Ethanol has a heat of vaporization of 38.56 kJ/mol and a normal boiling point of 78.4 °C. Using the Clausius-Clapeyron equation, what is the vapor pressure of ethanol at 15 °C?

Start by understanding the Clausius-Clapeyron equation, which relates the vapor pressure of a substance at two different temperatures. The equation is: ${\ln }_P=\frac{-\mathrm{\Delta H}/R}{(\frac{1}{T_2}-\frac{1}{T_1})}$, where $\mathrm{\Delta H}$ is the heat of vaporization, $R$ is the universal gas constant, and $T_1$ and $T_2$ are the temperatures in Kelvin.