MP203 Statistical and Thermal Physics. Problem set 7 - Solutions

Transcription

1 MP203 Statistical and Thermal Physics Problem set 7 - Solutions 1. For each of the following processes, decide whether or not they are reversible. If they are irreversible, explain how you can tell that they increase the total entropy of the universe. Finally, describe at least 3 of your favourite, or least favourite, irreversible processes, and explain how they increase the total entropy of the universe. a Mixing a cocktail. b Humpty Dumpty having a great fall. c Winding a watch. d Burning a log of firewood. e Sticking pins into a voodoo doll. f Teaching thermal physics to a class of undergraduate students. a Irreversible: The mixed coctail has more microstates and thus higher entropy. b Irreversible: All the kings horses and men couldn t put him back together again. There are many ways Humpty Dumpty could be in pieces, but only one way he could be in one piece. c Reversible: d Irreversible: The microstates of all matter released in burning process are greater than solid log and thus entropy increases. Alternatively, heat is created in the burning at high temperature, and transferred to a colder environment; this means entropy increases. e Irreversible: While the voodoo doll itself may return to its original state, the greater effects of the action will lead to an overall increase in entropy. f Reversible: This is currently an unsolved problem and any progress on this matter would be of interest. 2. An ice cube mass 30 g at 0 C is left sitting on the kitchen table, where it gradually melts. The temperature in the kitchen is 25 C. a Calculate the change in the entropy of the ice cube as it melts into water at 0 C. Don t worry about the fact that the volume changes somewhat. b Calculate the change in entropy of the water from the melted ice as its temperature rises from 0 C to 20 C. c Calculate the change in the entropy of the kitchen as it gives up heat to the melting ice and water. d Calculate the net change in the entropy of the universe during this process. Is the net change positive, negative, or zero? Is this what you would expect?

2 a The change in the entropy as the ice cube melts to water at 0 C is S = Q T = L mm T = J/kg0.03kg 273K = 36J/K b The change in entropy of the water as its temperature rises from 0 circ C to 20 circ C is 293 c v m 293 S = T dt = c vmln = 8.91J/K c Assuming no work is done, we only have to consider the heat the kitchen gives up to the ice cube to melt it and to heat the resulting water up to 20 C. This is Q = L m m+c v m T = 12.5kJ. The change in entropy of the kitchen then is S = Q T k = 2.08J/K where T k = 298 K is the temperature of the kitchen, which remains unchanged. d The net change in the entropy of the universe is the sum of all the entropies we just calculated S total = J/K = 2.83J/K This is positive which is what we would expect, since according to the second law of thermodynamics, the total entropy always increases. *3. a Suppose you flip four fair coins. i. Make a list off all the possible outcomes. ii. Make a list of all the different macrostates and their probabilities. iii. Compute the multiplicity of each macrostate using the combinatorial formula in the notes, and check that these results agree with what you got by brute-force counting. b i. Calculate the number of possible five-card poker hands, dealt from a deck of 52 cards. The order of cards in a hand does not matter. ii. A royal flush consists of the five highest-ranking cards ace, king, queen, jack, 10 of any one of the four suits. What is the probability of being dealt a royal flush on the first deal? a i. All possible outcomes of coins are as follows HHHH THHH THHT THTT HHHT HHTT THTH TTHT HHTH HTHT TTHH TTTH HTHH HTTH HTTT TTTT

3 ii. We can label macro states by the number of heads and number of tails. In the following table there is a row for each macro state which gives the numberofheads,thenumberoftailsandtheprobabilityforthemacrostate. #H #T P iii. To find the number of microstates with N H heads and N N H tails, we use the formula N N H = N! N H!N N H! For each macrostate we use this expression to work out the multiplicity Ω #H #T Ω 0 0 = = = = 0 = 1 We see that the multiplicity in each case is equal to the number of microstates from brute-force counting. b i. The number of possible five-card power hands from a deck of 52 cards is the number of ways of choosing 5 from = 52! 5 5!7! = ii. The probability of getting a royal flush on the first deal is the number of ways of making up a royal flush over the total number of possible hands. There are only ways of making up a royal flush the suits, so the probability is = An insulated box contains N identical molecules. We will consider the different possibilities of dividing these molecules up between the two halves left and right of the box. a If N = 8, how many possible ways are there of having i. 7 molecules in the left half and 1 molecule in the right half of the box? ii. 3 molecules in the left half and 5 in the right half?

4 b For a general value of N, show that the multiplicity of a macrostate with N 1 molecules in the left half and N 2 molecules in the right half of the box is ΩN,N 1,N 2 = N! N 1!N 2!. c Using Stirling s formula, with N = 500, find the entropy of i. the state with N 1 = 220,N 2 = 280, and ii. the state with N 1 = 29,N 2 = 251. iii. How much more probable is the second state than the first one? d A container is divided into two sections, with a wall between them. Initally, N molecules of a gas are in one section, and the other one is empty. After a hole is punched in the wall, the gas fills the entire container uniformly, so that N/2 molecules are in each section. Compute the multiplicity of the final state, and the increase in entropy from the initial to the final state. a i. The molecule in the left half can be any one of the 8. Therefore there are 8 possibilities. ii. There are 8 choices for the first of the molecules in the left half, 7 for the second and 6 for the third, giving possibilities. But this overcounts by a factor of 3!=6 since the order of the three molecules does not matter. Hence the total number of possibilities is 8 7 = 56. b Following the argument above, we have NN 1N 2...N N = N!/N N 1! = N!/N 2! ways of choosing the N 1 molecules on the left in order. Since their order does not matter, this overcounts by a factor of N 1!, which gives us the desired expression. c Applying Stirling s formula to the expression derived in the previous part gives us ΩN,N 1,N 2 N N e N 2πN = N N 1 1 e N 1 2πN1 N N 2 2 e N 2 2πN2 Using this approximation for the entropy gives N N N N N 1 1 N N 2 2 2π N1 N 2 SN,N 1,N 2 k B lnω = k B [ N lnn N lnn 1 N lnn ln2π ]. i. Using the approximation above we get: S500, 220, 280 k B 500.5ln ln ln ln2π = 339.6k B.

5 ii. S500, 29, 251 k B 500.5ln ln ln ln2π = 33.2k B iii. The entropy of a macrostate i is S i. The probability of this macrostate is P i = e Si k B j e S j k B For the two states in question, the second state is more probable by a factor of P S2 S 2 1 k = e B = e = P 1 d The multiplicity of the final state is given by: ΩN,N/2,N/2 = N! N 2!N 2! Assuming N is large, we can approximate the factorials with Stirling s formula and get an expression for the entropy as S = k B N lnn N ln N 2 = k B N lnn N lnn ln2 = k B N ln2 This is also the increase as the initial entropy associated with the arrangement of the molecules is zero this is ignoring the motion of the molecules, which will give the same contribution to the entropy in both cases.