Question

A certain liquid has a vapor pressure of 92.0 Torr at 23.0 °C and 228.0 Torr at 45.0 °C. Calculate the value of AHjap for this liquid. AHjap = KJ/mol Calculate the normal boiling point of this liquid. boiling point: : . . c

Answer 1

Here you go:

R * We can use the Clusius Clapeyson equation to determine this: Fireff part The Clausius. Clapeyson equationis Papan Animno (-) where P Praze pressures of a vapons at temperatures Ty In respectively. - Dhirap enthalpy of raporization - Rainivensal gas constant 38.314 J/molk. Given. P,= 92.0 Torr, T, 2 23°- 3 296k - P22228.0Torr, T2 = 45° - 318k .:. We get; In 220 - pe le zone) - or 0.90752 - Ahivap x (296-318) os, Atrap • 32.283 *103 kJ/mol. - Enthalpy of vaporizationa 32.283X103 1 kJ/mol. 2nd Part! We know that the temperature at which the pressure of liquid the pressure of the atmosphere, that temperature is called the boiling point. - The atmospheric pressure 1P2 2 760 Toss. Again.com 760 2 - 32-283X1035 ot, T22 352.78 K-701.7oc 8.314 318 x 296 - 8.2448 ( 79 lo

The boiling point is 79.7 degree Celsius