Question

The vapor pressure of liquid benzene is given by ln p* = 15 – 1280/T, and that of vapor benzene is given by ln p* = 23 – 2560/T, what is the boiling point of benzene?

(a) 160 K

(b) 200 K

(c) 240 K

(d) 300 K

This question was posed to me by my school principal while I was bunking the class.

I'd like to ask this question from Modeling and Predicting Vapor Pressure topic in division Real Gases: Equations of State, Single Component Two Phase Systems, Saturation & Condensation of Basic Chemical Engineering

Answer 1

The correct option is (a) 160 K

To explain: At boiling point pressure and temperature of both the phases will be same, => 23 – 2560/T = 15 – 1280/T, => 8 = 1280/T, => T = 160K.