Solution 22: Rate of Gas Leakage
Concept:
The rate of leakage (effusion) $R$ is proportional to the product of number density and average velocity. $$ R \propto \left( \frac{N}{V} \right) \bar{v} $$ Using $P = \frac{N}{V}kT$ and $\bar{v} \propto \sqrt{T}$: $$ R \propto \left( \frac{P}{T} \right) \sqrt{T} \implies R \propto \frac{P}{\sqrt{T}} $$
The rate of leakage (effusion) $R$ is proportional to the product of number density and average velocity. $$ R \propto \left( \frac{N}{V} \right) \bar{v} $$ Using $P = \frac{N}{V}kT$ and $\bar{v} \propto \sqrt{T}$: $$ R \propto \left( \frac{P}{T} \right) \sqrt{T} \implies R \propto \frac{P}{\sqrt{T}} $$
Calculation:
Given: Final Pressure $P_2 = 8 P_1$ Final Temperature $T_2 = 4 T_1$ $$ \frac{R_2}{R_1} = \frac{P_2 / \sqrt{T_2}}{P_1 / \sqrt{T_1}} $$ $$ \frac{R_2}{R_1} = \left( \frac{P_2}{P_1} \right) \sqrt{\frac{T_1}{T_2}} $$ $$ \frac{R_2}{R_1} = (8) \sqrt{\frac{1}{4}} = 8 \times \frac{1}{2} = 4 $$ Answer: The rate of leakage increases by 4 times.
Given: Final Pressure $P_2 = 8 P_1$ Final Temperature $T_2 = 4 T_1$ $$ \frac{R_2}{R_1} = \frac{P_2 / \sqrt{T_2}}{P_1 / \sqrt{T_1}} $$ $$ \frac{R_2}{R_1} = \left( \frac{P_2}{P_1} \right) \sqrt{\frac{T_1}{T_2}} $$ $$ \frac{R_2}{R_1} = (8) \sqrt{\frac{1}{4}} = 8 \times \frac{1}{2} = 4 $$ Answer: The rate of leakage increases by 4 times.