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Kinetic Theory
Ideal Gas Equation
1 question from 1 paper, each with its answer.
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Concepts tested
- Gauge Pressure and Gas Moles
All Questions
2025 1 question
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An oxygen cylinder of volume $30\text{ litre}$ has $18.20\text{ moles}$ of oxygen. After some oxygen is withdrawn from the cylinder, its gauge pressure drops to $11\text{ atmospheric pressure}$ at temperature $27^\circ\text{C}$. The mass of the oxygen withdrawn from the cylinder is nearly equal to:
[Given: $R = \frac{100}{12}\text{ J mol}^{-1}\text{K}^{-1}$, molecular mass of $O_2 = 32$, $1\text{ atm pressure} = 1.01 \times 10^5\text{ N/m}^2$]Absolute pressure is $P_{abs} = 11 + 1 = 12\text{ atm}$. Moles remaining $\approx 14.55\text{ mol}$. Moles withdrawn $\Delta n = 3.65\text{ mol}$, corresponding to mass $3.65 \times 32\text{ g} \approx 0.116\text{ kg}$.
Absolute pressure inside cylinder: $P_f = P_{gauge} + P_{atm} = 11 + 1 = 12\text{ atm} = 12 \times 1.01 \times 10^5\text{ Pa}$. Volume $V = 30\text{ L} = 30 \times 10^{-3}\text{ m}^3$, temperature $T = 27 + 273 = 300\text{ K}$. Moles remaining: $n_f = \frac{P_f V}{R T} = \frac{12 \times 1.01 \times 10^5 \times 0.03}{(100/12) \times 300} = \frac{36360}{2500} \approx 14.55\text{ mol}$. Moles withdrawn: $\Delta n = n_i - n_f = 18.20 - 14.55 = 3.65\text{ mol}$. Mass of oxygen withdrawn: $m = \Delta n \times M_{O_2} = 3.65 \times 32\text{ g} = 116.8\text{ g} \approx 0.116\text{ kg}$.