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Homework Week 7 Statistical MolecularThermodynamics University of Minnesota Homework Week 7 1. Given the following experimentally determined standard molar entropies at 298.15 K and 1 bar, ◦ −1 −1 S [CH3OH] = 126:8 J · K · mol S◦[CO] = 197:7 J · K−1 · mol−1 ◦ −1 −1 S [H2] = 130:7 J · K · mol ◦ calculate the value of ∆rS for the reaction CO(g) + 2H2(g) −! CH3OH(l): (a) 332.3 J · K−1 · mol−1 (b) 197.7 J · K−1 · mol−1 (c) -332.3 J · K−1 · mol−1 (d) -201.6 J · K−1 · mol−1 (e) 322.4 J · K−1 · mol−1 (f) 201.6 J · K−1 · mol−1 (g) -322.4 J · K−1 · mol−1 (h) -233.2 J · K−1 · mol−1 Answer: ◦ ◦ ◦ ∆rS =S (products) − S (reactants) =126:8 J · K−1 · mol−1 − 197:7 J · K−1 · mol−1 − 2(130:7 J · K−1 · mol−1) = − 332:3 J · K−1 · mol−1 2. The molar heat capacity of a certain gas can be expressed as C¯ 1000 K V = 10 − R T over the temperature range 100 < T < 500 K. Calculate ∆S if one mole of this gas is heated from 100 K to 200 K at constant volume. (a) −8 J · K−1 (b) −2 J · K−1 (c) 1 J · K−1 (d) 2 J · K−1 (e) 8 J · K−1 (f) 10 J · K−1 (g) 12 J · K−1 (h) 16 J · K−1 (i) 20 J · K−1 (j) 24 J · K−1 Answer: Z T2 C¯ ∆S¯ = V dT (see video 7.2) T1 T Z 200 K ¯ 10 1000 K ∆S = R − 2 dT 100 K T T ! 200 K 1000 K200 K ∆S¯ = R 10 ln + 100 K T 100 K 1000 K 1000 K ∆S¯ = R (10 · ln 2) + − = R ((10 · 0:7) + (5 − 10)) 200 K 100 K ∆S¯ = R (7 − 5) = 8 J · mol−1 · K−1 (2) = 16 J · mol−1 · K−1 3. The following equation gives S in terms of the partition function, Q, @ ln Q S = kBT + kB ln Q @T N;V The partition function for a monatomic ideal gas where all of the atoms are in their ground electronic state is given by, 1 2πmk T 3N=2 Q = B V N gN N! h2 e1 Derive an expression for the molar entropy, S¯ for this system. You may need Stirling's approximation, ln X! = X ln X − X. h 3=2 ¯ i ¯ 3 2πmkBT V ge1 (a) S = R + R ln 2 2 h NA h 3=2i ¯ 3 2πmkBT (b) S = 2 R + R ln h2 h ¯ i ¯ 5 2πmkBT V ge1 (c) S = R + R ln 2 2 h NA h 3=2 ¯ i ¯ 7 2πmkBT V ge1 (d) S = R + R ln 2 2 h NA h 3=2 ¯ i ¯ 5 2πmkBT V ge1 (e) S = R + R 2 2 h NA h 3=2 i ¯ 3 2πmkBT (f) S = 2 R + R ln h2 − R ln NA h 3=2 ¯ i ¯ 5 2πmkBT V ge1 (g) S = R + R ln 2 2 h NA Answer: " #N qN 1 2πmk T 3=2 Q = = B V g N! N! h2 e1 3 @ ln Q 3 1 ln Q = N ln T + ln (terms not depending on T ) −! = N 2 @T N;V 2 T 8 9 " 3=2 #N 3 1 < 1 2πmkBT = Used in the formula for S: S = kBT N + kB ln 2 V ge1 2 T :N! h ; Manipulating: " 3=2 # 3 2πmkBT S = NkB + NkB ln 2 V ge1 + kB ln 1 − kB ln N! 2 h (=0) Converting to per mole with N ! NA and NAkB = R: " # 3 2πmk T 3=2 S¯ = R + R ln B V¯ g − k ln N ! 2 h2 e1 B A Applying Stirling's approximation, ln N! = N ln N − N, to the last term: −kB ln NA! = −kBNA ln NA + kBNA = −R ln NA + R " # 3 2πmk T 3=2 S¯ = R + R ln B V¯ g − R ln N + R 2 h2 e1 A " 3=2 ¯ # ¯ 5 2πmkBT V ge1 S = R + R ln 2 2 h NA 4. Starting with the total derivative of the entropy as a function of T and P , @S @S dS = dT + dP; @T P @P T the definition of the first law for a reversible process, dU = δqrev + δwrev and the the second law, δq dS = rev T derive an expression for @S : @P T @S 1 @H (a) @P T = T @P T − V @S 1 @H (b) @P T = P @T P − V @S 1 @H (c) @P T = T @P T + V @S @H (d) @P T = T @P T − V @S Cp (e) @P T = T @S @H (f) @P T = T @P T + V Answer: First, write the total differential of H as a function of T and P : @H @H dH = dT + dP @T P @P T We know from week 5 that dH = T dS + V dP . Equate this to the total differential above: @H @H T dS + V dP = dT + dP @T P @P T Rearrange terms to obtain an equation for dS: @H @H T dS = dT + dP − V dP @T P @P T @H @H T dS = dT + − V dP @T P @P T 1 @H 1 @H dS = dT + − V dP T @T P T @P T Now write the total differential of S as a function of T and P : @S @S dS = dT + dP @T P @P T Compare the total differential to the other equation we obtained for dS: 1 @H 1 @H dS = dT + − V dP T @T P T @P T Equating the coefficients of dT from the two equations above gives, @S 1 @H = @T P T @T P @S C = P @T P T @H where we use CP = @T P to obtain an equation we've seen before. Equating the coefficients of dP gives, @S 1 @H = − V @P T T @P T 5. The molar constant volume heat capacity of ethylene is, 6085:929 K 822; 826 K2 C (T )=R = 16:4105 − + V T T 2 Use this equation to determine the change in entropy, ∆S, of one mole of ethylene heated from 300 K to 600 K at constant pressure. Assume ethylene behaves ideally (other than it's temperature-dependent heat capacity). (a) 742 J·K−1 (b) 2.689 ×1025 J·K−1 (c) 44.51 J·K−1 (d) 5.353 J ·K−1 (e) 47.33 J ·K−1 Answer: Assuming that ethylene behaves ideally, we can find the molar heat capacity at constant pressure using the following equation: CP − CV = R CP = R + CV CP =R = 1 + CV =R Now insert the equation for the molar heat capacity, 6085:929 K 822; 826 K2 C =R = 1 + 16:4105 − + P T T 2 Now we can calculate ∆S¯: Z T2 nC ∆S¯ = P dT T1 T Z 600 K 17:4105 6085:929 K 822; 826 K2 = nR − 2 + 3 dT 300 K T T T 6085:929 K 411; 413 K2 600 K = nR 17:4105 ln T + − 2 T T 300 K = nR(5:353) = (1 mol)(8:3145 J · K−1 · mol−1)(5:353) = 44:51 J · K−1 ¯ ¯ ¯ 6. For carbon monoxide, CO, the residual entropy defined as Sresidual = Scalc − Sexp is ::: (a) positive (b) negative (c) zero Answer: Carbon dioxide has a very weak dipole moment. As the temperature of solid carbon dioxide approaches zero kelvin, some of the dipoles will tend to be trapped in random orientations before they are able to line up in an energetically favorable way. Each carbon dioxide molecule will settle into the lattice with either dipole up, or dipole down. In this situation the lowest entropy state attainable with W = 1 is not attainable. The molecules are trapped and can not ”flip” into their lowest energy state because there is not enough energy near 0 K for this to happen. Thus, the magnitude of the experimental entropy of CO will be found to be smaller than that which is calculated because the reference experimental entropy near 0 K will not be as near to 0 J·K−1 as is assumed in calculations (so less input heat will be required to reach the entropy associated with a non-zero temperature). 7. The general formula for entropy, @lnQ S = kBlnQ + kBT @T N;V was discussed in the lecture videos of week 6. In the lecture videos of week 4, 2πmk T 3=2 q(V; T ) = B V · g h2 el was presented as the partition function for a monatomic ideal gas. Which of the following is the correct expression for the change in molar entropy of an ideal gas going isobarically from (T1;V1;P1) to (T2;V2;P1). (a) 3 Rln T2 + 3 ln V2 2 T1 2 V1 (b) 5 Rln T2 + Rln V2 2 T1 V1 (c) 3 Rln T2 + Rln V2 − 2R 3N 2 T1 V1 4T1T2 (d) Rln V2 V1 (e) 3 Rln T2 + Rln V2 2 T1 V1 (f) 3 Rln T2 2 T1 Answer: The general plan to derive an expression for the change in molar entropy of an ideal gas is to start first with the expression for the partition function Q = qN =N!, determine the derivative with respect to temperature, substitute relevant terms into the expression for the entropy given in the question, and then determine ∆S by subtracting the entropy of the latter state from that of the former; thus, " #N 1 2πmk T 3=2 Q = B V · g : N! h2 el and therefore " # 2πmk T 3=2 lnQ = Nln B V · g − lnN!: h2 el If we designate the partition function for the initial state of the system as Q1 and that for the final state of the system as Q2, 3 T2 V2 lnQ2 − lnQ1 = Nln + Nln : 2 T1 V1 Taking the derivative of the general logarithm of Q with respect to T gives, @lnQ 3N = @T N;V 2T We can now use the general formula for the entropy, @lnQ S = kBlnQ + kBT @T N;V to determine the entropy difference between the initial state and final state, ∆S¯ = ¯ ¯ S2 − S1: ¯ @lnQ2 @lnQ1 ∆S = kBlnQ2 + kBT2 − kBlnQ1 − kBT1 @T N;V @T N;V 3 T2 V2 3N 3N = NkBln + NkBln + kBT2 − kBT1 2 T1 V1 2T2 2T1 3 T V = Rln 2 + Rln 2 2 T1 V1 where the final line derives from taking N = NA.
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