Ideal Gas Law & Gas Stoichiometry

Ideal Gas Law

P V = n R TP V = n R T• 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 = nRTPV = 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 = nRTPV = nRT

• 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

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

Example

Permutations of the Ideal Gas Law

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

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

• 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 mol22.4 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