My question concerns Dalton's law of partial pressures.

Question

In a sample of 2.55 g of propane, C3H8, is collected in a container at a pressure of 2.55 bar. In a separate container 1.01 g of methane, CH4, are collected. the two gases are then combined into the container the propane is in originally. Determine the mole fraction and partial pressure for each gas in the mixture, and the pressure in the container after the two gasses are combined.

Answer 1

moles C₃H₈ = 2.55 g x 1 mol C₃H₈ / 44.097 g = 0.05783 moles

moles CH₄ = 1.01 g x 1 mol CH₄ / 16.043 g = 0.06296 moles

Total moles of gas = 0.05783 mols + 0.06296 mols = 0.12079 mols of gas

X_C3H8 = 0.05783 mols / 0.12079 mols = 0.479 = mol fraction of C₃H₈

X_CH4 = 0.06296 mols / 0.12079 mols = 0.521 = mol fraction of CH₄

To find the pressure in the container after combining the 2 gases, we assume that the volume of the container is constant, as is the temperature. Then...

P1/n1 = P2/n2 since under these conditions pressure (P) will be proportional to moles.

P1 = initial pressure = 2.55 bar

n1 = initial mols = 0.05783 mols C₃H₈

P2 = final pressure = ?

n2 = final moles = 0.12079 mols of total gas after combining C₃H₈ and CH4

2.55 bar/0.05783 mol = P2/0.12079 mol

P2 = 5.33 bar = total pressure after combining gases = 5.26 atm