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CBSE Class 11-science Answered

On the basis of the kinetic theory of gases derive an expression for the pressure exerted by a gas in terms of density and velocity of the molecules.
Asked by Topperlearning User | 04 Jun, 2014, 01:23: PM
answered-by-expert Expert Answer

Consider a cubical vessel (of side l) with walls perfectly elastic as in the figure.  Let the vessel have one gram molecule of a gas with its molecules in random motion.

Consider a molecule (mass m) moving with a velocity c, that can be resolved into three components u, v, and w in the direction of the edges of the cube along x, y, z ases.

Therefore, c2 = u2 + v2 + w2 ................(i)

Let us consider two faces of the cube, say P and Q, normal to x-axis.   When the molecule collides with the side Q with a velocity u, it rebounds with -u, while its other components remain unchanged.

 

Change in momentum of the molecule due to this collision = -2mu, which is imparted to the wall per collision.

The time taken by molecule to cover distance l = l/u

Therefore, after every interval of time 2l/u, the molecule will again collide with the wall Q, and the number of collisions per unit time with the wall Q is equal to u/2l.

Therefore, Momentum imparted to the wall per unit time = 2mu x u/2l = mu2/l (ignoring the negative sign).

The pressure exerted on the wall Q due to one molecule =

If the vessel contains N molecules with velocities u1, u2, u3.........un along x-axis, then the pressure exerted by them on Q, say

Answered by | 04 Jun, 2014, 03:23: PM
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