Question: 14. 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 After some oxygen is withdrawn from the cylinder, its gauge pressure drops to 11 atmospheric pressure at temperature 27^{\circ} \mathrm{C}. The mass of the oxygen withdrawn from the cylinder is nearly equal to :
(1) 0.125 kg
(2) 0.144 kg
(3) 0.116 kg
(4) 0.156 kg
Answer: Option (3)
Explanation:
Gauge pressure of 11 atmospheric means pressure above atmosphere is 11\,\text{atm},
so the absolute final pressure is
P_f=(11+1)\,\text{atm}=12\,\text{atm}Convert this to SI units using
1\,\text{atm}=1.01\times 10^{5}\,\text{N/m}^2:
P_f=12\times 1.01\times 10^{5}\,\text{Pa}Given volume and temperature:
V=30\,\text{L}=30\times 10^{-3}\,\text{m}^3,\qquad T=27^{\circ}\text{C}=300\,\text{K}Use ideal gas law to find number of moles remaining after withdrawal:
n_f=\dfrac{P_f V}{R T}Substitute R=\dfrac{100}{12}\,\text{J mol}^{-1}\text{K}^{-1} and the values above:
n_f=\dfrac{(12\times 1.01\times 10^{5})(30\times 10^{-3})}{\dfrac{100}{12}\times 300}Calculate n_f
(showing intermediate simplification):
n_f=\dfrac{12\times 1.01\times 10^{5}\times 30\times 10^{-3}}{\tfrac{100}{12}\times 300} =\dfrac{12\times 1.01\times 10^{5}\times 0.03}{\tfrac{100}{12}\times 300} \approx 14.544\ \text{mol}Initial moles were n_i=18.20, so moles withdrawn are
\Delta n=n_i-n_f=18.20-14.544\approx 3.656\ \text{mol}Molar mass of
O_2 is 32\ \text{g mol}^{-1} = 0.032\ \text{kg mol}^{-1}.
Mass withdrawn is
\Delta m=\Delta n\times 0.032\ \text{kg mol}^{-1}=3.656\times 0.032\approx 0.11699\ \text{kg}\approx 0.116\ \text{kg}Therefore the mass withdrawn is approximately 0.116\ \text{kg},
matching Option (3).