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KINETIC THEORY 249 their velocities. The collisions are considered to the wall. Thus, the number of molecules with be elastic. We can derive an expression for the velocity (vx, y ,zv ) hitting the wall in time ∆t is pressure of a gas based on the kinetic theory. ½A v ∆x n, where n is the number of molecules We begin with the idea that molecules of a per unit volume. The total momentum gas are in incessant random motion, colliding transferred to the wall by these molecules in against one another and with the walls of the time ∆t is: container. All collisions between molecules Q = (2mv x (½ n A v xt ) (12.10) among themselves or between molecules and the The force on the wall is the rate of momentum walls are elastic. This implies that total kinetic transfer Q/∆t and pressure is force per unit energy is conserved. The total momentum is area : conserved as usual. P = Q /(A ∆t) = n m v x2 (12.11) Actually, all molecules in a gas do not have 12.4.1 Pressure of an Ideal Gas the same velocity; there is a distribution in velocities. The above equation, therefore, stands Consider a gas enclosed in a cube of side l. Take for pressure due to the group of molecules with the axes to be parallel to the sides of the cube, as shown in Fig. 12.4. A molecule with velocity speed v xn the x-direction and n stands for the number density of that group of molecules. The (vx, y , z ) hits the planar wall parallel to yz- plane of area A(= l ). Since the collision is elastic,total pressure is obtained by summing over the the molecule rebounds with the same velocity; contribution due to all groups: 2 its y and z components of velocity do not change P = n m vx (12.12) in the collision but the x-component reverses where v2 is the average of v 2 . Now the gas x x sign. That is, the velocity after collision is is isotropic, i.e. there is no preferred direction (-vx, y , z ) . The change in momentum of the of velocity of the molecules in the vessel. molecule is: –mv – (mv ) = – 2mv . By the Therefore, by symmetry, x x x principle of conservation of momentum, the 2 2 2 momentum imparted to the wall in the collision v x = v y = v z = 2mv . x = (1/3) [vx + v y + v z] = (1/3) v2 (12.13) where v is the speed and v 2 denotes the mean of the squared speed. Thus P = (1/3) n m v2 (12.14) Some remarks on this derivation. First, though we choose the container to be a cube, the shape of the vessel really is immaterial. For a vessel of arbitrary shape, we can always choose a small infinitesimal (planar) area and carry through the steps above. Notice that both A and ∆t do not appear in the final result. By Pascal’s law, given in Ch. 9, pressure in one portion of Fig. 12.4 Elastic collision of a gas molecule with the the gas in equilibrium is the same as anywhere wall of the container. else. Second, we have ignored any collisions in To calculate the force (and pressure) on the the derivation. Though this assumption is wall, we need to calculate momentum imparted difficult to justify rigorously, we can qualitatively to the wall per unit time. In a small time interval see that it will not lead to erroneous results. The ∆t, a molecule with x-component of velocity v x number of molecules hitting the wall in time ∆t will hit the wall if it is within the distance x ∆t was found to be ½ n Av ∆t.xNow the collisions from the wall. That is, all molecules within the are random and the gas is in a steady state. volume Av ∆x only can hit the wall in time ∆t. Thus, if a molecule with velocity (v , v x vy) z But, on the average, half of these are moving acquires a different velocity due to collision with towards the wall and the other half away from some molecule, there will always be some other Reprint 2026-27