Xenon has a molar heat 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 = 8.31 J/K.mol)

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To find the vapor pressure of xenon at a new temperature, we can use the Clausius-Clapeyron equation in its integrated form:$$ \ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{vap}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) $$Given:- $\Delta H_{vap} = 12.6 \, \text{kJ/mol} = 12600 \, \text{J/mol}$ (conversion from kJ to J)- $R = 8.31 \, \text{J/K}\cdot\text{mol}$- $P_1 = 1.00 \, \text{atm}$- $T_1 = -108.0°C = 165K$ (conversion from °C to K: $T(K) = T(°C) + 273$)- $T_2 = -148.0°C = 125K$Rearranging the equation to solve for $P_2$:$$ \ln(P_2) = -\frac{12600}{8.31} \left(\frac{1}{125} - \frac{1}{165}\right) + \ln(1.00) $$Solving this equation gives the natural logarithm of $P_2$. To find $P_2$, we take the exponential of both sides.After calculating, you will find that $P_2$ is approximately 0.053 atm, which corresponds to choice A.

Xenon Has a Molar Heat of Vaporization of 12