Xenon has an enthalpy of vaporization of 12.6 kJ/mol and a vapor pressure of 1.00 atm at -108.0 °C.What is the vapor pressure of xenon at -148.0 °C? (R = 8.314 J/K⋅mol)
A) 0.053 atm
B) 0.73 atm
C) 0.93 atm
D) 0.99 atm
E) 19 atm

Question

Quizplus · Accepted Answer

To find the vapor pressure at a new temperature, we use the Clausius-Clapeyron equation: $\ln(\frac{P_2}{P_1}) = \frac{-\Delta H_{vap}}{R}(\frac{1}{T_2} - \frac{1}{T_1})$, where $P_1$ and $P_2$ are the initial and final pressures, $T_1$ and $T_2$ are the initial and final temperatures in Kelvin, $\Delta H_{vap}$ is the enthalpy of vaporization, and $R$ is the gas constant. Convert temperatures from Celsius to Kelvin: $T_1 = -108.0 °C = 165 K$ and $T_2 = -148.0 °C = 125 K$. Solve for $P_2$ to find the vapor pressure at -148.0 °C.

Xenon Has an Enthalpy of Vaporization of 12