Pressure Of An Ideal Gas MCQ - Practice Questions with Answers

Consider an ideal gas (consisting of N molecules each of mass m) enclosed in a cubical box of side L as shown in the below figure.

1. Instantaneous velocity-

Any molecule of gas moves with velocity $\overrightarrow{v}$ in any direction where $\overrightarrow{v}=v_x\hat{i}+v_y\hat{j}+v_z\hat{k}$

And Due to the random motion of the molecule

$$\begin{array}{rl}& v_x=v_y=v_z\\ & \text{ As }v=\sqrt{v_x^2+v_y^2+v_z^2}\\ & \Rightarrow v=3v_x^2=3v_y^2=3v_z^2\end{array}$$

2. The time during a collision-Time between two successive collisions with the wall $A_1$

$$\begin{array}{rl}& \text{ I.e }\mathrm{\Delta }t=\frac{\text{ Distance travelled by molecule between two successive collision }}{\text{ Velocity of molecule }}\\ & \text{ or }\mathrm{\Delta }t=\frac{2L}{v_x}\end{array}$$

3. Collision frequency ( $\mathbf{n}$ ): It means the number of collisions per second.
I.e $n=\frac{1}{\mathrm{\Delta }t}=\frac{v_x}{2L}$
4. Change in momentum: This molecule collides with $A_1$ wall (A1) with velocity ${\vee }_x$ and rebounds with velocity $(-v_{\mathrm{x}})$
The change in momentum of the molecule is given by

$$\begin{array}{rl}& \mathrm{\Delta }p=(-m{v}_x)-\left(m{v}_x\right)=-2m{v}_x\\ & \mathrm{\Delta }{p}_{\text{system }}=0\\ & \mathrm{\Delta }{p}_{\text{system }}=\mathrm{\Delta }{p}_{\text{molecule }}+\mathrm{\Delta }{p}_{\text{wall }}=0\\ & \mathrm{\Delta }{p}_{\text{wall }}=-\mathrm{\Delta }{p}_{\text{molecule }}\end{array}$$

As the momentum remains conserved in a collision, the change in momentum of the wall $A_1$ is

$$\mathrm{\Delta }p=2m{v}_x$$

5. Force on the wall: Force exerted by a single molecule on the $A_1$ wall is equal to the rate at which the momentum is transferred to the wall by this molecule.
i.e. $F_{\text{Single molecule }}=\frac{\mathrm{\Delta }p}{\mathrm{\Delta }t}=\frac{2m{v}_x}{\left(2L/v_x\right)}=\frac{m{v}_x^2}{L}$

The total force on the wall $A_1$ due to $N$ molecules

$$F_x=\frac{m}{L}\sum v_x^2=\frac{m}{L}(v_{x_1}^2+v_{x2}^2+v_{x3}^2+\dots )=\frac{mN}{L}\overline{v_x^2}$$

where $\overline{v_x^2}=$ mean square of $x$ component of the velocity.
6. Pressure-As pressure is defined as force per unit area, hence the pressure on $A_1$ wall

$$\begin{array}{rl}& P_x=\frac{F_x}{A}=\frac{mN}{AL}\overline{v_x^2}=\frac{mN}{V}\overline{v_x^2}\\ & \text{ As }\overline{v_x^2}=\overline{v_y^2}=\overline{v_z^2}\\ & \text{ So }\overline{v^2}=\overline{v_x^2}+\overline{v_y^2}+\frac{v_z^2}{\overline{v^2}}\\ & \Rightarrow \overline{v_x^2}=\overline{v_y^2}=\overline{v_z^2}=\frac{m}{3}\end{array}$$

So Total pressure inside the container is given by 

$$P=\frac{1}{3}\frac{mN}{V}\overline{v^2}=\frac{1}{3}\frac{mN}{V}{v}_{ms}^2(\text{ where }{v}_{ms}=\sqrt{\overline{v^2}})$$

Using total mass= $\mathrm{M}=\mathrm{mN}$
Pressure due to an ideal gas is given as

$$P=\frac{1}{3}\rho v_{rms}^2=\frac{1}{3}\left(\frac{M}{V}\right)\cdot v_{rms}^2$$

where
$m=$ mass of one molecule
$N=$ Number of the molecule

$$\begin{array}{rl}& v_{rms}^2=\frac{v_1^2+v_2^2+\dots \dots \dots .}{n}\\ & v_{rms}=\text{ RMS velocity }\end{array}$$