Question

The vapor pressure of Y is 4.590 atm at 330K and 9.180 atm at 380K. What are the standard free energies of vaporization at the two temperatures? Hint: vaporization reaction is Y(I) = Y(g). AG330º = kJ/mol AG380º = kJ/mol Assume that the standard entropy and enthalpy of vaporization of Y are independent of temperature and determine their values ASO = J/(mol-K) AH = kJ/mol What is the normal boiling point of Y in degrees Celcius? t = Ос What is the vapor pressure of Y at 410 K? po = atm

Answer 1

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Page 1 If Gas ja ideal in the vapoolization railion Y (2) = Y(A) "Then gibbs free Energy change DG = 2.303 HRT logement) nal R - 8-314 Jkt molt T = 330 R P2 = 9.180 atm P2 = 4.590 atm 16220 = 2.303 X 1 X 8-314 X 330R x log (4.5%) = 2.303 XI X 8.314 x 330 x log c) - 2.303 X1 X 8,314 x 330 x 0.300 = 1901.885 5 mol = 1.901 Rs mol-) AG 280 = 2.303 XI X 8.314 X 380 R X log (2.80) = 2.303 XI X 80314 380 x log (2) = 2303 X) X 8.314 X 380 X 0.3016 = 2190.0501 Smol/ = 2.190 KJ molt CS Scanned with CamScanner

Page 2 The molag Enthalpy c4 vaposúzatian: - Rin (P2) I A Hvap Г. Т. Given Vales "R" is the Gas Constant = 8.3144 Jx-'mol =) Initial tempegiature (1): 33012 final temperature (12): 380 K Vapan Pressure at temperature T (B): 465081.8 Pascal vapor pressure at temperatur T2 (P2): 930163.5 peral AHON - 8.3144 In (150163.5) i i 330 380 D Hvap = 14453.985 J/mol Duvap = 14.453 KJ/mol CS Scanned with CamScanner

Page 3 -D5° = AGP – DHO T osº - DHO – 06° Since, It is tempescatore independent there is no effect of T. Asº = 91° – ( DG7330 + $63380) 1 – 14.45 - (1.901 +2.140) = 14.45 - (4.091) = 10.359 Continue...n for the deteamination of nasimal boiling point Dhuap = Rin (3) onvqр - opas 2.303 R log (en) -t 14453.98 – 2.303 X 8.3148 dog (101325) CS Scanned with CamScanner 14650818 298

। 7373 T2 = -- = 0.00268-1 २.303 x 8.314 x log (4-5१) 1 4453.98 1222 x log (4-59) 14453,98 =) - - - १.२२२ x 0.6 6 18. + 0.000 268०१) 14453.98 ___ - - - ०.०००87554 + 0.0००२68097 CSScahnearwith Laro.oossos - 281.174 = 8-17 degse