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Question
Oxygen is filled in a closed metal jar of volume 1.0 × 10−3 m3 at a pressure of 1.5 × 105Pa and temperature 400 K. The jar has a small leak in it. The atmospheric pressure is 1.0 × 105 Pa and the atmospheric temperature is 300 K. Find the mass of the gas that leaks out by the time the pressure and the temperature inside the jar equalise with the surrounding.
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Solution
Here,
V1 = 1.0 × 10-3 m3
T1 = 400K
P1 = 1.5 × 105 Pa
P2 = 1.0 × 105 Pa
T2 = 300
M = 32 g
Number of moles in the jar before \[n_1 =\frac{P_1 V_1}{R T_1} \]
Volume of the gas when pressure becomes equal to external pressure is given by
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
\[ \Rightarrow V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} \]
\[ \Rightarrow V_2 = \frac{1.5 \times {10}^5 \times 1.0 \times {10}^{-3} \times 300}{1.0 \times {10}^5 \times 400} = 1.125 \times {10}^{-3 }\]
Net volume of leaked gas = V2 - V1
= 1.125 × 10-3 - 1.0 × 10-3
= 1.25 × 10-4 m3
Let n2 be the number of moles of leaked gas. Applying equation of state on this amount of gas, we get
\[ n_2 = \frac{P_2 V_2}{R T_2} = \frac{1.0 \times {10}^5 \times 1.25 \times {10}^{-4}}{8.3 \times 300} = 0.005 \]
Mass of leaked gas = 32 × 0.005 = 0.16 g
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