Homework 5 Solutions1. The substance with the lower molar Gibbs energy is the more stable; therefore, rhombicsulfur is more stable. The application of pressure tends to favor the substance with thesmaller molar volume (higher density); therefore, rhombic sulfur becomes even more stablerelative to monoclinic sulfur as the pressure increases.ΔG ≅ VΔpΔG is less positive for rhombic sulfur than for monoclinic sulfur, so relative to monoclinicsulfur, the Gibbs energy of rhombic sulfur becomes more negative.For the transition S(rhombic) → S(monoclinic), ΔGm = +0.33 kJ mol-1ΔSm = (32.6 – 31.8) = 0.8 J K-1 mol-1ΔGm = ΔHm – TΔSmWe assume that ΔHm and ΔSm are roughly independent of temperature. We need the changein ΔGm to be –0.33 kJ mol-1 as a result of the change in temperature:-(Tf – Ti)ΔSm = -0.33 kJ mol-1Tf = (330 J mol-1)/(0.8 J K-1 mol-1) + 298.15 K = 710.65 K2. dGm = -SmdT (at constant pressure)For this relatively small temperature change we may assume that Sm is constant.Then we may write after integration€ dGmTiTf∫= −Sm( )dTTiTf∫ ΔGm = -Sm(Tf – Ti) = - SmΔT = -173.3 J K-1 mol-1 × 30 K = -5.2 kJ mol-13. At these pressures, water vapor may be considered a perfect gas; thereforepiVi = pfVfand pf/pi = Vi/VfΔGm = RT ln(pf/pi) = RT ln(Vi/Vf) = 8.315 J K-1 mol-1 × 473 K × ln(300 mL/100 mL)= +4.3 kJ mol-14. Gibbs-Helmholtz equation:€ ∂∂TΔrGT      p= −ΔrHT2€ dΔrGT      = −ΔrHT2dT€ dΔrGT      T1T2∫= −ΔrHdTT2T1T2∫€ ΔrG2T2−ΔrG1T1= ΔrH1T2−1T1      € ΔrG2= ΔrHT21T2−1T1      + ΔrG1T2T1ΔrH = ΔrH∅(298 K) = 2ΔfH(CO2,g) - 2ΔfH(CO,g) = 2×(-393.51) – 2×(-110.53) =https://www.coursehero.com/file/11467753/Homework-5-Solution-on-Physical-Chemistry/This study resource wasshared via CourseHero.com-565.96 kJ mol-1ΔrG1 = ΔrG∅(298 K) = 2ΔfG(CO2,g) - 2ΔfG(CO,g) = 2×(-394.36) – 2×(-137.17) =-514.38 kJ mol-1ΔrG2 = ΔrG∅(375 K) = (-565.96 kJ mol-1)×(375 K)×{1/(375 K) – 1/(298 K)} + (-514.38 kJ mol-1)×(375 K / 298 K) = -501.05 kJ mol-15. Real gas: Gm(real) = Gm∅ + RT ln(f/p∅) f = φp Perfect gas Gm(perfect) = Gm∅ + RT ln(p/p∅)Gm(real) - Gm(perfect) = RT{ln(f/p∅) - ln(p/p∅)} = RT ln(f/p) = RT ln φGm(real) - Gm(perfect) = (8.3145 J K-1 mol-1)×(200 K)×ln(0.72) = -546.3 J mol-1 ≈-0.55 kJ mol-16. Consider differential dH:H = U + pV dH = dU + pdV + VdpdU = TdS – pdV the fundamental equationdH = TdS – pdV + pdV + Vdp = TdS + VdpH is a state function, so dH is an exact differential. The mathematical criterion of exactdifferential: if df = gdx + hdy, then€ ∂g∂y      x=∂h∂x      yUsing this, we can conclude that€ ∂T∂p      S=∂V∂S      pConsider differential dA:A = U – TSdA = dU – TdS – SdT = TdS – pdV – TdS – SdT = –pdV – SdTTherefore,€ ∂p∂T      V=∂S∂V      T7. A Maxwell relation: € ∂S∂V      T=∂p∂T      VFor a perfect gas, pV = nRT p = nRT/V€ ∂p∂T      V=nRV€ ∂S∂V      T=nRV€ dS = nRdVV€ dSS0S∫= nRdVVV0V∫€ S ∝ R lnV8. (a) The gaseous sample expands. (b) The sample contracts but remains gaseous because320 K is greater than the critical temperature. (c) The gas contracts and forms a liquid-likehttps://www.coursehero.com/file/11467753/Homework-5-Solution-on-Physical-Chemistry/This study resource wasshared via CourseHero.comsubstance without the appearance of a discernible surface. As the temperature lowersfurther to the solid phase boundary line, solid carbon dioxide forms in equilibrium withliquid. At 210 K the sample has become all solid. (d) The solid expands slightly as thepressure is reduced and sublimes when the pressure reaches about 5 atm. (e) The gasexpands as it is heated at constant pressure.9. ln(p’/p) = (ΔvapH/R)(1/T – 1/T’)p’ = 1 atm = 101.3 kPa, p = 50.0 kPa, T = 365.7 K, and T’ = 388.4 Kln(101.3/50.0) = (ΔvapH/8.31451 J K-1 mol-1)(1/(365.7 K) – 1/(388.4 K))0.706 = 1.922×10-5 J-1 mol × ΔvapH ΔvapH = 36.7 kJ mol-110. € p = p*+ΔfusHΔfusVlnTT*€ lnTT*=ΔfusVΔfusHp − p*( )ΔfusV = M/ρ(l) – M/ρ(s) = (78.1074 g mol-1)/(0.879 g cm-3) – (78.1074 g mol-1)/(0.891 g cm-3) = 1.197 cm3 mol-1 = 1.197×10-6 m3 mol-1ln(T/T*) = (1000 atm – 1 atm) × (1.197×10-6 m3 mol-1) / (10.59 kJ mol-1)ln(T/T*) = (999×1.01325×105 Pa) × (1.197×10-6 m3 mol-1) / (10.59×103 J mol-1) = 0.011T/T* = 1.0115 T* = 278.65 K T = 281.86 K = 8.7°Chttps://www.coursehero.com/file/11467753/Homework-5-Solution-on-Physical-Chemistry/This study resource wasshared via CourseHero.comPowered by TCPDF