FIU CHM 3410 - Homework 6 Solution on Physical Chemistry (1) (4 pages)

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Homework 6 Solution on Physical Chemistry (1)



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Homework 6 Solution on Physical Chemistry (1)

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Pages:
4
School:
Florida International University
Course:
Chm 3410 - Physical Chemistry I

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Solutions 6 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 volume of the laboratory is V 5 0 5 0 3 0 75 0 m3 a water p 24 Torr n pV RT 24 Torr 75 0 103 L 62 364 L Torr K 1 mol 1 298 15 K 96 855 mol m nM 96 855 mol 18 02 g mol 1 1745 3 g 1 7453 kg b benzene p 98 Torr n pV RT 98 Torr 75 0 103 L 62 364 L Torr K 1 mol 1 298 15 K 395 49 mol m nM 395 49 mol 78 11 g mol 1 30890 8 g 30 89 kg c mercury p 1 7 10 3 Torr n pV RT 1 7 10 3 75 0 103 L 62 364 L Torr K 1 mol 1 298 15 K 0 00686 mol m nM 0 00686 mol 200 59 g mol 1 1 376 g 2 a p1 10 Torr T1 358 95 K and p2 40 Torr T2 392 45 K Use the Clausius Clapeyron equation to calculate vapH p2 vap H 1 1 ln p1 R T1 T2 vap H p3 vap H 1 1 ln p1 R T3 T2 R ln p3 p1 T3 1 1 T1 H vap R ln p2 p1 1 T1 1 T2 vapH 8 3145 J K 1 mol 1 ln 40 10 1 358 95 K 1 392 45 K 48 469 kJ mol 1 b p3 760 Torr T3 Th T3 1 1 358 95 K 8 3145 J K 1 mol 1 ln 760 10 48 469 103 J mol 1 489 47 K 216 32 C c vapS vapH Tb 48 469 103 J mol 1 489 47 K 99 0 J K 1 mol 1 3 fusH 2 292 kJ mol 1 T 234 3 K fusV 0 517 cm3 mol 1 p 101 325 kPa The additional pressure on the bottom of mercury column p p p gh 13 6 103 kg m3 9 81 m s 2 10 m 1334160 Pa https www coursehero com file 11467740 Homework 6 Solution on Physical Chemistry fus H T p p ln fusV T p p fusV T ln T fus H ln T T 1334160 Pa 0 517 10 6 m3 mol 1 2 292 103 J mol 1 3 0 10 4 T 234 3 K exp 3 0 10 4 234 37 K 4 We assume that a liter of seawater contains roughly 1000 g of water Then nwater 1000 g 18 02 g mol 1 55 5 mol nsolutes 2 0 50 mol 1 00 mol xwater nwater nwater nsolutes 55 5 56 5 0 982 pwater xwater pwater 0 982 2 338 kPa 2 296 kPa sh is ar stu ed d vi y re aC s o ou urc rs e



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