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 $\vec{v}$ in any direction where $\vec{v}=v_x \hat{i}+v_y \hat{j}+v_z \hat{k}$

And Due to the random motion of the molecule

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

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

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

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

$$
\begin{aligned}
& \Delta p=\left(-m v_x\right)-\left(m v_x\right)=-2 m v_x \\
& \Delta p_{\text {system }}=0 \\
& \Delta p_{\text {system }}=\Delta p_{\text {molecule }}+\Delta p_{\text {wall }}=0 \\
& \Delta p_{\text {wall }}=-\Delta p_{\text {molecule }}
\end{aligned}
$$

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

$$
\Delta p=2 m v_x
$$

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. $F_{\text {Single molecule }}=\frac{\Delta p}{\Delta t}=\frac{2 m v_x}{\left(2 L / v_x\right)}=\frac{m v_x^2}{L}$

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

$$
F_x=\frac{m}{L} \sum v_x^2=\frac{m}{L}\left(v_{x_1}^2+v_{x 2}^2+v_{x 3}^2+\ldots\right)=\frac{m N}{L} \overline{v_x^2}
$$

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

$$
\begin{aligned}
& P_x=\frac{F_x}{A}=\frac{m N}{A L} \overline{v_x^2}=\frac{m N}{V} \overline{v_x^2} \\
& \text { As } \overline{v_x^2}=\overline{v_y^2}=\overline{v_z^2} \\
& \text { So } \overline{v^2}=\overline{v_x^2}+\overline{v_y^2}+\frac{v_z^2}{\overline{v^2}} \\
& \Rightarrow \overline{v_x^2}=\overline{v_y^2}=\overline{v_z^2}=\frac{m}{3}
\end{aligned}
$$

So Total pressure inside the container is given by 

$$
P=\frac{1}{3} \frac{m N}{V} \overline{v^2}=\frac{1}{3} \frac{m N}{V} v_{m s}^2 \quad\left(\text { where } v_{m s}=\sqrt{\overline{v^2}}\right)
$$

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

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

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

$$
\begin{aligned}
& v_{r m s}^2=\frac{v_1^2+v_2^2+\ldots \ldots \ldots .}{n} \\
& v_{r m s}=\text { RMS velocity }
\end{aligned}
$$