Ideal Gas Law
The equality for the four variables involved
in Boyle’s Law, Charles’ Law, Gay-Lussac’s
Law and Avogadro’s law can be written
PV = nRT
R = ideal gas constant
Ideal Gases
 Behave as described by the ideal gas
equation; no real gas is actually ideal
 Within a few %, ideal gas equation describes
most real gases at room temperature and
pressures of 1 atm or less
 In real gases, particles attract each other
reducing the pressure
 Real gases behave more like ideal gases as
pressure approaches zero.
PV = nRT
R is known as the universal gas constant
Using STP conditions
P
V
R
=
PV
nT
= (1.00 atm)(22.4 L)
(1mol) (273K)
n
T
= 0.0821 L-atm
mol-K
Learning Check G15
What is the value of R when the STP value
for P is 760 mmHg of 1 mole of gas?
Solution G15
What is the value of R when the STP value
for P is 760 mmHg?
R
=
PV
nT
= (760 mm Hg) (22.4 L)
(1mol)
(273K)
= 62.4 L-mm Hg
mol-K
Learning Check G16
Dinitrogen monoxide (N2O), laughing gas,
is used by dentists as an anesthetic. If 2.86
mol of gas occupies a 20.0 L tank at 23°C,
what is the pressure (mmHg) in the tank in
the dentist office?
Solution G16
Set up data for 3 of the 4 gas variables
Adjust to match the units of R
V = 20.0 L
20.0 L
T = 23°C + 273
296 K
n = 2.86 mol
2.86 mol
P =
?
?
Rearrange ideal gas law for unknown P
P = nRT
V
Substitute values of n, R, T and V and
solve for P
P = (2.86 mol)(62.4L-mmHg)(296 K)
(20.0 L)
(K-mol)
= 2.64 x 103 mm Hg
Learning Check G17
A 5.0 L cylinder contains oxygen gas
at 20.0°C and 735 mm Hg. How many
grams of oxygen are in the cylinder?
Solution G17
Solve ideal gas equation for n (moles)
n = PV
RT
= (735 mmHg)(5.0 L)(mol K)
(62.4 mmHg L)(293 K)
= 0. 20 mol O2 x 32.0 g O2 = 6.4 g O2
1 mol O2
Molar Mass of a gas
What is the molar mass of a gas if 0.250 g of
the gas occupy 215 mL at 0.813 atm and
30.0°C?
n = PV = (0.813 atm) (0.215 L) = 0.00703 mol
RT
(0.0821 L-atm/molK) (303K)
Molar mass =
g
mol
=
0.250 g = 35.6 g/mol
0.00703 mol
Density of a Gas
Calculate the density in g/L of O2 gas at STP.
From STP, we know the P and T.
P = 1.00 atm
T = 273 K
Rearrange the ideal gas equation for moles/L
PV = nRT
PV = nRT
P = n
RTV
RTV
RT
V
Substitute
(1.00 atm ) mol-K
=
(0.0821 L-atm) (273 K)
0.0446 mol O2/L
Change moles/L to g/L
0.0446 mol O2
1L
x
32.0 g O2
1 mol O2
=
1.43 g/L
Therefore the density of O2 gas at STP is
1.43 grams per liter
Formulas of Gases
A gas has a % composition by mass of
85.7% carbon and 14.3% hydrogen. At
STP the density of the gas is 2.50 g/L.
What is the molecular formula of the
gas?
Formulas of Gases
Calculate Empirical formula
85.7 g C x 1 mol C = 7.14 mol C/7.14 = 1 C
12.0 g C
14.3 g H x 1 mol H = 14.3 mol H/ 7.14 = 2 H
1.0 g H
Empirical formula = CH2
EF mass = 12.0 + 2(1.0) = 14.0 g/EF
Using STP and density ( 1 L = 2.50 g)
2.50 g
1L
x
22.4 L
1 mol
n = EF/ mol
=
=
56.0 g/mol
56.0 g/mol = 4
14.0 g/EF
molecular formula
CH2 x 4
=
C 4H 8
Gases in Chemical Equations
On December 1, 1783, Charles used 1.00 x 103
lb of iron filings to make the first ascent in a
balloon filled with hydrogen
Fe(s) + H2SO4(aq)  FeSO4(aq) + H2(g)
At STP, how many liters of hydrogen
gas were generated?
Solution
lb Fe  g Fe  mol Fe  mol H2  L H2
1.00 x 103 lb x 453.6 g x 1 mol Fe x 1 mol H2
1 lb
55.9 g
1 mol Fe
x 22.4 L H2 = 1.82 x 105 L H2
1 mol H2
Charles generated 182,000 L of hydrogen to fill his
air balloon.
Learning Check G18
How many L of O2 are need to react 28.0 g
NH3 at 24°C and 0.950 atm?
4 NH3(g) + 5 O2(g)
4 NO(g) + 6 H2O(g)
Solution G18
Find mole of O2
28.0 g NH3 x 1 mol NH3 x 5 mol O2
17.0 g NH3
4 mol NH3
= 2.06 mol O2
V = nRT =
P
(2.06 mol)(0.0821)(297K) = 52.9 L
0.950 atm
Summary of Conversions with
Gases
Volume A
Grams A
Atoms or
molecules A
Volume B
Moles A
Moles B
Grams B
Atoms or
molecules B
Daltons’ Law of Partial Pressures
The % of gases in air
Partial pressure (STP)
78.08% N2
593.4 mmHg
20.95% O2
159.2 mmHg
0.94% Ar
7.1 mmHg
0.03% CO2
0.2 mmHg
PAIR = PN + PO + PAr + PCO = 760 mmHg
2
2
2
Learning Check G19
A.If the atmospheric pressure today is 745
mm Hg, what is the partial pressure (mm
Hg) of O2 in the air?
1) 35.6
2) 156
3) 760
B. At an atmospheric pressure of 714, what is
the partial pressure (mm Hg) N2 in the air?
1) 557
2) 9.14
3) 0.109
Solution G19
A.If the atmospheric pressure today is 745
mm Hg, what is the partial pressure (mm
Hg) of O2 in the air?
2) 156
B. At an atmospheric pressure of 714, what is
the partial pressure (mm Hg) N2 in the air?
1) 557
Partial Pressure
Partial Pressure
Pressure each gas in a mixture would exert
if it were the only gas in the container
Dalton's Law of Partial Pressures
The total pressure exerted by a gas mixture
is the sum of the partial pressures of the
gases in that mixture.
PT = P1 +
P2 + P3 + .....
Partial Pressures
The total pressure of a gas mixture depends
on the total number of gas particles, not on
the types of particles.
STP
P = 1.00 atm
P = 1.00 atm
1.0 mol He
0.50 mol O2
+ 0.20 mol He
+ 0.30 mol N2
Health Note
When a scuba diver is several hundred feet
under water, the high pressures cause N2 from
the tank air to dissolve in the blood. If the
diver rises too fast, the dissolved N2 will form
bubbles in the blood, a dangerous and painful
condition called "the bends". Helium, which is
inert, less dense, and does not dissolve in the
blood, is mixed with O2 in scuba tanks used for
deep descents.
Learning Check G20
A 5.00 L scuba tank contains 1.05 mole of
O2 and 0.418 mole He at 25°C. What is the
partial pressure of each gas, and what is
the total pressure in the tank?
Solution G20
P = nRT
V
PT
=
PT = PO + PHe
2
1.47 mol x 0.0821 L-atm x 298 K
=
5.00 L
7.19 atm
(K mol)