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PhysicsNEET 2025Ideal gas equation and kinetic interpretation of pressure

Kinetic Theory

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 After some oxygen is withdrawn from the cylinder, its gauge pressure drops to 11 atmospheres at a temperature of 27∘27^\circC. The mass of the oxygen withdrawn from the cylinder is nearly equal to: [Given R=10012 J mol−1 K−1R = \dfrac{100}{12}\ \mathrm{J\,mol^{-1}\,K^{-1}}, molecular mass of O2\mathrm{O_2} = 32, 1 atm pressure =1.01×105 N/m2= 1.01 \times 10^{5}\ \mathrm{N/m^2}]

  1. A.0.125 kg
  2. B.0.144 kg
  3. C.0.116 kg✓
  4. D.0.156 kg

Solution

Final moles n=PVRT=12×1.01×105×0.030(100/12)×300≈14.5n = \frac{PV}{RT} = \frac{12 \times 1.01\times10^{5} \times 0.030}{(100/12) \times 300} \approx 14.5 mol (gauge 11 atm means absolute 12 atm). Withdrawn ≈18.2−14.5=3.7\approx 18.2 - 14.5 = 3.7 mol ≈0.116\approx 0.116 kg.

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