Question

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?

Answer 1

Answer:

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

$ln(\frac{P_2}{P_1}) = \frac{d_{vap}H}{R} (\frac{1}{T_1} -\frac{1}{T_2} )$

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

$ln(\frac{5*P_1}{P_1}) = \frac{70.83}{8.3145} (\frac{1}{331} -\frac{1}{T_2} )$ = $ln(5) = 8518.853 (\frac{1}{331} -\frac{1}{T_2} )$ or T₂ = $\frac{1}{\frac{1}{331} -\frac{ln(5)}{8518.853} }$ = 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.