Thermodynamics Problems: Energy Distribution and Entropy in Statistical Mechanics, Assignments of Thermal Physics

Download Thermodynamics Problems: Energy Distribution and Entropy in Statistical Mechanics and more Assignments Thermal Physics in PDF only on Docsity! 1    Fall 2008 Phys 461 Solution Set 4 Schroeder 2.8 (a) Solid A can have anywhere between 0 and 20 units of energy. Therefore, 21 different macrostates are available to the system of weakly coupled A and B solids sharing 20 units of energy. (b) The combined system has 20 oscillators and 20 units of energy. Therefore 101089.6 !19!20 !39 )!1(! )!1(),( ×= × = − −+ =Ω Nq NqqN (c) For the macrostate with all energy in solid A, 7101 !9!20 !29)0,10()20,10( ×= × =ΩΩ=ΩΩ=Ω BABA Probability of this macrostate is 410 7 1045.1 1089.6 101 −×= × × (d) For the macrostate with half the energy in A and half in B, 9 2 1053.8 !9!10 !19)10,10()10,10( ×=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × =ΩΩ=Ω BA Probability = 124.0 1089.6 1053.8 10 9 = × × (e) If the energy was initially all in A, the system would evolve toward half energy in A and half energy in B, and this distribution of energy would be irreversible. In other words, once the energy is evenly distributed, it is highly unlikely that the system would find itself with all the energy back in A. 2    Schroeder 2.17 !! )!( )!1(! )1().( Nq qN Nq qNqN +≈ − −+ =Ω and NNqqqNqN lnln)ln()(ln −−++≈Ω In the limit q << N N qN N qNqN +≈⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +≈+ ln1ln)ln( Therefore, q q NqNNqq N qqNqNN +⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ≈−−+++≈Ω lnlnlnlnlnln 2 where we have dropped the term Nq /2 . Exponentiate both sides to get q q q q eNe q Nq q Nq ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =Ω )exp(lnexp Schroeder 2.27 The probability of finding any one gas molecule in the leftmost 99% of the container is 0.99. Therefore, the probability of finding all N molecules in the leftmost 99% of the container is (0.99)N. For N=100, (0.99)100 = 0.366 i.e. the probability that the rightmost 1% of the container will be empty is ~37% for N=100. For N=10,000, (0.99)10000 = 2.25×10−44. For N=1023, (0.99)N= 20104.410 ×− , which is an extremely small probability. Schroeder 3.5 From problem 2.17, q q eNqN ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =Ω ),( for q << N The entropy [ ] [ ]qNkqqNekq q eNkqkS lnln1lnlnlnlnln −+=−+=⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =Ω=