Question

Predict the boiling point of water at a pressure of 1.5 atm.

How to solve this question by using Clausius-clapeyron equation?

Answer 1

100.8 °C

Step-by-step explanation:

The Clausius-clapeyron equation is:

`ln(P_(1) )/(P_(2)) =`-Δ
`(H_(vap))/(r) ((1)/(T_(2))-(1)/(T_(1))  )`

Where 'ΔHvap' is the enthalpy of vaporization; 'R' is the molar gas constant (8.314 j/mol); 'T1' is the temperature at the pressure 'P1' and 'T2' is the temperature at the pressure 'P2'

Isolating for T2 gives:

`T_(2)=((1)/(T_(1)) -(Rln(P_(2))/(P_(1)) )/(Delta H_(vap))`

(sorry for 'deltaHvap' I can not input symbols into equations)

thus T2=100.8 °C