IDEAL GAS LAW
Ideal Gas Law Derivation
 Recall that
P1V1 = P2V2
n1T1 n2T2
 Also we learned that
 At OoC, and 1 atm (101.3 kPa) that 1 mole of
gas occupies 22.4 L (molar volume)
Ideal Gas Law Derivation
 We can set one side of the equation under




standard conditions ie.
n = 1 mole
V = 22.4 L
P = 101. 3 kPa
T = 0oC
Ideal Gas Law Derivation
P1V1 = P2V2
n1T1 n2T2
So….
(101.3kPa) (22.4L) = P2V2
(1mole) (273K)
n2T2
Ideal Gas Law Derivation
8.314 kPa x L = P V
mol x K
nT
Overall….
PV = nRT
8.314 kPaL/mol K = R
= Ideal Gas Law 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 other STP conditions
P
V
R
=
PV
nT
= (1.00 atm)(22.4 L)
(1mol) (273K)
n
T
= 0.0821 L-atm
mol-K
Additional R Values
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 (kPa) 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)(8.314)(296 K)
(20.0 L)
= 351.91 kPa
What is the volume (in liters) occupied by 49.8
g of HCl at STP?
T = 0 0C = 273.15 K
P = 101.3 kPa
PV = nRT
nRT
V=
P
1 mol HCl
n = 49.8 g x
= 1.37 mol
36.45 g HCl
1.37 mol x 8.314 x 273.15 K
V=
V = 30.6 L
1 atm
Learning Check G17
A 5.0 L cylinder contains oxygen gas
at 20.0°C and 98 kPa. How many
grams of oxygen are in the cylinder?
Solution G17
Solve ideal gas equation for n (moles)
n = PV
RT
= (98 kPa)(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