Question

A gas mixture contains 0.150 mol of O2 gas, 0.116 mol of N2 gas, and 0.211 mol of Ar gas in a 0.500 L flask at 298 K. What is the partial pressure of N2 the mixture?

Answer 1

The partial pressure of N₂ in the mixture is 5.67 atm

Step-by-step explanation:

According to Dalton's law, the partial pressure of a gas in a mixture of gases is given by the formula;

Partial pressure of gas A = Mole fraction of gas A × Total pressure of the mixture

We are required to calculate the partial pressure of N₂ in the mixture;

We are going to use the following simple steps;

Step 1. Calculating the total pressure of the mixture

We are given;

Moles of O₂ = 0.150 mol

Moles of N₂ = 0.116 mol

Moles of Ar = 0.211 mol

Thus , total moles of the mixture = 0.150 mol +0.116 mol + 0.211 mol

= 0.477 mol

Volume of the flask = 0.5 L

Temperature = 298 K

But; PV = nRT , where n is the number of moles and R is the ideal gas constant,0.082057 L.atm/mol.K

Thus; P = nRT ÷ V

= (0.477 mol × 0.082057 × 298 )÷ 0.5

= 23.328 atm

Step 2: Mole fraction of N₂

Mole fraction of a gas = Moles of a gas ÷ Total moles of the mixture

Mole fraction of N₂ = 0.116 mol ÷ 0.477 mol

= 0.243

Step 3: Partial pressure of N₂

P(N₂) = Mole fraction of N₂ × P(total)

= 0.243 × 23.328 atm

= 5.669 atm

= 5.67 atm ( 2 d.p.)

Therefore, the partial pressure of N₂ in the mixture is 5.67 atm