Given, volume of the box, $V=1.00 \mathrm{~m}^3$

$$\begin{aligned} \text { Area } & =a=0.010 \mathrm{~mm}^2 \\ = & 8.01 \times 10^{-6} \mathrm{~m}^2 \\ & =10^{-8} \mathrm{~m}^2 \end{aligned}$$

Temperature outside $=$ Temperature inside

Initial pressure inside the box $=1.50 \mathrm{~atm}$.

Final pressure inside the box $=0.10 \mathrm{~atm}$.

Assuming,

$v_{i x}=$ Speed of nitrogen molecule inside the box along $x$-direction.

$n_i=$ Number of molecules per unit volume in a time interval of $\Delta T$, all the particles at a distance $\left(v_{i x} \Delta t\right)$ will collide the hole and the wall, the particle colliding along the hole will escape out reducing the pressure in the box.

Let area of the wall, number of particles colliding in time

$$\Delta t=\frac{1}{2} n_i\left(v_{i x} \Delta t\right) A$$

$\frac{1}{2}$ is the factor because all the particles along $x$-direction are behaving randomly. Hence, half of these are colliding against the walls on either side.

Inside the box,

$$\begin{aligned} & v_{i x}^2+v_{i y}^2+v_{i z}^2=v_{\text {rms }}^2 \\ \therefore \quad & v_{i x}^2=\frac{v_{\text {rms }}^2}{3} \quad \left(\because v_{i x}=v_{i y}=v_{i z}\right) \end{aligned}$$

$$\text { or } \quad \frac{1}{2} m v_{\mathrm{rms}}^2=\frac{3}{2} k_B T \quad\left[\begin{array}{rl} V_{\mathrm{rms}} & =\text { Root mean square velocity } \\ K_B & =\text { Boltzmann constant } \\ T & =\text { Temperature } \end{array}\right]$$

$$\begin{aligned} &\begin{aligned} & v_{\mathrm{rms}}^2=\frac{3 k_B T}{m} \\ & v_{\mathrm{rms}}=\sqrt{\frac{3 k_B T}{m}} \end{aligned}\\ &\text { [According to kinetic theory of gases] } \end{aligned}$$

Now, $$v_{i x}^2=\frac{v_{\mathrm{rms}}^2}{3}=\frac{1}{3} \times \frac{3 k_B T}{m}$$

or $$v_{i x}^2=\frac{k_B T}{m}$$

$\therefore$ Number of particles colliding in time

$$\Delta t=\frac{1}{2} n_i \sqrt{\frac{k_B T}{m}} \Delta t A$$

If particles collide along hole, they move out. Similarly, outer particles colliding along hole will move in.

If $a=$ area of hole

Then, net particle flow in time $\Delta t=\frac{1}{2}\left(n_1-n_2\right) \sqrt{\frac{k_B T}{m}} \Delta t a$

[Temperatures inside and outside the box are equal]

$$p V=\mu R T \Rightarrow \mu=\frac{p V}{R T}$$

$$\text { Let } n=\text { number density of nitrogen }=\frac{\mu N_A}{V}=\frac{p N_A}{R T} \quad\left[\because \frac{\mu}{V}=\frac{p}{R T}\right]$$

Let $N_A=$ Avogardro's number

If after time $\tau$ pressure inside changes from $p$ to $p_1{ }^1$

$$\therefore \quad n_1^{\prime}=\frac{v N_A}{R T}$$

$$\begin{aligned} \text { Now, number of molecules gone out } & =n_1 V-n_1^{\prime} V \\ & =\frac{1}{2}\left(n_1-n_2\right) \sqrt{\frac{k_B T}{m}} \tau a \end{aligned}$$

$$\therefore \quad \frac{p_1 N_A}{R T} V-\frac{V N_A}{R T} V=\frac{1}{2}\left(p_1-p_2\right) \frac{N_A}{R T} \sqrt{\frac{k_B T}{m}} \tau a$$

$$ \begin{aligned} \text{or}\quad \frac{p_1 N_A}{R T} V-\frac{v N_A}{R T} V & =\frac{1}{2}\left(p_1-p_2\right) \frac{N_A}{R T} \sqrt{\frac{k_B T}{m}} \tau a \\ \therefore \quad \tau & =2\left(\frac{p_1-p v_1}{p_1-p_2}\right) \frac{V}{a} \sqrt{\frac{m}{k_B T}} \end{aligned}$$

Putting the values from the data given,

$$\begin{aligned} \tau & =2\left(\frac{1.5-1.4}{1.5-1.0}\right) \frac{1 \times 1.00}{0.01 \times 10^{-6}} \sqrt{\frac{46.7 \times 10^{-27}}{1.38 \times 10^{-23} \times 300}} \\ & =2\left(\frac{0.1}{0.5}\right) \frac{1}{10^{-8}} \sqrt{\frac{4.7}{1.38 \times 3} \times 10^{-6}} \\ & =2\left(\frac{1}{5}\right) 1 \times 10^8 \times 10^{-3} \times \sqrt{\frac{46.7}{4.14}}=\frac{2}{5} \times 10^5 \sqrt{\frac{45.7}{4.14}} \\ & =\frac{2}{5} \times 10^5 \sqrt{11.28} \\ & =\frac{2}{5} \times 3.358 \times 10^5=\frac{6.717}{5} \times 10^5=1.343 \times 10^5 \mathrm{~s} \end{aligned}$$

Consider the diagram

Let $n=$ number of molecules per unit volume

$v_{\mathrm{rms}}=\mathrm{rms}$ speed of the gas molecules

When block is moving with speed $v_o$, relative speed of molecules w.r.t. front face $=v+v_o$

Coming head on, momentum transferred to block per collision $=2 m\left(v+v_o\right)$, where, $m=$ mass of molecule.

Number of collission in time $\Delta t=\frac{1}{2}\left(v+v_o\right) n \Delta t A$, where, $A=$ area of cross-section of block and factor of $1 / 2$ appears due to particles moving towards block.

$\therefore$ Momentum transferred in time $\Delta t=m\left(v+v_0\right)^2 n A \Delta t$ from front surface.

Similarly, momentum transferred in time $\Delta t=m\left(v-v_0\right)^2 n A \Delta t$ (from back surface)

$\therefore$ Net force

$$\begin{aligned} (\text { drag force }) & =m n A\left[\left(v+v_0\right)^2-\left(v-v_0\right)^2\right] \quad \text{[from front]}\\ & =m n A\left(4 \vee v_0\right)=(4 m n A v) v_0 \\ & =(4 \rho A v) v_0 \quad \text{... (i)}\\ \text { where we have assumed} \quad \rho & =\frac{m n}{V}=\frac{M}{V} \end{aligned}$$

If $v=$ velocity along $x$-axis

Then, we can write

$$\mathrm{KE}=\frac{1}{2} m v^2=\frac{1}{2} k_B T

$$\Rightarrow \quad v=\sqrt{\frac{k_B T}{m}} \quad \left[\begin{array}{rl} K_B & =\text { Boltzmann constant } \\ K E & =\text { Kinetic energy } \\ T & =\text { Temperature } \end{array}\right]$$

$\therefore$ From Eq. (i), $\quad$ Drag force $=(4 \rho A v) v_0=4 \rho A \sqrt{\frac{k_B T}{m}} v_0$.