Respuesta :
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
[tex]ln(\frac{P_2}{P_1}) = \frac{d_{vap}H}{R} (\frac{1}{T_1} -\frac{1}{T_2} )[/tex]
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
[tex]ln(\frac{5*P_1}{P_1}) = \frac{70.83}{8.3145} (\frac{1}{331} -\frac{1}{T_2} )[/tex] = [tex]ln(5) = 8518.853 (\frac{1}{331} -\frac{1}{T_2} )[/tex] or T₂ = [tex]\frac{1}{\frac{1}{331} -\frac{ln(5)}{8518.853} }[/tex] = 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.
