An oxygen cylinder of volume 30 litres has 18.20 moles of oxygen. After some oxygen is withdrawn fro

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Question

An oxygen cylinder of volume 30 litres has 18.20 moles of oxygen. After some oxygen is withdrawn from the cylinder, its gauge pressure drops to 11 atmospheric pressures at temperature 27°C. The mass of the oxygen withdrawn from the cylinder is nearly equal to: [Given R = 100/12 J mol⁻¹ K⁻¹, molecular mass of O₂ = 32, 1 atm = 1.01 × 10⁵ N/m²]

Accepted Answer

Correct Answer: D — 0.116 kgStep-by-step Solution:After withdrawal: Gauge pressure = 11 atm → Absolute pressure = 11 + 1 = 12 atmP = 12 × 1.01 × 10⁵ N/m² = 12.12 × 10⁵ N/m²V = 30 L = 30 × 10⁻³ m³, T = 27°C = 300 K, R = 100/12 J mol⁻¹ K⁻¹Number of moles remaining: n = PV/RTn = (12 × 1.01 × 10⁵ × 30 × 10⁻³) / [(100/12) × 300]n = (12 × 1.01 × 10⁵ × 30 × 10⁻³ × 12) / (100 × 300)n = (12 × 1.01 × 12 × 30) / (100 × 300 × 10³/10³)n = (12 × 1.01 × 12) / (1000) = 145.44/1000 × 30 ... simplifying:n = (12.12 × 10⁵ × 0.03) / (8.33 × 300) = 36360 / 2499 ≈ 14.55 molesMoles withdrawn = 18.20 − 14.55 = 3.65 molesMass withdrawn = 3.65 × 32 = 116.8 g ≈ 0.116 kgNote on gauge vs absolute pressure: Gauge pressure = absolute pressure − atmospheric pressure. The cylinder's gauge pressure is 11 atm, so absolute pressure = 12 atm.Key Concepts from Kinetic Theory of Gases (NCERT Class 11, Chapter 13):Ideal Gas Law: PV = nRT. Always use absolute pressure (not gauge pressure) and temperature in Kelvin.Gauge pressure: Pgauge = Pabsolute − Patmospheric. Atmospheric pressure ≈ 1 atm = 1.01 × 10⁵ Pa.Number of moles n = PV/RT. Moles withdrawn = initial moles − final moles.Mass of gas = moles × molar mass.

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