$A$ box of $1 \ m^{3}$ is filled with nitrogen at $1.5 \ atm$ at $300 \ K$. The box has a hole of an area $0.010 \ mm^{2}$. How much time is required for the pressure to reduce by $0.10 \ atm$,if the pressure outside is $1 \ atm$?

(D) The rate of effusion of gas molecules through a small hole is given by the kinetic theory of gases. The number of molecules $dN$ escaping in time $dt$ is $dN = \frac{1}{4} n \langle v \rangle A dt$,where $n$ is the number density,$\langle v \rangle$ is the average speed,and $A$ is the area of the hole.
Using $PV = N k_B T$,we have $n = \frac{N}{V} = \frac{P}{k_B T}$.
The average speed $\langle v \rangle = \sqrt{\frac{8 k_B T}{\pi m}}$.
The rate of change of pressure is $\frac{dP}{dt} = -\frac{1}{4} \frac{P}{k_B T} \sqrt{\frac{8 k_B T}{\pi m}} \frac{k_B T}{V} A = -\frac{A}{V} \sqrt{\frac{k_B T}{2 \pi m}} P$.
Integrating this from $P_i = 1.5 \ atm$ to $P_f = 1.4 \ atm$ (since pressure reduces by $0.10 \ atm$):
$\int_{P_i}^{P_f} \frac{dP}{P} = -\frac{A}{V} \sqrt{\frac{k_B T}{2 \pi m}} \int_0^t dt$.
$\ln\left(\frac{P_f}{P_i}\right) = -\frac{A}{V} \sqrt{\frac{k_B T}{2 \pi m}} t$.
Substituting the values $m = \frac{28 \times 10^{-3}}{6.022 \times 10^{23}} \ kg$,$T = 300 \ K$,$A = 10^{-8} \ m^2$,$V = 1 \ m^3$,we can solve for $t$.