Ideal Gas Law & Gas
Stoichiometry
Ideal Gas Law
PV=nRT
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P = Pressure (atm)
V = Volume (L)
T = Temperature (K)
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
PV = nRT
• Calculate the number of moles of a gas
contained in a 3.0 L vessel at 300.0K
with a pressure of 1.50 atm
PV = nRT
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n=?
V = 3.0 L
T = 300.0 K
P = 1.50 atm
PV = nRT
• (1.50 atm)(3.0 L) = n (0.0821L atm / mol K)(300.0 K)
• n = 0.18 mol
Example
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?
Note: 1atm = 760 mm Hg
Example
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n = 2.86 mol
V = 20.0 L
T = 23 °C = 273 + 23 = 296 K
P=?
PV = nRT
• (P)(20.0 L) = (2.86) (0.0821L atm / mol K)(296 K)
• P = 3.5 atm
• P = 2600 mm Hg
Permutations of the Ideal Gas
Law
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PV = mRT
M
P = Pressure (atm)
V = volume (L)
m = mass of the gas (g)
R = 0.0821 L atm / mol K
T = Temperature (K)
M = molecular mass
Example
• What is the pressure 2.0 g of nitrogen
gas in a 5.0 L container at 300.0 K?
• P=?
• m = 2.0 g
• V = 5.0 L
• T = 300.0K
• M = 28.02 g/mol
Example
PV = mRT
M
P(5.0) = (2.0)(0.0821)(300.0)
28.04
P = 2.8 atm
Permutations of the Ideal Gas
Law
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P = DRT
M
P = pressure (atm)
D = density (g/L)
R = 0.0821 L atm / mol K
T = temperature (K)
M = molecular mass
Example
• What is the molar mass of a gas that
has a density of 1.40 g/L at STP?
– NOTE – STP is standard temperature and
pressure
– At STP temperature is 273 K and pressure
is 1.00 atm
Example
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P = 1.00 atm
D = 1.40 g/L
R = 0.0821 L atm / mol K
T = 273 K
M=?
P = DRT
M
1.00 = (1.40)(0.0821)(273)
M
M = 31.4 g/mol
Avogadro’s Principle
• Avogadro’s Principle – equal volumes of
gases at equal temperature and pressure
contain the same number of particles
• Molar volume – the volume of gas that 1 mole
of a substance occupies at STP
• At STP 1 mol of a gas = 22.4 L
• New conversion factor at STP ONLY!
1 mol
22.4 L
Example
• Calculate the volume 0.881 mol of a gas
will occupy at STP.
• 0.881 mol x 22.4 L = 19.7 L
1 mol
(You could also have worked this out with
the ideal gas law equation)
Example
• Calculate the volume that 2.000 kg of
methane would occupy at STP.
• 2.000 kg x 1x10 3g x 1 mol x 22.4 L =
1kg
16.05g 1 mol
• 2791 L CH4