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 After some oxygen is withdrawn from the cylinder, its gauge pressure drops to $11$ atmospheric pressure at temperature $27°$C. The mass of the oxygen withdrawn from the cylinder is nearly equal to :
[Given, $R = \dfrac{100}{12}\ \text{J mol}^{-1}\text{K}^{-1}$, and molecular mass of $O_2 = 32$, $1$ atm pressure $= 1.01 \times 10^5\ \text{N/m}^2$]
Answer: (C) $0.116$ kg
Concept. Absolute pressure $=$ gauge $+$ atmospheric; $n = \dfrac{PV}{RT}$ gives the moles left.
Step 1. Pressure. $P = 12$ atm $= 12.12 \times 10^5$ Pa.
Step 2. Moles left. $V = 30 \times 10^{-3}\ \text{m}^3$, $T = 300$ K:
$$n_2 = \frac{12.12 \times 10^5 \times 30 \times 10^{-3}}{\frac{100}{12} \times 300} = \frac{36360}{2500} = 14.54\ \text{mol}$$
Step 3. Mass withdrawn. $(18.20 - 14.54) \times 32 \approx 117$ g $\approx 0.116$ kg.
Answer. $0.116$ kg (Option C).
Solution by Sreeraj P, M.Sc Physics