What number of moles of oxygen would exert a pressure of 10 atm at 320 K in an 8.2 dm$^{3}$ cylinder? [R=0.082 atm dm$^{-3}$ mol$^{-1}$ K$^{-1}$]

Analyze the question: We're given pressure, temperature, volume and R, and we need the number of moles. This is a job for the ideal gas equation.

Concept: The ideal gas law is

PV = nRT, so n = PV / RT

Solution:

• P = 10 atm
• V = 8.2 dm³
• T = 320 K
• R = 0.082 atm dm³ mol⁻¹ K⁻¹

n = (10 × 8.2) / (0.082 × 320)

Numerator: 10 × 8.2 = 82

Denominator: 0.082 × 320 = 26.24

n = 82 / 26.24 ≈ 3.125 ≈ 3.13 mol

Answer: The correct option is C (3.13).

Tip: Before calculating, look for numbers that cancel. Here 8.2 / 0.082 = 100, so n = (10 × 100) / 320 = 1000/320 = 3.125. Spotting this saves time in the exam. Also, always check that your units match those of R (atm, dm³, K) before substituting.