Question

A certain liquid has a vapor pressure of 92.0 Torr at 23.0°C and 306.0 Torr at 45.0°C. Calculate the value of AH ap for this liquid. kJ/mol AHÇap Calculate the normal boiling point of this liquid. boiling point:

Answer 1

Auto ex Piaga toor Ti =23°c = 296k P2 = 306 for T2=48°C = 318k R= 8.314 J Kt molt using stays clauerun- claeson equation In 2 alluck It tol en e sta - olen pass she] 064X16 1201 - Arab 8314 Jetmolt | 064X163) AHaab = 120188:3145 molt 0064 X103 Harap = 156017.4063 J/m1 = 156.017 KJ K1/mol Normal botting point → cohen presser u 760 toor Pizga toor To= 296 wing claux'y daypron equation P2 = 760 tor T2=7 Atrob= 156017 - 40 J/mol. R=2314 J K molt che non e te thote In (760 ) = 156017.40 I foot 8.314 X ki mott 1 9.10 = 18765.62 K 13.37% 103- pelaX 104 = 3.378103- 1

te 3:37 x16 3:112700-3)*7 23.958X103 ky Tz z 1 x103k 3:258 T2 = 306.93K T2 = 33.93600 this in the normal boiling point