# FIU CHM 3410 - Homeowrk 5 Solution on Physical Chemistry (1) (3 pages)

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## Homeowrk 5 Solution on Physical Chemistry (1)

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- School:
- Florida International University
- Course:
- Chm 3410 - Physical Chemistry I

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Homework 5 Solutions sh is ar stu ed d vi y re aC s o ou urc rs e eH w er as o co m 1 The substance with the lower molar Gibbs energy is the more stable therefore rhombic sulfur is more stable The application of pressure tends to favor the substance with the smaller molar volume higher density therefore rhombic sulfur becomes even more stable relative to monoclinic sulfur as the pressure increases G V p G is less positive for rhombic sulfur than for monoclinic sulfur so relative to monoclinic sulfur the Gibbs energy of rhombic sulfur becomes more negative For the transition S rhombic S monoclinic Gm 0 33 kJ mol 1 Sm 32 6 31 8 0 8 J K 1 mol 1 Gm Hm T Sm We assume that Hm and Sm are roughly independent of temperature We need the change in Gm to be 0 33 kJ mol 1 as a result of the change in temperature Tf Ti Sm 0 33 kJ mol 1 Tf 330 J mol 1 0 8 J K 1 mol 1 298 15 K 710 65 K 2 dGm SmdT at constant pressure For this relatively small temperature change we may assume that Sm is constant Then we may write after integration Tf T i dGm Tf T Sm dT i Gm Sm Tf Ti Sm T 173 3 J K 1 mol 1 30 K 5 2 kJ mol 1 3 At these pressures water vapor may be considered a perfect gas therefore piVi pfVf and pf pi Vi Vf Gm 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 1 rG H r2 T T p T T2 G rG r H d d r r H 2 dT T1 T T T 1 1 r G2 r G1 r H T2 T1 T2 T1 1 1 T r G2 r HT2 r G1 2 T1 T2 T1 Th 4 Gibbs Helmholtz equation T 1 dT T2 rH rH 298 K 2 fH CO2 g 2 fH CO g 2 393 51 2 110 53 https www coursehero com file 11467753 Homework 5 Solution on Physical Chemistry T2 565 96 kJ mol 1 rG1 rG 298 K 2 fG CO2 g 2 fG CO g 2 394 36 2 137 17 514 38 kJ mol 1 rG2 rG 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 1 5 Real gas Gm real Gm RT ln f p f p 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 Gm real Gm perfect 8 3145 J K 1 mol 1 200 K ln 0 72 546 3 J mol 1 0 55 kJ mol 1 sh is ar stu ed d vi y re aC s o ou urc rs e eH w er as o co m 6 Consider differential dH H U pV dH dU pdV Vdp dU TdS pdV the fundamental equation dH TdS pdV pdV Vdp TdS Vdp H is a state function so dH is an exact differential The mathematical criterion of exact differential g h y x x y T V p S S p if df gdx hdy then Using this we can conclude that Consider differential dA A U TS dA dU TdS SdT TdS pdV TdS SdT pdV SdT Therefore Th 7 A Maxwell relation S p V T T V For a perfect gas S nR V T V p S T V V T pV nRT dV V S R lnV dS nR p nR T V V S V dV dS nR S0 V0 V p nRT V 8 a The gaseous sample expands b The sample contracts but remains gaseous because 320 K is greater than the critical temperature c The gas contracts and forms a liquid like https www coursehero com file 11467753 Homework 5 Solution on Physical Chemistry substance without the appearance of a discernible surface As the temperature lowers further to the solid phase boundary line solid carbon dioxide forms in equilibrium with liquid At 210 K the sample has become all solid d The solid expands slightly as the pressure is reduced and sublimes when the pressure reaches about 5 atm e The gas expands as it is heated at constant pressure 10 p p fus H fusV ln T T ln T fusV p p T fus H https www coursehero com file 11467753 Homework 5 Solution on Physical Chemistry Powered by TCPDF www tcpdf org fusV M l M 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 1 ln 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 0 011 T T 1 0115 T 278 65 K T 281 86 K 8 7 C Th ln p p vapH R 1 T 1 T p 1 atm 101 3 kPa p 50 0 kPa T 365 7 K and T 388 4 K ln 101 3 50 0 vapH 8 31451 J K 1 mol 1 1 365 7 K 1 388 4 K 0 706 1 922 10 5 J 1 mol vapH vapH 36 7 kJ mol 1 sh is ar stu ed d vi y re aC s o ou urc rs e eH w er as o co m 9

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