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Unformatted text preview: CHAPTER 22 THE SECOND LAW OF THERMODYNAMICS ActivPhysics can help with these problems: Activity 8.14. Section 221: Reversibility and Irreversibility Problem 1. The egg carton shown in Fig. 2227 has places for one dozen eggs. (a) How many distinct ways are there to arrange six eggs in the carton? (b) Of these, what fraction correspond to all six eggs being in the left half of the carton? Treat the eggs as distinguishable, so an interchange of two eggs gives rise to a new state. figure 2227 Problem 1. Solution (a) There are twelve places for first egg, eleven for the second, etc., so the number of arrangements (states) of six distinguishable eggs into twelve places is 12 × 11 × 10 × 9 × 8 × 7 = 12! / 6! ≈ 6 . 65 × 10 5 . (b) If limited to the left side of the carton only, there are 6! arrangements, so the fraction is 6! / (12! / 6!) = 1 / 924 . Problem 2. A gas consists of four distinguishable molecules, all moving with the same speed v either to the left or to the right. A microscopic state is specified by telling which molecules are moving in which direction. (a) How many possible microscopic states are there? (b) How many of these correspond to all four molecules moving in the same direction? (c) What is the probability of finding the gas with all its molecules moving in one direction? Repeat parts (a–c) for a gas of (d) 10 molecules and (e) a more typical value, 10 23 molecules. Solution (a) There are 2 directions for each molecule, so the number of states is 2 4 = 16 . (b) Two of these, all to the right or all to the left, have all molecules moving in the same direction. (c) If all states are equally probable, the probability of a particular property equals the number of states with that property divided by the total number of states, which is 2 / 2 4 for the property in part (b). (d) For N molecules, there are 2 N states, just 2 of which have all the molecules moving in the same direction. The probability of this is 2 / 2 N =1 / 2 N 1 . For N = 10 , 2 10 = 1024 and 1 / 2 9 = 1 / 512 . (e) For N = 10 23 , the number of states is huge (2 10 23 ) and the probability all molecules are moving in the same direction (2 1 10 23 ) is negligible. Problem 3. Estimate the energy that could be extracted by cooling the world’s oceans by 1 ◦ C. How does your estimate compare with humanity’s yearly energy consumption of about 2 . 5 × 10 20 J? Solution The volume of the oceans is about 1 . 35 × 10 18 m 3 (the average depth is 3.73 km over 71% of the earth’s surface). The heat extracted by cooling this volume of water by 1 ◦ C is Q = ρ Vc Δ T = (10 3 kg / m 3 ) × (1 . 35 × 10 18 m 3 )(4184 J / kg · K)(1 K) ’ 5 . 65 × 10 24 J , or about 23,000 times the world’s annual energy consumption....
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This note was uploaded on 08/28/2009 for the course PHYS 2330 taught by Professor Wilke during the Spring '08 term at Rensselaer Polytechnic Institute.
 Spring '08
 WILKE
 Second Law Of Thermodynamics

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