a certain substance has a heat of vaporization of 26.70 kj/mol. at what kelvin temperature will the vapor…

Explanation:

Step1: Recall Clausius - Clapeyron equation

The Clausius - Clapeyron equation is $\ln\left(\frac{P_2}{P_1}\right)=\frac{\Delta H_{vap}}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$, where $P_1$ and $P_2$ are vapor - pressures, $T_1$ and $T_2$ are temperatures, $\Delta H_{vap}$ is the heat of vaporization, and $R = 8.314\ J/(mol\cdot K)$. Given that $\frac{P_2}{P_1}=5.50$, $\Delta H_{vap}=26.70\ kJ/mol = 26700\ J/mol$, and $T_1 = 323\ K$.

Step2: Rearrange the Clausius - Clapeyron equation to solve for $T_2$

Starting from $\ln\left(\frac{P_2}{P_1}\right)=\frac{\Delta H_{vap}}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$, we first substitute the known values: $\ln(5.50)=\frac{26700\ J/mol}{8.314\ J/(mol\cdot K)}\left(\frac{1}{323\ K}-\frac{1}{T_2}\right)$. $1.7047=\frac{26700}{8.314}\left(\frac{1}{323}-\frac{1}{T_2}\right)$. $\frac{1.7047\times8.314}{26700}=\frac{1}{323}-\frac{1}{T_2}$. $0.000531=\frac{1}{323}-\frac{1}{T_2}$. $\frac{1}{T_2}=\frac{1}{323}- 0.000531$. $\frac{1}{T_2}=\frac{1}{323}-\frac{0.000531\times323}{323}=\frac{1 - 0.171513}{323}=\frac{0.828487}{323}$. $T_2=\frac{323}{0.828487}\approx389\ K$.

Answer:

$389\ K$