An oxygen cylinder of volume 30 litre 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 After some oxygen is withdrawn from the cylinder, its gauge pressure drops to 11 atmospheric pressures at temperature 27. The mass of the oxygen withdrawn from the cylinder is nearly equal to: [Given, R = 100/12 J mol⁻¹ K⁻¹, and molecular mass of O₂ = 32, 1 atm pressure = 1.01 × 10⁵ N/m² ] (a) 0.116 kg (b) 0.156 kg (c) 0.125 kg (d) 0.144 kg

Answer: (a). Absolute pressure = gauge + 1 atm = 12 atm = 12 × 1.01 × 10⁵ = 1.212 × 10⁶ Pa. Moles left: n = PV/RT = (1.212 × 10⁶ × 0.030)/(100/12 × 300) = (36360)/(2500) = 14.54 mol Moles withdrawn = 18.20 - 14.54 = 3.66 mol Mass = 3.66 × 32 ≈ 117 g ≈ 0.116 kg So the answer is (a) 0.116 kg.

An oxygen cylinder of volume 30 litre has 18.20 moles of oxygen… | NEET 2025

Q25NEET 2025Medium

Gas Laws & Ideal Gas Equation

An oxygen cylinder of volume 30 litre has 18.20 moles of oxygen. After some oxygen is withdrawn from the cylinder, its gauge pressure drops to 11 atmospheric pressures at temperature $27^\circ\text{C}$ . The mass of the oxygen withdrawn from the cylinder is nearly equal to:

[Given, $R = \frac{100}{12}\text{ J mol}^{-1}\text{ K}^{-1}$ , and molecular mass of $\text{O}_2 = 32,\ 1$ atm pressure $= 1.01 \times 10^{5}\text{ N/m}^{2}$ ]

Show answer and solution

Answer: (a)

Absolute pressure = gauge + 1 atm $= 12$ atm $= 12 \times 1.01 \times 10^5 = 1.212 \times 10^6$ Pa.
Moles left: $n = \dfrac{PV}{RT} = \dfrac{1.212 \times 10^6 \times 0.030}{\frac{100}{12} \times 300} = \dfrac{36360}{2500} = 14.54$ mol
Moles withdrawn $= 18.20 - 14.54 = 3.66$ mol
Mass $= 3.66 \times 32 \approx 117$ g $\approx 0.116$ kg
So the answer is (a) 0.116 kg.

Report a mistake

Gas Laws & Ideal Gas Equation question from NEET 2025

More questions on Gas Laws & Ideal Gas Equation