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 KThis 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

The vapor pressure of liquid benzene is given by ln p* = 15 – 1280/T

1.	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 KThis 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» 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.

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