a certain substance has a heat of vaporization of 48.60 kJ/mol. At what Kelvin temperature will the vapor pressure be 3.50 times higher than it was at 399 K?

To calculate the temperature of a substance at its given vapour pressure, we can use The Clausius-Clapeyron equation. The Clausius-Clapeyron equation is given by :

[tex] \sf \ln \bigg(\dfrac{p_1}{p_2 } \bigg)= \dfrac{\Delta_{vap} H}{R} \bigg( \dfrac{1}{T_1} -\dfrac{1}{T_2} \bigg )[/tex]

where :

[tex]\sf p_1 [/tex] and [tex]\sf p_2 [/tex] are the pressures at the temperatures [tex]\sf T_1 [/tex] and [tex]\sf T_2 [/tex]
[tex]\sf \Delta_{vap} H[/tex] is Heat of vaporization
R is universal gas constant (8.314 J K/mol)

We are given : a certain substance has a heat of vaporization of 48.60 kJ/mol. We need to calculate at what Kelvin temperature [tex]\sf(T_2 )[/tex] will the vapor pressure be 3.50 times higher than it was at 399 K [tex]\sf (T_1 )[/tex]

Here,

[tex]\sf p_2 [/tex] = 3.50 [tex] \sf p_1[/tex] at Temperature [tex]\sf T_1 [/tex]
[tex]\sf T_1 [/tex] = 399 K
[tex]\sf \Delta_{vap} H[/tex] = 48600 J/K
[tex]\sf T_2 [/tex] = ?

Substitute the given values in the Clausius-Clapeyron equation:

[tex] \sf \ln \bigg(\dfrac{3.50p_1}{p_1 } \bigg)= \dfrac{48600}{8.314} \bigg( \dfrac{1}{399} -\dfrac{1}{T_2} \bigg )[/tex]

[tex]\sf \: \: \: \: \:5.858= 5846 \bigg( 0.0026-\dfrac{1}{T_2} \bigg )[/tex]

[tex]\sf \: \: \: \: \:5.858= 15.200 -\dfrac{5846}{T_2}[/tex]

[tex]\sf \: \: \: \: \:5.858 - 15.200 -\dfrac{5846}{T_2}[/tex]

[tex]\sf \: \: \: \: \: - 9.342 = -\dfrac{5846}{T_2}[/tex]

[tex] \sf \: \: \: \: \: \: \: \: {T_2} = 625 \: K[/tex]

Hence, at 625 K temperature, the vapor pressure be 3.50 times higher than it was at 399 K.