Name: Kinetic Theory and Critical Point
Uploaded: 2019-07-19
Duration: 9 min 8 s
Description: It is the set of conditions under which a liquid and its vapour become identical. Let's watch the video to learn more about this concept.

Question

Derive the expression of pressure exerted by the gas on the walls of the container.

Accepted Answer

Expression for pressure exerted by a gas: Consider a mono-atomic gas of $N$ molecules each having a mass $m$ inside a cubical container of side $l$ .
The molecules of the gas are in random motion. They collide with each other and also with the walls of the container. As the collisions are elastic in nature, there is no loss of energy, but a change in momentum occurs.

The molecules of the gas exert pressure on the walls of the container due to collision on it. During each collision, the molecules impart certain momentum to the wall. Due to transfer of momentum, the walls experience a continuous force. The force experienced per unit area of the walls of the container determines the pressure exerted by the gas. It is essential to determine the total momentum transferred by the molecules in a short interval of time.
A molecule of mass m moving with a velocity $v$ having components $(v_{x}, v_{y}, v_{z})$ hits the right side wall. Since we have assumed that the collision is elastic, the particle rebounds with same speed and its $x - c o m p o n e n t$ is reversed. The components of velocity of the molecule after collision are $(- v_{x}, v_{y}, v_{z})$ .

The $x - c o m p o n e n t$ of momentum of the molecule before collision = $m v_{x}$
The $x - c o m p o n e n t$ of momentum of the molecule after collision = $- m v_{x}$
$T h e c h a n g e i n m o m e n t u m o f t h e m o l e c u l e i n x d i r e c t i o n; F i n a l m o m e n t u m - I n i t i a l m o m e n t u m = - m v_{x} - m v_{x} = - 2 m v_{x} A c c o r d i n g t o l a w o f c o n s e r v a t i o n o f l i n e a r m o m e n t u m, t h e c h a n g e i n m o m e n t u m o f t h e w a l l = 2 m v_{x}$
The number of molecules hitting the right side wall in a small interval of time $∆ t$ .
The molecules within the distance of $v_{x} ∆ t$ from the right side wall and moving towards’ the right will hit the wall in the time interval $∆ t$ . The number of molecules that will hit the right side wall in a time interval, At is equal to the product of volume $(A v_{x} ∆ t)$ and number density of the molecules (n). Here $A$ is area of the wall and $n$ is number of molecules per unit volume $N / V$ .

We have assumed that the number density is the same throughout the cube.
Not all the $n$ molecules will move to the right, therefore on an average only half of the n molecules move to the right and the other half moves towards left side.
The number of molecules that hit the right side wall in a time interval $∆ t$
$= \frac{n}{2} A v_{x} ∆ t . . . . . (1)$
In the same interval of time $∆ t$ , the total momentum transferred by the molecules
$∆ P = \frac{n}{2} A v_{x} ∆ t \times 2 m v_{x} = A v_{x}^{2} m n ∆ t . . . . . (2)$
From Newton’s second law, the change in momentum in a small interval of time gives rise to force.
The force exerted by the molecules on the wall (in magnitude)
$F = \frac{∆ P}{∆ t} = n m A v_{x}^{2} . . . . . (3)$
Pressure, P = force divided by the area of the wall
$P = \frac{F}{A} = n m v_{x}^{2} . . . . . (4)$

Since all the molecules are moving completely in random manner, they do not have same speed. So we can replace the term $v_{x}^{2}$ by the average $\bar{v_{x}^{2}}$ in equation (4).
$P = n m \bar{v_{x}^{2}} . . . . . (5)$
Since the gas is assumed to move in random direction, it has no preferred direction of motion (the effect of gravity on the molecules is neglected). It implies that the molecule has same average speed in all the three direction. So, $\bar{v_{x}^{2}}, \bar{v_{y}^{2}}, \bar{v_{z}^{2}}$ The mean square speed is written as

$\bar{v^{2}} = \bar{v_{x}^{2}} + \bar{v_{y}^{2}} + \bar{v_{z}^{2}} = 3 \bar{v_{x}^{2}} \bar{v_{x}^{2}} = \frac{1}{3} \bar{v^{2}} Using this in equation (5), we get P = \frac{1}{3} m n \bar{v^{2}} o r P = \frac{1}{3} \frac{N}{V} m \bar{v^{2}} . . . . . [a s n = \frac{N}{V}]$

Derive the expression of pressure exerted by the gas on the walls of the container.

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