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KINETIC THEORY 251 of a molecule of the gas. Therefore, 2 (vrms)Ar ( )Cl ( ) Cl 70.9 2 = = = =1.77 (vrms)Cl (m )Ar (M )Ar 39.9 where M denotes the molecular mass of the gas. (For argon, a molecule is just an atom of argon.) Taking square root of both sides, (v ) rms Ar (vrms) = 1.33 Cl You should note that the composition of the mixture by mass is quite irrelevant to the above calculation. Any other proportion by mass of argon and chlorine would give the same answers to (i) and (ii), provided the temperature remains Fig. 12.5Molecules going through a porous wall. unaltered. ⊳ ⊳ ⊳Example 12.7 (a) When a molecule (or an Example 12.6 Uranium has two isotopes of masses 235 and 238 units. If both are elastic ball) hits a ( massive) wall, it rebounds with the same speed. When a ball present in Uranium hexafluoride gas which hits a massive bat held firmly, the same would have the larger average speed ? If thing happens. However, when the bat is atomic mass of fluorine is 19 units, moving towards the ball, the ball rebounds estimate the percentage difference in speeds at any temperature. with a different speed. Does the ball move faster or slower? (Ch.5 will refresh your Answer At a fixed temperature the average memory on elastic collisions.) energy = ½ m <v > is constant. So smaller the (b) When gas in a cylinder is compressed mass of the molecule, faster will be the speed. by pushing in a piston, its temperature The ratio of speeds is inversely proportional to rises. Guess at an explanation of this in the square root of the ratio of the masses. The terms of kinetic theory using (a) above. masses are 349 and 352 units. So v / v = ( 352/ 349) 1/2= 1.0044 . (c) What happens when a compressed gas 349 352 pushes a piston out and expands. What V would you observe ? Hence difference = 0.44 %. (d) Sachin Tendulkar used a heavy cricket V bat while playing. Did it help him in [35Uis the isotope needed for nuclear fission. To separate it from the more abundant isotope anyway ? 23U, the mixture is surrounded by a porous Answer (a) Let the speed of the ball be u relative cylinder. The porous cylinder must be thick and narrow, so that the molecule wanders through to the wicket behind the bat. If the bat is moving individually, colliding with the walls of the long towards the ball with a speed V relative to the pore. The faster molecule will leak out more than wicket, then the relative speed of the ball to bat the slower one and so there is more of the lighter is V + u towards the bat. When the ball rebounds (after hitting the massive bat) its speed, relative molecule (enrichment) outside the porous cylinder (Fig. 12.5). The method is not very to bat, is V + u moving away from the bat. So efficient and has to be repeated several times relative to the wicket the speed of the rebounding for sufficient enrichment.]. ⊳ ball is V + (V + u) = 2V + u, moving away from the When gases diffuse, their rate of diffusion is wicket. So the ball speeds up after the collision with the bat. The rebound speed will be less than inversely proportional to square root of the masses (see Exercise 12.12 ). Can you guess the u if the bat is not massive. For a molecule this explanation from the above answer? would imply an increase in temperature. Reprint 2026-27