Subject: Physics

A gas consists of a large number of molecular which are in continuous random motion. These molecules are far apart from each other that their average separation is much greater than the diameter of each molecule. Even though molecules posses continuous random motion and moves freely, the velocity of molecules changes due to intermolecular forces between them. Such changes may be considered as collisions at a molecular level.

**Postulates**

- All gases consist of molecules and the molecules of a gas are all alike but differ with the molecules of other gases.
- The volume of molecules is negligible compared to the volume occupied by the gas.
- During the motion, the molecules collide with each other and to the walls of the containing vessel. The collision of molecules is perfectly elastic.
- The molecules are in random motion and the velocities in all directions are ranging from zero to a maximum.
- The molecules behave as a perfectly elastic sphere.
- The duration of a collision between two molecules is negligible compared to the time between two successive collisions on the wall.
- The attractive or repulsive force between the molecules is negligible, and no force is exerted by the molecules.

Let us consider a cubical vessel of length ‘l’ containing ‘n’ number of gas molecule. Each of mass ‘m’. let C_{1} , C_{2}, …….C_{n} be the velocity of each gas molecule in any direction. When we resolve their velocities along X, Y and Z direction. Then we can write

$$ C_1^2 = U_1^2 + V_1^2 + W_1^2 $$

$$C_2^2 = U_2^2 + V_2^2 + W_2^2$$

$$\vdots = \vdots + \vdots+ \vdots $$

$$C_n^2 = U_n^2 + V_n^2 + W_n^2 $$

The first molecule moves with velocity u_{1} along x-axis and collides on the face of the wall of the vessel in x-axis with change in momentum \(= mu_1 = m(-u_1) = 2mu_1\)

Time between two successive collision

$$Time\;taken = \frac {\text{total distance}}{\text{velocity}}$$

$$= \frac{2l}{u_1}$$

\(\therefore\)force exerted by a gas molecule when moving along x-axis is given by the rate of change in momentum \(=\frac{\text{change in momentum}}{\text{time taken}}\)

$$= \frac{2mu_1}{\frac{2l}{u_1}}$$

$$= \frac{mu_1^2l}{l}$$

\(\therefore\)force exerted by all small 'n' molecules moving along x-axis is given by

$$F_x = \frac{mu_1^2l}{l} + \frac{mu_2^2l}{l} + \frac{mu_3^2l}{l} + \dots +\frac{mu_n^2l}{l}$$

So. total pressure exerted by all gas molecules moving along x-axis is only given by

$$p_x = \frac{F_n}{A}$$

$$p_x= \frac{\frac{mu_1^2l}{l} + \frac{mu_2^2l}{l} + \frac{mu_3^2l}{l} + \dots +\frac{mu_n^2l}{l}}{l^2}$$

$${\frac{m}{l^3}(u_1^2 + u_2^2 + u_3^2 +\dots + u_n^2 )}$$

Similarly, the total pressure exerted by all gas molecules moving along y and z-axis is given by

$$p_y= {\frac{m}{l^3}(v_1^2 + v_2^2 +v_3^2 +\dots + v_n^2 )}$$

$$p_z= {\frac{m}{l^3}(w_1^2 + w_2^2 + w_3^2 +\dots + w_n^2 )}$$

\(\therefore\) The average pressure exerted by all gas molecules inside the cubical vessel is

$$p = \frac{p_x +p_y + p_z}{3}$$

$$= \frac{\frac{m}{l^3}(u_1^2 + u_2^2 + u_3^2 +\dots + u_n^2 )(v_1^2 + v_2^2 +v_3^2 +\dots + v_n^2 )(w_1^2 + w_2^2 + w_3^2 +\dots + w_n^2 )}{3}$$

$$= \frac{m}{l^3}(u_1^2 + u_2^2 + u_3^2 +\dots + u_n^2 )(v_1^2 + v_2^2 +v_3^2 +\dots + v_n^2)(w_1^2 + w_2^2 + w_3^2 +\dots + w_n^2 ) $$

we know,

$$p= \frac{m}{l^3}(c_1^2 + c_2^2 + c_3^2 +\dots + c_n^2)\dots(i)$$

If \(\bar c^2\) be the mean square root of the gas molecules then,

$$\bar c^2 =\frac {c_1^2 + c_2^2 + c_3^2 +\dots + c_n^2} {n})$$

$$\therefore c_1^2 + c_2^2 + c_3^2 +\dots + c_n^2 = n\bar c^2 $$

So, equation (ii) becomes

$$p = \frac{m}{3l^3} nc^2$$

$$p= \frac{mn\bar c^2}{3l^3} $$

$$p= \frac{M\bar c^2}{3V} \dots (ii)$$

where M = mn is the total mass of all gas molecules inside the cubical vessel and l^{3} = V is the total volume of the vessel.

$$\therefore p = \frac{1}{3}\rho \bar c^2(\because \rho = \frac{m}{V})$$

which is the required expression of average pressure exerted by all gas molecules placed inside the cubical vessel.

From equation (ii) we have

$$ p = \frac{1}{3}\frac{M}{V} \bar c^2 =\frac{2}{3}\frac{1}{2}\frac{M\bar c^2}{V} $$

$$p = \frac{2}{3}E$$

where E is the total kinetic energy of the gas molecules per unit volume.

$$\therefore E = \frac{3}{2}p$$

This shows pressure exerted by gas molecule is two third of the kinetic energy.

A gas consists of a large number of molecular which are in continuous random motion.

All gases consist of molecules and the molecules of a gas are all alike but differ with the molecules of other gases.

The molecules are in random motion and the velocities in all directions are ranging from zero to a maximum.

The duration of a collision between two molecules is negligible compared to the time between two successive collisions on the wall.

The attractive or repulsive force between the molecules is negligible, and no force is exerted by the molecules.

Pressure exerted by gas molecule is two third of the kinetic energy.

- It includes every relationship which established among the people.
- There can be more than one community in a society. Community smaller than society.
- It is a network of social relationships which cannot see or touched.
- common interests and common objectives are not necessary for society.

© 2019-20 Kullabs. All Rights Reserved.