The Ideal Gas Law
PV = nRT
Adds in the factor of number of moles of gas “n”.
P = Pressure (atm)
V = Volume (Liters)
T = Temperature (Kelvin)
n = number of moles
Remember when dealing with moles:
# moles (n) = grams of substance
Gram Formula Mass
What is “R”?
R is a constant, called the “Universal Gas Constant”
It is derived by plugging in the values for 1 mole of
gas at STP. (22.4 Liters, 273K, 1 atm)
R = 0.0821
L • atm
Mol • K

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.
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
Example

# 1 Honors Packet

A 9.81 L cylinder contains 23.5 moles of
nitrogen at 23° C. What pressure is
exerted by the gas?
Example

# 2 Honors Packet

A pressure of 850 mmHg is exerted by
28.6 grams of sulfur dioxide at a temp of
40 °C. Calculate the volume of the vessel
holding the gas.
Density is Hidden in this
Formula

Density =
Mass (g)
Volume
# moles = Mass (g)
Gram formula mass
Can you rework PV = nRT to solve for
Density?
Example

# 7 in Honors Packet

What is the density of neon at 40 °C and
1.23 atmospheres?
Dalton’s Law of Partial
Pressures
Ptotal =
P1+P2+….

Total pressure of a mixture of gases in a
container is the sum of the individual
pressures (partial pressures) of each gas,
as if each took up the total space alone.

This is often useful when gases are
collected “over water”
Collecting Gas over Water
Pressureatm = Pressure (O2) + Pressure (H2O vapor)
http://www.kentchemistry.com/links/GasLaws/dalton.htm
Crash Course: Partial Pressure and Vapor Pressure
http://www.youtube.com/watch?v=JbqtqCunYzA&safe=active
Dalton’s Law

GIVEN:
PH2 = ?
Ptotal = 94.4 kPa
PH2O = 2.72 kPa
Look up water-vapor pressure
on for 22.5°C.
H2 gas is collected over
water at 22.5°C. Find
the pressure of the dry
gas if the atmospheric
pressure is 94.4 kPa.
WORK:
Ptotal = PH2 + PH2O
94.4 kPa = PH2 + 2.72 kPa
PH2 = 91.7 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?
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
Examples
of Vapor
Pressure
Tables
You can even use Table H to
find VP of Water
Using Mole
Fraction
(Honors)
Moles gas (X)
x P (total) = P (X)
Total Moles Gas
Example

# 4 Honors Packet

A mixture of 2.00 moles H2, 3.00 moles
NH3, 4.00 moles CO2, and 5.00 moles N2
exert a total pressure of 800 mmHg. What
is the partial pressure of each gas?
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
Smaller and lighter gas particles
do this faster!
https://www.youtube.com/watch?feature=player_embedded&v=H7QsDs8ZRMI
https://www.youtube.com/watch?feature=player_embedded&v=L41KhBPBymA

Crash Course: Grahams Law

http://www.youtube.com/watch?v=TLRZAFU_9Kg&safe=active
Graham’s Law

Speed of diffusion/effusion

At the same temp & KE, heavier molecules
move more slowly.

Ex: Which of the following gases will diffuse
most rapidly?
A.) N2
B.) CO2
C.) CH4
Graham’s Law Formula

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 Br2
m Br2
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

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?
vA

vB
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
vH 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
mB
mA
vA

v O2
mO2
mA

32.00
g/mol
 4.0 

m AA

32.00 g/mol
16 
mA
Square both
sides to get rid
of the square
root sign.




32.00 g/mol
 2.0 g/mol
mA 
16
2
Example

# 1 Honors Packet

Under the same conditions of temp and
pressure, how many times faster will
hydrogen effuse compared to carbon dioxide?
Example

#7 Honors Packet

If CO2 diffuses from a flask in 30 seconds
and an unknown gas diffuses from the
same flask in 5 minutes, calculate the
molecular weight of this unknown gas.
Avogadro’s Principle

Volume of a gas is directly proportional to
the number of moles of gas particles
present.
V
n
 Equal
volumes of gases contain equal
numbers of moles of particles
at constant temp & pressure
 true for any gas

Avogadros Law: https://www.youtube.com/watch?feature=player_embedded&v=fexEvn0ZOpo
Molar Volume of a Gas: https://www.youtube.com/watch?feature=player_embedded&v=4b852VIEkHQ
Practice Questions

A.
At the same temperature and pressure, 1.0
liter of CO(g) and 1.0 liter of CO2(g) have
equal masses and the same number of molecules
B. different masses and a different number of molecules
C. equal volumes and the same number of molecules
D. different volumes and a different number of molecules

Which rigid cylinder contains the same
number of gas molecules at STP as a 2.0liter rigid cylinder containing H2(g) at STP?
(1) 1.0-L cylinder of O2(g)
(2) 2.0-L cylinder of CH4(g)
(3) 1.5-L cylinder of NH3(g)
(4) 4.0-L cylinder of He(g

Which two samples of gas at STP contain
the same total number of molecules?
(1) 1 L of CO(g) and 0.5 L of N2(g)
(2) 2 L of CO(g) and 0.5 L of NH3(g)
(3) 1 L of H2(g) and 2 L of Cl2(g)
(4) 2 L of H2(g) and 2 L of Cl2(g)
Gas Stoichiometry

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 1.017atm & 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 1.017atm & 25ºC?
GIVEN:
WORK:
P = 1.017atm
PV = nRT
n = 1.26 mol
(1.017atm)V
T = 25°C = 298 K =(1mol)(.0821 Latm/molK)(298K)
R = .0821 Latm/molK
V = 1.26 L CO2
Gas Stoichiometry Problem

How many grams of Al2O3 are formed from 15.0 L of
O2 at 0.96 atm & 21°C?
4 Al
GIVEN:
P = 0.96 atm
+
3 O2

2 Al2O3
15.0 L
non-STP
?g
WORK:
Given liters: Start with
Ideal Gas Law and
calculate moles of O2.
PV = nRT
(0.96 atm ) (15.0 L)
= n (.0821 Latm/molK) (294K)
V = 15.0 L
n=?
NEXT 
T = 21°C = 294 K
n = 0.597 mol O2
R = .0821 Latm/molK
Gas Stoichiometry Problem

How many grams of Al2O3 are formed from
15.0 L of O2 at 0.96 atm & 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
Example

# 10 Honors Multiple Choice
Ni(CO)4 (l) → Ni (s) + 4CO (g)

What volume of CO is formed from the
complete decomposition of 444g of
Ni(CO)4 at 752 torr and 22.0 °C
Do stoich for 444g to find moles of CO.
 Plug the moles into Ideal Gas Law to get V at
non standard conditions.

Some Cool Videos






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