Three gases (8.00 g of methane, ch4, 18.0 g of ethane, c2h6, and an unknown amount of propane, c3h8) were added to the same 10.0-l container. at 23.0 ∘c, the total pressure in the container is 5.50 atm. calculate the partial pressure of each gas in the container.

Assume ideal gas behavior, then solve for the total number of moles:

PV = nRT

(5.50 atm) (10 L) = n (0.0821 L-atm/mol-K) (23+273 K)

n = 2.263 mol

Moles methane: 8 g : 16.04 g/mol = 0.499 mol

Moles ethane: 18 g : 30.07 g/mol = 0.599 mol

Moles propane: 2.263 - (0.499+0.599) = 1.165 mol

Applying Raoult's Law:

Partial pressure = Mole fraction * Total Pressure

Partial Pressure of Methane = (0.499/2.263) (5.5 atm) = 1.21 atm

Partial Pressure of Ethane = (0.599/2.263) (5.5 atm) = 1.46 atm

Partial Pressure of Propane = (1.165/2.263) (5.5 atm) = 2.83 atm