A certain substance has a heat of vaporization of 70.83 kJ / mol. 70.83 kJ/mol. At what Kelvin temperature will the vapor pressure be 5.00 5.00 times higher than it was at 331 K?

The answer to the question is

The temperature at which the vapor pressure will be 5.00 times higher than it was at 331 K is 353.0797 K.

Explanation:

To solve the question, we make use of the Clausius-Clapeyron equation as follows

Where P₁ = Initial pressure

P₂ = Final pressure

T₁ = Initial temperature = 331 K

T₂ = Final temperature

dvapH = ΔvapH = Heat of vaporization = 70.83 kJ / mol.

R = Universal gas constant = 8.3145. J K⁻¹ mol⁻¹

We are required to find the temperature when P₂ = 5 × P₁

Therefore we have

= or T₂ = = 353.0797 K

The vapor pressure be 5.00 times higher than it was at 331 K when the temperature is raised to 353.0797 K.