so there is more of the lighter molecule (enrichment) outside the porous cylinder (Fig. . ). The method is not very efficient and has to be repeated several times for sufficient enrichment.].
When gases diffuse, their rate of diffusion is inversely proportional to square root of the masses (see Exercise . ). Can you guess the explanation from the above answer? Fig.
. Molecules going through a porous wall. Example . (a) When a molecule (or an elastic ball) hits a ( massive) wall, it rebounds with the same speed.
When a ball hits a massive bat held firmly, the same thing happens. However, when the bat is moving towards the ball, the ball rebounds with a different speed. Does the ball move faster or slower? (Ch.
will refresh your memory on elastic collisions.) (b) When gas in a cylinder is compressed by pushing in a piston, its temperature rises. Guess at an explanation of this in terms of kinetic theory using (a) above. (c) What happens when a compressed gas pushes a piston out and expands. What would you observe ?
(d) Sachin Tendulkar used a heavy cricket bat while playing. Did it help him in anyway ? Answer (a) Let the speed of the ball be u relative to the wicket behind the bat. If the bat is moving towards the ball with a speed V relative to the wicket, then the relative speed of the ball to bat is V + u towards the bat.
When the ball rebounds (after hitting the massive bat) its speed, relative to bat, is V + u moving away from the bat. So relative to the wicket the speed of the rebounding ball is V + ( V + u ) = V + u , moving away from the wicket. So the ball speeds up after the collision with the bat. The rebound speed will be less than u if the bat is not massive.
For a molecule this would imply an increase in temperature. You should be able to answer (b) (c) and (d) based on the answer