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The vapor pressure of a substance is 0.0500 bars at 300 K. If Δ<sub>vap</sub>H = 32.6 kJ/mol, at what temperature will the substance have a vapor pressure of 0.200 bars?

Expert Solution

Step 1

Given data:

Pressure p₁ = 0.0500 bars

Enthalpy of vaporization $∆ H_{v a p}$ = 32.6 kJ/mol

Pressure p₂= 0.200 bars

Here the temperature T₁ = 300k

Solution:

The Temperature pressure relation for the Enthalpy of vapourization is determined by the following expression that is,

$\ln (\frac{p_{2}}{p 1}) = \frac{- ∆ H_{v a p}}{R} (\frac{1}{T_{2}} - \frac{1}{T_{1}})$

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