Using PV = nRT
(Honors)
P = Pressure
V = Volume
T = Temperature
N = number of moles
R is a constant, called the Ideal Gas Constant
Instead of learning a different value for R for all the possible unit
combinations, we can just memorize one value and convert the
units to match R.
R = 0.0821
L • atm
Mol • K
Ideal Gas Law: P V = n R T
P = pressure in atm
V = volume measured in Liters
n =# of moles
T = temperature K
R=Universal gas constant
=0.0821 L atm/(mol K)
Ideal
Gas
Law
(Honors)
 Calculate the pressure in atmospheres of
0.412 mol of He at 16°C & occupying 3.25 L.
IDEAL GAS LAW
GIVEN:
WORK:
P = ? atm
PV = nRT
n = 0.412 mol
P(3.25)=(0.412)(0.0821)(289)
L
mol Latm/molK K
T = 16°C = 289 K
V = 3.25 L
P = 3.01 atm
R = 0.0821Latm/molK
Some Cool Videos (Honors)








Crash Course: Ideal Gas Laws
http://www.youtube.com/watch?v=BxUS1K7xu30&safe=active
Crash Course: Ideal Gas Law Problems
http://www.youtube.com/watch?v=8SRAkXMu3d0
Crash Course: Real Gases
http://www.youtube.com/watch?v=GIPrsWuSkQc&safe=active
Crash Course: Grahams Law
http://www.youtube.com/watch?v=TLRZAFU_9Kg&safe=active
Dalton’s Law of
Partial Pressures

Total pressure of a mixture of gases in a
container equals the sum of the
individual partial pressures of each gas.
Ptotal = P1 + P2 + ...
Patm = PH2 +
PH2O
This is often useful when gases are
collected “over water”
Crash Course: Partial Pressure
and Vapor Pressure
http://www.youtube.com/watch?
v=JbqtqCunYzA&safe=active
You can use Table H
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 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
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.
DALTON’S LAW
GIVEN:
Pgas = ?
Ptotal = 742.0 torr
PH2O = 42.2 torr
Look up water-vapor pressure
for 35.0°C.
WORK:
Ptotal = Pgas + PH2O
742.0 torr = PH2 + 42.2 torr
Pgas = 699.8 torr
Sig Figs: Round to least number
of decimal places.
Using Mole Fraction (Honors)
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.
Graham’s Law Formula (Honors)

Graham’s Law

Rate of diffusion of a gas is inversely related
to the square root of its molar mass.
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 mean find the ratio “vA/vB”.
vA
vB
v Kr
v Br 2

m Br 2
m Kr


mB
mA
159.80 g/mol
 1.381
83.80 g/mol
Kr diffuses 1.381 times faster than Br2.
Graham’s Law

vA
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?

vB
mB
vH2
mA
12.3 m/s

32.00 g/mol
2.02 g/mol
vH2
vH2
vO2

m O2
mH2
Put the gas with
the unknown
speed as
“Gas A”.
 3.980
12.3 m/s
v H 2  49.0 m/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
vO2
mB
mA

Square both
sides to get rid
of the square
root sign.
16 
m O2
mA

 4.0 


32.00 g/mol 


mA

32.00 g/mol
mA
mA 
32.00 g/mol
16
 2.0 g/mol
2
Avogadro’s Principle

Equal volumes of gases contain
equal numbers of moles


at constant temp & pressure
true for any gas
Equal volumes of gases at the same T and P have the
same number of molecules.
V
V
n
n
k
Gas Stoichiometry (Honors)

Moles  Liters of a Gas



STP - use 22.4 L/mol
Non-STP - use ideal gas law
Non-STP Problems

Given liters of gas?


start with ideal gas law
Looking for liters of gas?

start with stoichiometry conv.

Gas Stoichiometry Problem
What volume of CO2 forms from 5.25 g of
CaCO3 at 103 kPa & 25ºC?
CaCO3
5.25 g

CaO
+
Looking for liters: Start with stoich
and calculate moles of CO2.
5.25 g 1 mol
CaCO3 CaCO3
1 mol
CO2
100.09g 1 mol
CaCO3 CaCO3
CO2
?L
non-STP
= 1.26 mol CO2
Plug this into the Ideal
Gas Law to find liters.

Gas Stoichiometry Problem
What volume of CO2 forms from 5.25 g
of CaCO3 at 103 kPa & 25ºC?
GIVEN:
WORK:
P = 103 kPa
V=?
n = 1.26 mol
T = 25°C = 298 K
R = 8.315 dm3kPa/molK
PV = nRT
(103 kPa)V
=(1mol)(8.315dm3kPa/molK)(298K)
V = 1.26 dm3 CO2

Stoichiometry
Problem
How Gas
many grams
of Al2O3 are formed
from 15.0 L of
O2 at 97.3 kPa & 21°C?
4 Al
+
3 O2

2 Al2O3
15.0 L
non-STP
?g
GIVEN:
WORK:
P = 97.3 kPa
V = 15.0 L
n=?
T = 21°C = 294 K
R = 8.315 dm3kPa/molK
Given liters: Start with
Ideal Gas Law and
calculate moles of O2.
PV = nRT
(97.3 kPa) (15.0 L)
= n (8.315dm3kPa/molK) (294K)
NEXT 
n = 0.597 mol O2

Gas Stoichiometry Problem
How many grams of Al2O3 are formed from
15.0 L of O2 at 97.3 kPa & 21°C?
3 O2 
15.0L
Use stoich to convert moles
of O to grams Al O .
non-STP
0.597 2 mol 101.96 g
mol O2 Al2O3
Al2O3
4 Al
2
2
+
2 Al2O3
?g
3
3 mol O2
1 mol
Al2O3
= 40.6 g Al2O3