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Unformatted text preview: Physics 105 Problem Set 11 Solutions Problem 1 (3 pts) (a) When bouncing off a wall perpendicular to the xaxis, only the xcomponent of the atoms velocity changes its sign. Collisions with the other walls do not affect the xcomponent of the atoms velocity. Hence the atom transfers to the wall orthogonal to the xaxis the amount of momentum equal to p = 2 mv x during each collision; and it happens exactly once during the time interval equal to t = 2 L v x . Consequently, the time averaged force is F = p t = mv 2 x L . (b) Using the result from the previous part, the pressure of the gas on the wall orthogonal to the xaxis is p = N i =1 F i L 2 = 1 L 3 N X i =1 m i v 2 ix . Given that the direction of the atoms velocity is equally likely to point in any direction, we find: 1 2 N X i =1 m i v 2 ix = 1 2 N X i =1 m i v 2 iy = 1 2 N X i =1 m i v 2 iz = N 2 k B T . This yields for the pressure pV = N N A ( k B N A ) T . Recalling that the ideal gas constant is R = k B N A , we recover the ideal gas law....
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 '08
 LYMANA.PAGE
 Physics, mechanics

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