NEET (UG) 2025 — Question 13

Use the ideal gas law $PV = nRT$. Initial conditions: $V = 30\,\text{L} = 30 \times 10^{-3}\,\text{m}^3$, $n_i = 18.20\,\text{mol}$, $T = 27^\circ\text{C} = 300\,\text{K}$. Final gauge pressure = 11 atm, so absolute pressure $P_f = 11 + 1 = 12\,\text{atm} = 12 \times 1.01 \times 10^5\,\text{Pa}$. Final number of moles: $n_f = \frac{P_f V}{RT} = \frac{12 \times 1.01 \times 10^5 \times 30 \times 10^{-3}}{8.314 \times 300}$. Calculate: numerator = $12 \times 1.01 \times 10^5 \times 0.030 = 36360$, denominator = $8.314 \times 300 = 2494.2$, so $n_f \approx 14.58\,\text{mol}$. Moles withdrawn: $\Delta n = n_i - n_f = 18.20 - 14.58 = 3.62\,\text{mol}$. Mass withdrawn: $\Delta m = \Delta n \times M = 3.62 \times 32\,\text{g/mol} = 115.84\,\text{g} = 0.116\,\text{kg}$. Thus the mass withdrawn is approximately 0.116 kg, matching option C.