Question

The vapor pressure of trichloromethane (chloroform) is 41.5 Torr at -7.6 ∘C. Its enthalpy of vaporization is 29.2 kJ⋅mol−1. Calculate its normal boiling point. Express your answer using three significant figures.

Answer 1

Using Classius Clapeyron equation let us solve this
question

ln [P2/P1] = -dH/R [1/T2 -1/T1]

following datas are given
initial pressueP1= 41.5 Torr
initial temperature T1= -7.6C =273 +[-7.6C] = 265.4 K

final temperatute is the normal boiling point of Chloroform
T2= ??

Final pressure P2 = 760 torr
gas constan R= 8.314 j/mol-k
enthalpy of vaporization dH = 29200 j/mol
Plug in these numbers in the above equation

ln [P2/P1] = -dH/R [1/T2 -1/T1]

ln[760 torr /41.5 Torr = -29200 j/mol/8.314 j/mol-k[[1/T2 -1/T1]
let us solve both side of the equation using calculator
2.9076 =-3512.15[[1/T2 -1/T1]
[1/T2 -1/T1]=2.9076/-3512.15
= -0.000828

1/T2 = -0.000828 +1/265.4 =0.00294

T2 = 1/0.00294 = 340.14 K

this is the normal boiling point of chloroform
= 340.14 K
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