Problem Set 2 Solutions - Thermal Physics | PHYS 224, Assignments of Thermal Physics

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224 problem set 2 solutions

224 problem set 2 solutions Problem 1.33. The work done on the gas is positive when its volume decreases (step C), negative when its volume increases (step A), and zero when its volume stays the same (step B). The net work done during the whole cycle is positive, since the average pressure is higher during step C than during step A. To determine the sign of the change in energy, note that for an ideal gas, U = ENkT = fPV, so any increase in P or V indicates an increase in U'. Therefore the energy increases during steps A and B, but decreases during step C. For the whole cycle the energy must be unchanged, since both P and V are unchanged. To determine the sign of Q for each step, just use the first law: Q = AU —W. Here are the results in tabular form: Ww AU Q stepA: - + + + ° ! Apparently, the net result of the cycle is to absorb energy as work and emit energy as heat. Furthermore, the heat emitted during step C could go to a different place than the source of heat during steps A and B, so this “machine” could move heat around. Similar cycles can act as refrigerators (see Chapter 4), but this particular cycle would not make a practical refrigerator because the absorption of heat does not always take place at a lower temperature than the emission of heat Problem 1.38. Since the bubbles each contain the same number of molecules and each end up at the same pressure, the ideal gas law tells us that whichever one has the higher temperature at the surface also has the larger volume: V = NkT/P. And the one with the higher temperature is the one with the larger energy content, by the equipartition theorem: U = NkT/2. So the question is, which bubble has more energy at the surface? Since both bubbles expand, they both lose energy as they do work on the surrounding water. Apparently, bubble B absorbs heat to replace this lost energy and thus remains at the initial temperature. Since bubble A does not absorb any heat, its energy drops. So bubble B, the one that does absorb heat, ends up at the higher temperature and thus has the larger volume. Problem 1.41. (Measuring a heat capacity.) (a) The heat gained by the water is My Cu AP) = (250 g)(4.186 J/g°C)}(4°C) = 4186 J. (b) The heat lost by the metal must be the same as the heat lost by the water, 4186 J, since there are no other energy transfers going on and energy must be conserved. (c) The heat capacity of the chunk of metal must therefore be Q — ~4186 J On = AT, Tae = 55 JC. {d) The specific heat capacity is the heat capacity per unit mass, — Om — 86 TPC mm lWO0g = 0.55 IfgeC.