24.1 Entropy of Mixing of Ideal Solutions In an ideal solution: the sizes of the solute and solvent molecules are similar and the energies of interaction between and among the solute and solvent molecules are also similar. Does a mixture of ideal gases fit this definition of an ideal solution? Would you expect benzene, C 6 H 6, and toluene, C 7 H 8, to form an ideal solution? benzene CH 3 toluene Would you expect normal octane and water to form and ideal solution? CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 3 H 2 O

24.2 As a model of a solution consider a 3-dimensional lattice with N lattice sites, which we will fill with N 1 solvent and N 2 solute molecules: Note that the subscript 1 will imply a solvent molecule and the subscript 2 will imply a solute molecule (this is a typical convention). To make the visualization a little easier we will focus on a 2- dimensional slice of the 3-dimensional lattice. There are N ways that the 1st distinguishable molecule can be placed in the N unoccupied lattice sites: 1 solute molecule solvent molecule A distinguishable molecule can be distinguished or recognized among other molecules. For the moment we will treat all molecules as distinguishable. the number distinguishes this solvent molecule from other solvent molecules

24.3 There are now N-1 ways of placing the 2nd distinguishable molecule in the lattice. Why? How many ways can the 3rd distinguishable molecule be placed in the lattice? The total number of ways that N distinguishable molecules can be placed in N lattice sites is N factorial: N (N-1) (N-2) (N-3) ……. (2) (1) = N! In reality, while we can distinguish between a solvent and a solute molecule because they are different, we cannot distinguish one solvent molecule from another solvent molecule or one solute molecule from another solute molecule, i.e., they are indistinguishable. Or to put it another way, we could switch solvent molecules 1 and 2 in the above sketch and, if they were not labeled, we would not be able to tell that the molecules had been switched.

24.4 Thus N! overcounts the number of distinguishable arrangements and to get the correct number we have to divide out the number of ways of permuting the solvent molecules among themselves and also divide out the number of ways of permuting the solute molecules among themselves. The result is the number of distinguishable arrangements of N 1 solvent and N 2 solute molecules among N lattice sites:  = N! / (N 1 ! N 2 !) Why is the number of ways of permuting the solvent molecules among themselves equal to N 1 ! ? Ludwig Boltzmann recognized that there is a connection between the number of ways of arranging a system and its entropy and expressed this relationship in the equation (now inscribed on his tombstone in Vienna): S = k ln  S = k ln  = k ln [N! / (N 1 ! N 2 !)] = k [ln N! - ln N 1 ! - ln N 2 !] To evaluate the natural logarithm of factorials of the very large values that N, N 1, and N 2 will assume, we will use Stirling’s approximation: ln x!  x ln x - x For what value of x does the error in Stirling’s approximation fall below 5%?

24.5 Ludwig Boltzmann Ludwig Boltzmann was born in 1844 in Vienna. He was awarded a doctorate from Vienna in 1866 and then proceeded to teach physics at Graz, Heidelberg, Berlin, Vienna, and Leipzig. During this time he worked with Stefan, Bunsen, Kirchhoff, Helmholtz, and Ostwald. In 1871 he developed, independently of Maxwell, the Maxwell - Boltzmann distribution of molecular speeds,and in 1884 derived from thermodynamics a confirmation of Josef Stefan's 1879 empirical T 4 law for black body radiation. Boltzmann worked on statistical mechanics using probability to describe how the properties of atoms determine the properties of matter. Depressed and in bad health, Boltzmann committed suicide in 1906 just before experiments at with newly discovered forms of radiation at Berkley and Stanford verified his work. His grave in Vienna is inscribed with the relation he developed that links entropy and probability, S = k ln W.

24.6 Using Stirling’s approximation, the entropy of the mixed binary solution is thus: S = k [(N ln N - N) - (N 1 ln N 1 - N 1 ) - (N 2 ln N 2 - N 2 )] Rembering that N = N 1 + N 2 : S = k [(N 1 + N 2 ) ln (N 1 + N 2 ) - (N 1 + N 2 ) - (N 1 ln N 1 - N 1 ) - (N 2 ln N 2 - N 2 )] = k [(N 1 + N 2 ) ln (N 1 + N 2 ) - (N 1 ln N 1 ) - (N 2 ln N 2 )] = k [N 1 (ln (N 1 + N 2 ) - ln N 1 ) + N 2 (ln (N 1 + N 2 ) - ln N 2 )] = - k [N 1 ln [N 1 / (N 1 + N 2 )] + N 2 ln [N 2 / (N 1 + N 2 )]] Multiplying and dividing by Avogadro’s number, N o converts the numbers of molecules to moles: S = - k N o [(N 1 / N o ) ln [(N 1 / N o ) / ((N 1 / N o ) + (N 2 / N o ))] + (N 2 / N o ) ln [(N 2 / N o ) / ((N 1 / N o ) + (N 2 / N o ))]] = - R [ n 1 ln [(n 1 / (n 1 + n 2 )] + n 2 ln [(n 2 / (n 1 + n 2 )] ] = - R [ n 1 ln X 1 + n 2 ln X 2 ] where X i is the mole fraction of component i. If we divide both sides of the equation by the total number of moles, n 1 + n 2, then we calculate the molar entropy of the mixed solution: S molar = - R [ X 1 ln X 1 + X 2 ln X 2 ] Could you extend these results to the mixing of more than 1 solute with the solvent to form a more complex solution?

24.7 The mixing process to form an ideal binary solution can be viewed as: T, P pure solute + pure solvent > ideal solution   S mixing  = S mixed solution - S pure solute - S pure solvent = - R [ n 1 ln X 1 + n 2 ln X 2 ] - k ln  pure solute - k ln  pure solvent = - R [ n 1 ln X 1 + n 2 ln X 2 ] - k ln (  - k ln (1) = - R [ n 1 ln X 1 + n 2 ln X 2 ] Why are  pure solute and  pure solvent equal to one? Again the molar entropy of mixing to form the ideal solution at constant temperature and pressure would be given by:  S mixing, molar = - R [ X 1 ln X 1 + X 2 ln X 2 ] Why is this result only valid for forming an ideal solution while the temperature and pressure are held constant? Alloys are ideal solid solutions of metals. Wood’s metal, used for low temperature soldering, is low melting alloy of 50% Bi, 25.0% Pb, 12.5% Sn, and 12.5% Cd (all weight percents). What is  S to form Wood’s metal from its constituient metals at constant temperature and pressure?

24.8 N2N2 O2O2 Naturally occurring atomic chlorine is 75 atom percent the stable (non-radioactive) isotope, chlorine 35, 35 Cl, with the remainder being the stable isotope, chlorine 37, 37 Cl. What is the contribution per mole to the 3 rd law entropy of molecular chlorine, Cl 2, due to the mixing at constant temperature and pressure of the various stable isotope compositions of Cl 2 ? moles of N 2 and mole of O 2 are contained in coupled gas bulbs of equal volume. The stopcock separating the coupled gas bulbs is opened and the gases expand and mix isothermally to fill the entire volume: Correctly calculate  S by evaluating:  S = n N 2 R ln (V 2 / V 1, N 2 ) + n O 2 R ln (V 2 / V 1, O 2 ) Why doesn’t this value agree with?  S mixing = - R [ n 1 ln X 1 + n 2 ln X 2 ] Calculate  S when both gas bulbs initially contain N 2.

24.9 Consider 3 of the 7 possible ways that 6 gas molecules can be distributed between two gas bulbs: To the right of each distribution are listed the number of distinguishable arrangements or ways of achieving each distribution. Molecules in one bulb can be distinguished from a molecule in the other bulb by virtue of their positions, but molecules in the same bulb are indistinguishable. Also listed are the probabilities of achieving each distribution. probability of observing a given distribution Number of distinguishable ways a given distribution can be achieved Total number of arrangements for all distributions = # of arrangements = 1 probability = 1 / 64 # of arrangements = 6 probability = 6 / 64 # of arrangements = 15 probability = 15 / 64 Sketch the remaining four ways that 6 molecules can be distributed between two gas bulbs and for each list the number of ways that particular distribution can be realized and the probability that it will occur.

24.10 Now consider 1 mole or Avogadro’s number, N o, of molecules evenly distributed between two gas bulbs of equal volume: What is the probability of achieving this distribution? Is it possible that all these molecules could be found only in the left hand bulb at some point in time? What is the probability of achieving this distribution? Use Boltzmann’s formula for entropy, S = k ln , to calculate the entropy change,  S, for the process: Does this agree with the entropy change calculated using:  S = R ln (V 2 / V 1 )

24.11 a.Calculate the probability that Avogadro’s number, N o, of molecules would all be in the left hand bulb in when occupying two connected gas bulbs of equal volume. Consider the tube connecting the two gas bulbs to be of negligible volume. where in the initial state, Avogadro’s number, N o, of ideal gas molecules are evenly distributed between the two gas bulbs, i.e., initially there are N o /2 molecules in each gas bulb. In the final state all the gas molecules are in the left hand bulb. b.Use Boltzmann’s formula for entropy, S = k ln , to calculate the entropy change,  S, in J/ (mol K) for the isothermal process: Entropy of Mixing Bonus Problem This bonus problem is worth 10 points and is due in class 1 week after it is assigned and will be graded on the answer only.