K mol
nO2 = (0.964 x 0.650)
(0.0821 x 295)
nO2 = 0.0259 mol O2
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Dalton’s Law Practice Problems
1.
If a gaseous mixture is made of 2.41 g of He and 2.79 g of Ne in 1.04-L
container at 25oC, what will be the partial pressure of each gas and the
total pressure in the container?
2.
How many moles of helium gas would be required to fill a 2.41-L
container to a pressure of 759 mm Hg at 25oC?
3.
A sample of oxygen gas (O2) is saturated with water vapor (H2O) at 27oC.
The total pressure of the mixture is 772 torr, and the vapor pressure of
water is 26.7 torr at 27oC. What is the partial pressure of the oxygen
gas?
4.
Suppose a gaseous mixture of 1.15 g helium and 2.91 g argon is placed in
a 5.25-L container at 273oC. What pressure would exist in the container?
5.
A tank contains a mixture of 3.0 mol of N2, 2.0 mol of O2, and 1.0 mol of
CO2 at 25oC and a total pressure of 10.0 atm. Calculate the partial
pressure (in torr) of each gas in the mixture.
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