Question

An ideal gas is trapped in a stationary container with the help of a piston and there will be a small hole in the container through which the gas leaks. If by heating the gas and simultaneously pushing the piston in to the container, temperature and pressure of the gas is increased to 4 times and 8 times respectively of their initial values, how many times does the rate of leakage of the gas (i.e number of molecules leaking per second) change?

Answer 1

4 times

Answer 2

The rate of gas leakage is proportional to the flux of molecules incident on the hole, which is given by $J = \frac{1}{4} n \bar{v}$. Using the ideal gas law ($PV = NkT$), the number density is $n = \frac{P}{kT}$. The average speed of gas molecules is $\bar{v} = \sqrt{\frac{8kT}{\pi m}}$. Substituting these into the expression for flux: $J = \frac{1}{4} \left(\frac{P}{kT}\right) \sqrt{\frac{8kT}{\pi m}} = \frac{P}{\sqrt{kT}} \sqrt{\frac{8}{16\pi m}} = \frac{P}{\sqrt{kT}} \sqrt{\frac{1}{2\pi m}}$. The rate of leakage, $\Gamma$, is proportional to the flux $J$ and the area of the hole $A$. Since $A$ is constant, $\Gamma \propto J$. Therefore, $\Gamma \propto \frac{P}{\sqrt{kT}}$. For a given gas, $k$ and $m$ are constants, so $\Gamma \propto \frac{P}{\sqrt{T}}$. Let the initial values be $P_1, T_1$ and the final values be $P_2, T_2$. Given $T_2 = 4T_1$ and $P_2 = 8P_1$. The ratio of the final rate of leakage to the initial rate is: $\frac{\Gamma_2}{\Gamma_1} = \frac{P_2 / \sqrt{T_2}}{P_1 / \sqrt{T_1}} = \frac{P_2}{P_1} \sqrt{\frac{T_1}{T_2}} = (8) \sqrt{\frac{T_1}{4T_1}} = 8 \sqrt{\frac{1}{4}} = 8 \times \frac{1}{2} = 4$ The rate of leakage changes by 4 times.