Question

.6 The vapour pressure of a liquid at 8 °C is 2.7 kPa. Its enthalpy of vaporization is constant and equal to 42700 kJ/kmol. Take R = 8.314 kJ/kmol.K. The temperature (in °C) at a vapour pressure of 13.5 kPa is

Answer 1

The temperature (in °C) at a vapour pressure of 13.5 kPa is 34.98⁰C

Explanation :

Given ,

P₁ = 2.7

P₂ = 13.5

T₁ = 8°C= 273+8 = 281 K

ΔH = 42700

R = 8.314

T₂ = ?

we know that,

ln(P₂/P₁) = ΔH/R   x  [1/T₁ - 1/T₂]

=> ln(13.5/2.7) = 42700/8.314 [1/281 - 1/T₂]

=> ln5 = 5136[1/281 - 1/T₂]

=> 1.6 = 5136[1/281 - 1/T₂]

=> 1/T₂ = 1/281 - 1.6/5136

=> T₂ = 307.9k = 307.9 - 273 = 34.98 ⁰C

Hence the temperature (in °C) at a vapour pressure of 13.5 kPa is 34.98⁰C