Question

Oxygen is filled in a closed metal jar of volume 1.0 × 10⁻³ m³ at a pressure of 1.5 × 10⁵ Pa 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

$\begin{array}{l}\text{Here,}\\ V_1=1.0\times {10}^{-3}{m}^3\\ T_1=400K\\ P_1=1.5\times {10}^{5}Pa\\ P_2=1.0\times {10}^{5}Pa\\ T_2=300\\ M=32g\\ \text{Number of moles in the jar before}{n}_1\text{=}\frac{P_1{V}_1}{R{T}_1}\\ \text{Volume of the gas when pressure becomes equal to external pressure is given by}\\ \frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}\\ \Rightarrow V_2=\frac{P_1{V}_1{T}_2}{P_2{T}_1}\\ \Rightarrow V_2=\frac{1.5\times {10}^5\times 1.0\times {10}^{-3}\times 300}{1.0\times {10}^5\times 400}=1.125\times {10}^{-3}\\ \text{Net volume of leaked gas =}{V}_2-V_1\\ =1.125\times {10}^{-3}-1.0\times {10}^{-3}\\ =1.25\times {10}^{-4}{\text{m}}^3\\ {\text{Let n}}_2\text{be the number of moles of leaked gas.}\mathrm{Applying\; equation\; of\; state\; on\; this\; amount\; of\; gas,}\mathrm{we}\mathrm{get}\\ \\ n_2=\frac{P_2{V}_2}{R{T}_2}=\frac{1.0\times {10}^5\times 1.25\times {10}^{-4}}{8.3\times 300}=0.005\\ \text{Mass of leaked gas= 32}\times \text{0}\text{.005=0}\text{.16 g}\end{array}$