.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
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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
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