The normal boiling point of ether is 307.8 K. Calculate the temperature at which its vapor pressure is exactly half of that at its normal boiling point. The heat of vaporization for ether is 26.69 kJ/mol.
A)305 K
B)302 K
C)295 K
D)289 K
E)281 K

Question

Quizplus · Accepted Answer

The Clausius-Clapeyron equation relates the vapor pressure of a substance at different temperatures to its heat of vaporization. It can be expressed as:$$ \ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{vap}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) $$Given that $P_2 = \frac{1}{2}P_1$, $T_1 = 307.8 \, K$, and $\Delta H_{vap} = 26.69 \, kJ/mol = 26690 \, J/mol$, we can solve for $T_2$. The gas constant $R = 8.314 \, J/(mol \cdot K)$.Rearranging the equation to solve for $T_2$, we get:$$ \ln(0.5) = -\frac{26690}{8.314} \left(\frac{1}{T_2} - \frac{1}{307.8}\right) $$Solving this equation for $T_2$ gives a value close to 289 K, which corresponds to choice D.

The Normal Boiling Point of Ether Is 307