NEET 2025 — Physics solution

NEET2025Physics

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 After some oxygen is withdrawn from the cylinder, its gauge pressure drops to $11$ atmospheric pressure at temperature $2 7^{\circ}$ C. The mass of the oxygen withdrawn from the cylinder is nearly equal to:

[Given, $R = \frac{100}{12}$ J mol $^{- 1}$ K $^{- 1}$ , and molecular mass of O $_{2}$ = 32, 1 atm pressure $= 1.01 \times 1 0^{5}$ N/m $^{2}$ ]

A
0.125 kg
B
0.144 kg
C
0.116 kg
D
0.156 kg

Solution

Absolute pressure = gauge + atmospheric = $11 + 1 = 12$ atm.

Final moles from $P V = n R T$ : $n_{f} = \frac{P _{ab s} V}{R T} = \frac{12 \times 1.01 \times 1 0 ^{5} \times 30 \times 1 0 ^{- 3}}{( 100/12 ) \times 300} = \frac{36360}{2500} = 14.544$ mol.

Moles removed $= 18.20 - 14.544 = 3.656$ mol.

Mass removed $= 3.656 \times 32/1000 \approx 0.117$ kg, closest to $0.116$ kg.

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