Mark S. Cracolice
Edward I. Peters
www.cengage.com/chemistry/cracolice
Chapter 14
The Ideal Gas Law
and Its Applications
Mark S. Cracolice • The University of Montana
Avogadro’s Law
Avogadro’s Law
Equal volumes of gases at the same temperature and
pressure contain the same number of molecules.
Volume of a gas is proportional to number of moles:
V
µ
n
V=k×n
Molar Volume
The molar volume is the volume of one mole of
gas. The molar volume of an ideal gas
depends on the temperature and pressure.
One mole of any ideal gas occupies 22.7 liters
at 0 0C and 1 bar (STP).
The Ideal Gas Law
PV = n RT
P = pressure of the gas
V = volume of the gas
T = temperature, in Kelvin
n = number of moles
R = universal gas constant
R = 0.0821 L.atm/mol.K
R = 62.4 L torr/mol.K
The Ideal Gas Law
PV = n RT
Since the number of moles is equal to the mass m
divided by the molar mass MM
n = m / MM
PV = (m/MM) x RT
Determination of a single variable
What volume will be occupied by 0.393 mole of
nitrogen at 0.971 atm and 240C?
PV = nRT
V = nRT/P
=
0.393 mol x (0.0821 L.atm/mol.K) x
(24+273)K / 0.971 atm
= 9.87 L
Determination of molar mass
1.67 grams of an unknown liquid are vaporized
at a temperature of 125 0C. Its volume is
measured as 0.421 liter at 749 torr. Calculate
the molar mass.
PV = (m/MM) x RT
Molar mass= MM = mRT/PV
= 1.67g x (62.4 L.torr/mol.K) x
(125+273)K/(749 torr x 0.421L)
= 132g/mol
Gas Density
mass
Density º
volume
D 
m
V
m
PV =
RT
MM
mRT
m
RT
RT
=
´
= D ´
(MM) V
V
MM
MM
RT
P = D ´
MM
P =
D =
(MM) P
R T
Gas Density
Density depends on temperature, pressure and molar mass.
Hydrogen and helium have lowest densities.
What is the density of nitrogen at 44°C and 0.632 atm?
.
(MM) P
D =
RT
=
28.02 g
 0.632 atm 
mol
mol  K
0.0821 L  atm

1
(44 + 273) K
= 0.680
g/L
Molar Mass from Gas Density
Find the molar mass of an unknown gas if its
density is 3.97g/L at 1250C and 749 torr.
(MM) P
D =
RT
Molar mass= MM = D x RT/P
= (3.97g/L) x (62.4 L.torr/mol.K) x
(125+273)K/(749 torr)
•
= 132g/mol
Molar Volume
Molar Volume of a Gas
The volume occupied by one mole of gas molecules.
V
In symbols, MV º
n
PV = nRT
V =
nRT
P
V
RT
=
= MV
n
P
Molar Volume
Example:
What is the molar volume of a ideal gas at 11°C and 1.33 atm?
Solution:
Solve with algebra.
V
RT
0.0821 L × atm
MV =
=
=
´
n
P
mol × K
1
(11 + 273) K ´
= 17.5 L/mol
1.33 atm
Gas Stoichiometry at STP
Example:
What volume of hydrogen, measured at STP, is
released when a 42.7 g Zinc is added to a
hydrochloric acid solution?
Zn
+
2 HCl
Given: 42.7 g Zn
→
H2
+
ZnCl2
wanted: volume H2
Molar volume at STP is 22.7 L/mol.
Gas Stoichiometry at STP
Zn
+
2 HCl →
H2
+
ZnCl2
Mass Zn mol Zn mol of H2volume H2
42.7 g Zn
x ( 1 mol Zn/ 65.38 g Zn )
x (1 mol H2 / 1 mol Zn )
x ( 22.7 L H2 / 1 mol H2 )
= 14.8 L H2
Stoichiometry: Molar Volume Method
Solving a Gas Stoichiometry Problem using
Molar Volume Method
Step 1: Use the ideal gas equation to find the molar
volume at the given temperature and pressure: V/n =
RT/P.
Step 2: Use the molar volume to calculate the wanted
quantity by all three steps of the stoichiometry path.
Stoichiometry: Molar Volume Method
What volume of CO2, measured at 131°C and 744
torr, is produced when 16.2 g of C4 H10 is burned
completely?
Step 1 is to find the molar volume at the given
temperature and pressure.
Given: 131°C; 744 torr
Wanted: MV
Molar volume = RT/P = (62.4 L torr/mol. K) x
( 131 + 273) / 744 torr
= 33.9 L/mol
Stoichiometry: Molar Volume Method
Step 2 is to use the molar volume to calculate the
wanted quantity by the stoichiometry path.
2 C4H10 + 13 O2 →
10 H2O + 8 CO2
gram of C4H10 → mol C4H10→ mol CO2 → L CO2
16.2 gram of C4H10
x (1 mol.C4H10/ 58.12 g C4H10 )
x(8 mol.CO2/ 2 mol.C4H10 )
x (33.9 L CO2 / mol. CO2)
= 37.8 L CO2
Volume–Volume Gas Stoichiometry
Volume–Volume Gas Stoichiometry
• At constant temperature and pressure, the gas
volume is directly proportional to the number of
moles.
• This means that at the same temperature and
pressure the ratio of gas volumes is equal to
moles ratio.
Volume–Volume Gas Stoichiometry
Hydrogen and nitrogen gases react to form gaseous
ammonia. How many liters of ammonia can be
produced from 5.5 L of nitrogen? Both gases are
measured at same temperature and pressure.
3 H2
+ N2 → 2 NH3
3 mol of H2 react with 1 mol N2 to form 2 mol NH3
At same T and P, same volumes of gases will have
same number of moles.
3 L of H2 react with 1L N2 to form 2 L of NH3
Volume–Volume Gas Stoichiometry
N2
+
3 H2
GIVEN: 5.5 L N2
5.5 L N 2 
→
2 NH3
WANTED: L NH3
2 L NH 3
1L N2
= 11 L NH 3
Volume–Volume Gas Stoichiometry
It is found that 1.75 L of oxygen, measured
at 24 0C and 755 torr, is used in burning
sulfur. The sulfur dioxide produced is at
165 0C and 785 torr. Find the volume of
sulfur dioxide at those condition.
O2
+
24 0C,755 torr
S
→
SO2
165 0C,785 torr
Volume–Volume Gas Stoichiometry
First calculate the volume of O2 at same
T and P as SO2 (165 0C, 785 torr)
Volume of O2 at 165 0C and 785 torr
= 1.75 L x ( 755 torr/785 torr)
x ( 438 K / 297 K)
= 2.48 L O2
Volume–Volume Gas Stoichiometry
Next calculate the volume of sulfur dioxide
from the volume of oxygen
O2
+
S
→
SO2
165 0C & 785 torr
165 0C & 785 torr
Litters SO2 (165 0C & 785 torr)
= 2.48 L O2 (165 0C & 785 torr)
x (1L SO2/1 L O2 )
= 2.48 L SO2
Homework
• Homework: 3, 5, 19, 21, 23, 29, 35, 43,
48, 51, 53