Question

Oxygen is filled in a closed metal jar of volume $1.0\times {10}^{-3}{m}^3$ at a pressure of $1.5\times {10}^5$ Pa and temperature $400K$. The jar has a small leak in it. The atmospheric pressure is $1.0\times {10}^5$ Pa and the atmospheric temperature is $300K$. Find the mass of the gas that leaks out by the time the pressure and the temperature inside the jar equalize with the surrounding.

Answer 1

The volume of the metal jar is $V_1=1\times {10}^{-3}{m}^3$

The pressure of the gas is $P_1=1.5\times {10}^{5}Pa$

The temperature of the gas in container is $T_1=400k$

Using the ideal gas equation:

$P_1{V}_1=n_1{R}_1{T}_1$

$n_1=\frac{P_1{V}_1}{R_1{T}_1}=\frac{1.5\times {10}^5\times {10}^{-3}}{8.3\times 400}$

$n_1=\frac{1.5}{8.3\times 4}$

The mass of the gas present in container is:

$m_1=\frac{1.5}{8.3\times 4}\times m$

$=\frac{1.5}{8.3\times 4}\times 32=1.4457\cong 1.446$

After the gas is leaked,

$P_2=1\times {10}^{5}Pa,V_2=1\times {10}^{-3}{m}^3,T_2=300k$

Again using the ideal gas equation:

$P_2{V}_2=n_2{R}_2{T}_2$

$n_2=\frac{P_2{V}_2}{R_2{T}_2}$

$=\frac{{10}^5\times {10}^{-3}}{5.3\times 300}=\frac{1}{3\times 8.3}=0.040$

The mass present in after after leakage is:

$\therefore m_2=0.04\times 32=1.285$

The mass of the gas leaked from container is:

$\Delta m=m_1-m_2=1.441-1.285=0.160$

$\delta \cong 0.16g$