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Homework Week 8 Statistical MolecularThermodynamics University of Minnesota Homework Week 8 1. By comparing the formal derivative of G with the derivative obtained taking account of the first and second laws, use Maxwell relations to determine which of the following equations is valid. Start with the definition of the Gibbs energy, G = H − T S: @T @P (a) − @V S = @S V @V @T (b) @S P = @P S @S @P (c) @V T = @T V @S @V (d) − @P T = @T P Answer: Given that G = H − TS, using the first and second laws we can write, dG =d(H − TS) =dH − T dS − SdT =d(U + PV ) − T dS − SdT =dU + (P dV + V dP ) − T dS − SdT =(T dS − P dV ) + (P dV + V dP ) − T dS − SdT = − SdT + V dP and we can write the total derivative of G, @G @G dG = dP + dT @P T @T P in terms of its natural independent variables. We can compare terms in the two equa- tions and see that @G = V @P T and @G = −S @T P If we now take the cross derivatives of this result we have @V @S = − @T P @P T ¯ ◦ 2. Calculate ∆vapG for benzene at 77 C and 1 bar given that the molar enthalpy of vaporization of benzene at its boiling point of 80.09◦ C and 1 bar is 30.72 kJ·mol−1. ¯ ¯ Assume that the values of ∆vapH and ∆vapS are constant over the temperature range considered here. ¯ Hint: You will first need to calculate ∆vapS from the given information. (a) ∆Gvap(350 K) = −0:6512 kJ/mol (b) ∆Gvap(350 K) = 0:2687 kJ/mol (c) ∆Gvap(350 K) = 0:3414 J/mol (d) ∆Gvap(350 K) = 100:2 kJ/mol (e) ∆Gvap(350 K) = 0:4414 kJ/mol 4 (f) ∆Gvap(350 K) = −3:04 × 10 kJ/mol 4 (g) ∆Gvap(350 K) = 3:04 × 10 kJ/mol Answer: From lecture video 8.2, we know that ∆G = ∆H − T ∆S. For a given ∆H and ∆S (assuming they are independent of T ) we can compute ∆G at different temperatures. ¯ ¯ We are given only ∆vapH, so we first must determine ∆vapS. At the boiling point (and 1 bar) the liquid and gas phases are in equilibrium and thus ∆Gvap = 0. We can use ¯ this result to solve for ∆vapS as follows: ∆Gvap =∆Hvap − T ∆Svap 0 =30:72 kJ/mol − [(80:09 + 273:15)K]∆Svap 30:72 kJ/mol ∆S = vap (80:09 + 273:15) K −1 −1 −1 −1 ∆Svap =0:08697 kJ · mol · K = 86:97 J · mol · K o Now we can compute ∆Gvap at 77 C (which is 350.15) K: −1 −1 ∆Gvap(350:15 K) = 30:72 kJ/mol − 350:15 K · 0:08697 kJ · mol · K ∆Gvap(350:15 K) = 0:2687 kJ/mol (result carries all digits from the above calculation) 3. In lecture video 8.5, we derived for a thermoelastomer: @U @S = T + f: @L T @L T For a weight suspended on a tense rubber band and resting on a scale, assuming that the rubber band acts as an ideal elastomer and does not change volume when stretched, which of the following statements is TRUE if you increase the temperature of the rubber band? (a) The force exerted by the rubber band, f, will not change. (b) The force exerted by the rubber band, f, will decrease. (c) The force exerted by the rubber band, f, will increase. Answer: In lecture video 8.5, we learned that the Helmholtz energy for the rubber band can be expressed as dA = −SdT + fdL: We can write the total derivative of A in terms of the natural, independent variables T and L as @A @A dA = dT + dL: @T L @L T And thus if we compare terms, @A = −S @T L and @A = f: @L T If we determine the cross derivatives, i.e. @ @A @S = − @L @T L T @L T and @ @A @f = : @T @L T L @T L These two mixed partial derivatives must be equal to one another. And, we know that @S the entropy decreases with increasing length, so − @L T must be a positive value. @f Thus, @T L must be positive, which means f must increase with increasing tempera- ture. (One can further manipulate the equations to show that the increase should be linear in temperature, but we will not do that here.) A final, intuitive way to see the answer to this question is to recognize that increasing temperature will favor lengths having greater entropy, i.e., shorter lengths (where the polymer chains can twist and turn and are not pulled straight). So, there will be increased resistance to losing entropy by stretching, and this is reflected in a higher force constant. 4. One of the Maxwell relations derived in the lecture videos is, @P @S = : @T V @V T Using this relationship, find the correct expression for the isothermal change in the entropy of a gas that obeys the van der Waals equation of state when you change the ¯ ¯ molar volume from V1 to V2. ¯ (a) ∆S¯ = R ln V2−b V¯1−b ¯ (b) ∆S¯ = R ln V2 V¯1 ¯ ¯ V2−b 1 1 (c) ∆S = R ln ¯ + 3aR ¯ 3 − ¯ 3 V1−b V2 V1 ¯ (d) ∆S¯ = R ln V2−b − aR V¯ 2 − V¯ 2 V¯1−b 2 1 Answer: From lecture video 2.2, we know that the van der Waals equation of state is, RT a P = − V¯ − b V¯ 2 and therefore @P R = ¯ : @T V¯ V − b We can substitute into the Maxwell relation above to obtain, @S R ¯ = ¯ @V T V − b We can separate the variables and integrate this equation to yield R dS = dV V¯ − b ¯ V2 − b ∆S = R ln ¯ : V1 − b 5. Write the total derivative of H(P; T ) as @H @H dH = dP + dT (1) @P T @T P and determine another expression, derived with the use of Maxwell relations, that expresses dH in terms of readily measured quantities. (Hint: equations involving the differential of the Gibbs free energy are likely to be the most useful to you in seeking partial derivative equalities.) @V (a) dH = V − T @T P dP + CP dT @T (b) dH = T − V @V P dP + CP dT @V (c) dH = V − T @T V dV + CV dT @V (d) dH = V − T @T P dP + CV dT @V (e) dH = V + T @T V dV + CP dT @P (f) dH = P − T @T V dV + CP dT To solve this, we start with the definition of the Gibbs energy, G = H + TS and differ- entiate this with respect to pressure at constant temperature to obtain an expression @H that contains the first term of the total derivative as written, @P T . We can then @V re-arrange this term using Maxwell relations to include V and @T P . The second term of the original total derivative we can simply recognize as the definition of the constant pressure heat capacity. Taking the derivative of the expression for G, @G @H @S = − T (2) @P T @P T @P T We know we can combine the first and second laws to write, dG =d(H − TS) =dH − T dS − SdT =d(U + PV ) − T dS − SdT =dU + (P dV + V dP ) − T dS − SdT =(T dS − P dV ) + (P dV + V dP ) − T dS − SdT = − SdT + V dP: From writing the total derivative of G, @G @G dG = dP + dT @P T @T P in terms of its natural independent variables, we see that @G = V @P T and @G = −S: @T P If we now take the cross derivatives of this result we have @V @S = − @T P @P T Now we can substitute these results into Eq. 2 to obtain, @H @V V = + T @P T @T P which we can solve to get @H @V = V − T : @P T @T P This result can now be substituted into the original total differential for H to yield, @V dH = V − T dP + CP dT @T P @H using the definition that CP = @T P . 6. Determine the Gibbs-Helmholtz equation for A, i.e determine, @(A=T ) @T V Hint: It was shown in lecture video 7.1 that @S C = V @T V T G (a) − T 2 − H H (b) − T 2 h @(S=V ) i (c) @T V U (d) − T 2 (e) U (f) G (g) none of the above Answer: We can begin with the definition A = U − TS and determine the derivative after division by T , @ A @ U = − S @T T @T T and therefore, @(A=T ) U 1 @U @S = − 2 + − @T V T T @T V @T V From the definition of CV we know 1 @U C = V T @T V T and using the hint given in the question we can write @(A=T ) U CV CV U = − 2 + − = − 2 @T V T T T T 7. For the following spontaneous endothermic reaction, Ba(OH)2(s) · 8 H2O + 2 NH4NO3(s) ! Ba(NO3)2(s) + 10 H2O(l) + 2 NH3(aq) −1 at 300 K, ∆rH = 60 kJ and ∆rS = 400 J K . What is the change in the Gibbs free energy, ∆rG, for this reaction at 300 K? (a) 0 kJ (b) 30 kJ (c) -30 kJ (d) 60 kJ (e) -60 kJ (f) 120 kJ (g) -120 kJ (h) none of the above Answer: We can apply the formula ∆G = ∆H − T ∆S −1 ∆rG = 60 kJ − 300K · 0:4 kJ K = −60 kJ ¯ ◦ 8.
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