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.

Answer 1

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