What number of moles of oxygen would exert a pressure of 10 atm at 320 K in an 8.2 dm$^{2}$ cylinder? [R = 0.082 atm dm$^{3}$ mol$^{-1}$ K$^{-1}$]

Analyze the question: We're given pressure, temperature and volume of oxygen gas and asked for the number of moles. This is a direct use of the ideal gas equation. (The "8.2 dm²" in the question is a typo for 8.2 dm³, since volume is measured in dm³ and R is given in dm³ units.)

Concept: The ideal gas law is

PV = nRT, so n = PV / RT

Solution:

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

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 number of moles is about 3.13, which is option C.

Tip: Notice that 8.2 and 0.082 differ by a factor of 100, so you can simplify before calculating: n = (10 × 100) / 320 = 1000/320 = 3.125. Cancelling like this saves time in the exam. Also always check that your units match those of R (atm, dm³, K) before substituting.