Pressure Exerted by a Gas

Postulates of Kinetic theory of gases

(a) A gas consists of a very large number of molecules. Each one is a perfectly identical elastic sphere.

(b) The molecules of a gas are in a state of continuous and random motion. They move in all directions with all possible velocities.

(c) The size of each molecule is very small as compared to the distance between them. Hence, the volume occupied by the molecule is negligible in comparison to the volume of the gas.                  

(d) There is no force of attraction or repulsion between the molecules and the walls of the container.

(e) The collisions of the molecules among themselves and with the walls of the container are perfectly elastic. Therefore, momentum and kinetic energy of the molecules are conserved during collisions.

(f) A molecule moves along a straight line between two successive collisions and the average distance travelled between two successive collisions is called the mean free path of the molecules.

(g) The collisions are almost instantaneous (i.e) the time of collision of two molecules is negligible as compared to the time interval between two successive collisions.

Avogadro number

Avogadro number is defined as the number of molecules present in one mole of a substance. It is constant for all the substances. Its value is 6.023 × 10²³.

Pressure exerted by a gas

The molecules of a gas are in a state of random motion. They continuously collide against the walls of the container. During each collision, momentum is transfered to the walls of the container. The pressure exerted by the gas is due to the continuous  collision  of  the  molecules against the walls of the container. Due to this  continuous  collision, the  walls experience a continuous force which is equal to the total momentum imparted to the walls per second. The force experienced per unit area of the walls of the container determines the pressure exerted by the gas. Consider a cubic container of side containing n molecules of perfect gas moving with velocities C₁, C₂, C₃  ... C_n as shown in figure. A molecule moving with a velocity C₁, will have velocities u₁, v₁ and w₁ as components along the x, y and z axes respectively. Similarly u₂, v₂ and w₂ are the velocity components of the second molecule and so on. Let a molecule P as shown in figure having velocity C  collide against the wall marked I (BCFG) perpendicular to the x-axis. Only the x-component of the velocity of the molecule is relevant for the wall I. Hence momentum of the molecule before collision is mu₁ where m is the mass of the molecule. Since the collision is elastic, the molecule will rebound with the velocity u₁ in the opposite direction. Hence momentum of the molecule after collision is –mu₁.

Change in the momentum of the molecule

= Final momentum - Initial momentum

= –mu₁ – mu₁ = –2mu₁

During each successive collision on face I the molecule must travel a distance 2l from face I to face II and back to face I.

Time taken between two successive collisions is = 2l / u₁

∴ Rate of change of momentum = Change in the momentum/Time taken

= – 2mu₁/(2l/u₁) = -2mu₁²/2l = – mu₁²/l

(i.e) Force exerted on the molecule = – mu₁²/l

∴ According to Newton’s third law of motion, the force exerted by the molecule,

= – (– mu₁²)/l = mu₁²/l

Force exerted by all the n molecules is

F_x = mu₁²/l + mu₂²/l + …...+mu_n²/l

Pressure exerted by the molecules,

P_x = F_x/A

= 1/l² (mu₁²/l + mu₂²/l + …...+mu_n²/l)

= m/l³ (u₁²+ u₂² + …...+u_n²)

Similarly, pressure exerted by the molecules along Y and Z axes are,

P_y = m/l³ (v₁²+ v₂² + …...+v_n²)

P_z = m/l³ (ω₁²+ ω₂² + …...+ω_n²)

Since the gas exerts the same pressure on all the walls of the container

P_x = P_y = P_z = P

P = [P_x+P_y+P_z]/3

P = [(1/3) (m/l³)] [(u₁²+ u₂² + …...+u_n²) + (v₁²+ v₂² + …...+v_n²) + (ω₁²+ ω₂² + …...+ω_n²)]

P = [(1/3) (m/l³)] [(u₁² + v₁² +ω₁²) + (u₂² + v₂² +ω₂²) +.............+ (u_n² + v_n² +ω_n²) 

P = [(1/3) (m/l³)] [C₁²+C₂³+ …...+C_n²]

Here, C₁² = (u₁² + v₁² +ω₁²)

P = [(1/3) (mn/l³)] [C₁²+C₂³+ …...+C_n²/n]

P = (1/3) (mn/V) C²

So,

 

where C is called the root mean square (RMS) velocity, which is defined as the square root of the mean value of the squares of velocities of individual molecules.

That is, C = √[C₁²+C₂³+ …...+C_n²/n]

Refer this video to know more about “Pressure Exerted by a Gas”

Pressure is associated with collisions of the particles with the container walls.  Gases are made up of molecules: We can treat molecules as point masses that are perfect spheres. Molecules in a gas are very far apart, so that the space between each individual molecule is many orders of magnitude greater than the diameter of the molecule. Molecules are in constant random motion: There is no general pattern governing either the magnitude or direction of the velocity of the molecules in a gas. At any given time, molecules are moving in many different directions at many different speeds. The movement of molecules is governed by Newton’s Laws: In accordance with Newton’s First Law, each molecule moves in a straight line at a steady velocity, not interacting with any of the other molecules except in a collision. In a collision, molecules exert equal and opposite forces on one another. Molecular collisions are perfectly elastic: Molecules do not lose any kinetic energy when they collide with one another. Gases are made up of particles with no defined volume but with a defined mass. In other words their volume is miniscule compared to the distance between themselves and other molecules. Gas particles undergo no intermolecular attractions or repulsions. This assumption implies that the particles possess no potential energy and thus their total energy is simply equal to their kinetic energies. Gas particles are in continuous, random motion. Collisions between gas particles are completely elastic. In other words, there is no net loss or gain of kinetic energy when particles collide. The average kinetic energy is the same for all gases at a given temperature, regardless of the identity of the gas. Furthermore, this kinetic energy is proportional to the absolute temperature of the gas. We can gain a better understanding of pressure (and temperature as well) from the kinetic theory of gases, which assumes that atoms and molecules are in continuous random motion. Since the assumption is that the particles move in random directions, the average value of velocity squared along each direction must be same.

Problem (JEE Advanced):

A gas at pressure P₀ is contained in a vessel. If the masses of all the molecules are halved and their speeds doubled, the resulting pressure would be

(a) 4P₀                   (b) 2P₀

(c) P₀                     (d) P₀/2

Solution:

We know that, pressure,

P = (1/3) (mn/V) v_rms²

That is, P ∝ mv_rms²

m is halved and v_rms is doubled.

Thus, P will become two times.

Note: In equation (1), m is the mass of one gas molecule and n the total number of gas molecules.

From the above observation we conclude thta, option (b) is correct.

Question 1