FIU CHM 3410 - Homework 7 Solution on Physical Chemistry (1) (3 pages)

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Homework 7 Solution on Physical Chemistry (1)



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Homework 7 Solution on Physical Chemistry (1)

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3
School:
Florida International University
Course:
Chm 3410 - Physical Chemistry I
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Solutions 7 1 For an ideal solution A A RT lnxA 8 3145 J K 1 mol 1 353 25 K ln 0 30 3 536 kJ mol 1 aA pA pA A xA pA pA pA A xA pA At the normal boiling point pA 1 atm 760 Torr pA 0 93 0 30 760 Torr 212 Torr mix S nR x A ln x A xB ln xB xB 1 x A mix S nR x A ln x A 1 x A ln 1 x A d mix S x nR ln x A 1 ln 1 x A 1 nR ln A dx A 1 x A d mix S x 0 i e at A 1 The maximum mixS is reached when dx A 1 x A sh is ar stu ed d vi y re aC s o ou urc rs e eH w er as o co m 2 Therefore xA xB 0 5 a hexane and heptane should be mixed each with the mole fraction of 0 5 b to have the same chemical amounts of hexane and heptane their mass ratio should be equal to the ratio of their molar masses m hexane m heptane M C6H14 M C7H16 86 17 g mol 1 100 2 g mol 1 0 86 3 For ideal mixing mixG nRT x A ln x A xB ln xB xA 0 25 MCH xB 0 75 THF mixG ideal 4 0 mol 8 3145 J K 1 mol 1 303 15 K 0 25ln 0 25 0 75ln 0 75 5 67 kJ GE RTx 1 x 0 4857 0 1077 2x 1 0 0191 2x 1 2 x 0 25 GE 0 1021RT 0 257 kJ mol 1 note that GE is a molar quantity mixG real mixG ideal nGE 5 67 kJ 4 0 mol 0 257 kJ mol 1 4 6 kJ Th 4 Using the mole fraction in the vapor phase we can calculate partial vapor pressure of each component yA pA pA pB pA 101 3 kPa 0 314 pA 0 314 101 3 kPa 31 8 kPa pB p pA 101 3 kPa 31 8 kPa 69 5 kPa aA pA pA 31 8 kPa 73 0 kPa 0 436 aB pB pB 69 5 kPa 92 1 kPa 0 755 A aA xA 0 436 0 220 1 98 B aA xB 0 755 1 0 220 0 968 https www coursehero com file 11467744 Homework 7 Solution on Physical Chemistry 1 bi 2 zi 2 i b For an MpXq salt b b pb b b b qb b so 1 b I pz 2 qz 2 i 2 b 2 5 I I I K3 Fe CN 6 I KCl I NaBr 1 2 3 3 b K3 Fe CN 6 b b NaBr b 6 0 040 0 03 0 05 0 320 b KCl b 6 Th sh is ar stu ed d vi y re aC s o ou urc rs e eH w er as o co m 7 Denote the mole fraction of methylbenzene as xM and mole fraction of dimethylbenzene as xD xD 1 xM According to Raoult s law pM x M pM pD xDpD When the mixture the mixture bolis at 0 5 atm the total vapor pressure should be 0 50 atm 380 Torr p pM pD xMpM xDpD xMpM 1 xM pD xM p pD pM pD 380 Torr 150 Torr 400 Torr 150 Torr 0 92 xD 1 xM 0 08 Vapor composition according to Dalton s law https www coursehero com file 11467744 Homework 7 Solution on Physical Chemistry yM pM p xMpM p 0 92 400 Torr 380 Torr 0 968 yD 1 yM 0 032 8 At the boiling point the total vapor pressure of solution should be 1 atm 760 Torr For an ideal solution according to Raoult s law p pA pB xApA xBpB 0 6589 957 Torr 1 0 6589 379 5 Torr 760 01 Torr 760 Torr Therefore the solution is very close to ideal yA pA p xApA p 0 6589 957 Torr 760 Torr 0 830 yB 1 yA 0 170 sh is ar stu ed d vi y re aC s o ou urc rs e eH w er as o co m 9 Retrace the argument given in the derivation of van t Hoff s equation for osmotic pressure in the text At equilibrium A p A x A p For a real solution RT ln a A p A p RT ln a A A p Therefore p p p p The osmotic coefficient is defined as Vm RT ln a A xA ln a A xB nB nA because xB n RT B xA nA Vm nA RT nB Vm nA V V RT nB nB B molar concentration of solute Therefore we obtain V RT B Th https www coursehero com file 11467744 Homework 7 Solution on Physical Chemistry Powered by TCPDF www tcpdf org Vm dp RT ln a A Vm dp Vm so Vm RT p Vm dp RT ln a A For an incompressible solution p


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