Ten small planes are flying at a speed of $150 \mathrm{~km} / \mathrm{h}$ in total darkness in an air space that is $20 \times 20 \times 1.5 \mathrm{~km}^3$ in volume. You are in one of the planes, flying at random within this space with no way of knowing where the other planes are. On the average about how long a time will elapse between near collision with your plane. Assume for this rough computation that a saftey region around the plane can be approximated by a sphere of radius 10 m .
The situation can be considered as the time of relaxation, based on kinetic theory of gases. Mean free path is the distance between two successive collisions, which we will consider here as the distance travelled by the plane before it just avoids the collision safe radius is equivalent to radius of the atom.
Hence, the required time
$$\begin{aligned} t=\frac{l}{v}, l=\text { mean free path } & =\frac{1}{\sqrt{2} \pi d^2 n}, n=\text { number density }=\frac{N}{V} \\ n & =\frac{\text { Number of aeroplanes }(\mathrm{N})}{\text { Volume }(\mathrm{V})} \\ & =\frac{10}{20 \times 20 \times 1.5}=0.0167 \mathrm{~km}^{-3} \\ t & =\frac{1}{\sqrt{2} \pi d^2(\mathrm{~N} / \mathrm{V})} \times \frac{1}{\mathrm{~V}} \quad \quad[\mathrm{~V}=\text { velocity of aeroplane }] \end{aligned}$$
By putting the given data,
$$\begin{aligned} t & =\frac{1}{\sqrt{2} \times 3.14 \times(20)^2 \times 0.0167 \times 10^{-6} \times 150} \\ & =\frac{10^6}{1776.25 \times 2.505} \\ & =\frac{10^6}{4449.5}=224.74 \mathrm{~h} \\ & \approx 225 \mathrm{~h} \end{aligned}$$
A box of $1.00 \mathrm{~m}^3$ is filled with nitrogen at 1.50 atm at 300 K . The box has a hole of an area $0.010 \mathrm{~mm}^2$. How much time is required for the pressure to reduce by 0.10 atm, if the pressure outside is 1 atm .
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 a rectangular block of wood moving with a velocity $v_0$ in a gas at temperature $T$ and mass density $\rho$. Assume the velocity is along $x$-axis and the area of cross-section of the block perpendicular to $v_0$ is $A$. Show that the drag force on the block is $4 r A v_0 \sqrt{\frac{k T}{m}}$, where, $m$ is the mass of the gas molecule.
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 $=(4 \rho A v) v_0=4 \rho A \sqrt{\frac{k_B T}{m}} v_0$.