A gas mixture contains 2.0 moles N2 and 3.0 moles Ar in a 10.0 L container at 298 K; calculate the partial pressure of each using the ideal gas law.

• A. P(N2) ≈ 4.89 atm; P(Ar) ≈ 7.34 atm
• B. P(N2) ≈ 7.34 atm; P(Ar) ≈ 4.89 atm
• C. P(N2) ≈ 2.00 atm; P(Ar) ≈ 3.00 atm
• D. P(N2) ≈ 9.80 atm; P(Ar) ≈ 0.50 atm

Answer: (A)

In a gas mixture kept at the same temperature and volume, each gas contributes to the total pressure in proportion to how much of the total amount of gas it represents. The direct way to get the partial pressures is P_i = n_i RT / V, and the total pressure is P_total = (n_total RT) / V.

Here, n_N2 = 2.0, n_Ar = 3.0, so n_total = 5.0; V = 10.0 L; T = 298 K. Using R = 0.082057 L atm / (mol K):

P_total = 5.0 × 0.082057 × 298 / 10.0 ≈ 12.23 atm.

Then the partial pressures follow from the mole fractions:

P_N2 = (2.0/5.0) × 12.23 ≈ 4.89 atm,

P_Ar = (3.0/5.0) × 12.23 ≈ 7.34 atm.

So the partial pressures are about 4.89 atm for nitrogen and 7.34 atm for argon. This matches the stated values.