10.9 reif - Physics 211 (Moore) Spring 2010 Problem Set #5...

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Physics 211 (Moore) Spring 2010 Problem Set #5 Solution April 22, 2010 1. Reif 5.19 a. The equation of state reads ( p + av - 2 )( v - b ) = RT . We want to solve for the critical point in terms of a and b . The easiest way to do this is to first take a derivative with respect to v (treating T as a constant) of both sides, and set dp/dv to zero. We find that ( v - b ) 2 v 3 = RT 2 a Next we cross multiply to clear out the fraction and take another derivative with respect to v . (This works since we are only neglecting d 2 p/dv 2 , which we’re setting to zero anyway.) This leads to a simple formula for v which gives v c = 3 b . Plugging back into the above equation gives RT c = 8 a 27 b . Inverting these formulae give a = 9 8 v c RT c b = v c 3 b. Plugging the formula for a and b back into the equation of state gives p c = 3 8 RT c v c c. The original equation of state can be written ± p 0 + a p c v 2 c v 0 2 ²± v 0 - b v c ² = RT 0 T c p c v c Then plugging in the values for a and b and p c gives ± p 0 + 3 v 0 2 ²± v 0 - 1 3 ² = 8 3 T 0 2. Reif 5.24 The minimum amount of heat which the refrigerator rejects (while still, presumably, operating reversibly) is given when the total entropy change of the system is zero. This means any entropy lost by the ice must be gained by the reservoir, and this must furthermore be the only entropy gained. Clearly if the reservoir has an infinite heat capacity, the answer is Q = mL , so the problem is how to deal with the finite heat capacity case. Consider a generic moment in the heating process. The reservoir is at some T > T 0 and some mass of water has already been frozen. We consider freezing a small mass dm further. The ice-water mix must lose an entropy of dS ice = - Ldm/T 0 in the freezing process. Assuming optimality, the reservoir must gain an entropy dS r = + Ldm/T 0 . But dS r = CdT/T and equating these and integrating leads immediately to T = T 0 e Lm/CT 0
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2 Problem Set #5 Solution April 22, 2010 for the whole process. The heat gained by the reservoir is then Q = CT 0 ± e Lm/CT 0 - 1 ² In the limit of a large heat capacity, Q CT 0 Lm CT 0 + 1 2 ³ Lm CT 0 ´ 2 + ...
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10.9 reif - Physics 211 (Moore) Spring 2010 Problem Set #5...

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