Discussion
The vapour pressure of a liquid at 8 °C is 2.7 kPa. Its enthalpy of vaporization is constant and equal to 42700 kJ/kmol. Take R = 8.314 kJ/kmol.K. The temperature (in °C) at a vapour pressure of 13.5 kPa is
Answer: D
The temperature is 35.2 °C.
With a constant enthalpy of vaporization, the Clausius-Clapeyron equation gives
\[ \begin{aligned} \ln\frac{P_{2}}{P_{1}} &= \frac{\Delta h_{v}}{R} \\ &\quad \times \left(\frac{1}{T_{1}} - \frac{1}{T_{2}}\right) \end{aligned} \]
Here \(\ln(13.5/2.7) = \ln 5 = 1.609\), \(\Delta h_{v}/R = 42700/8.314 = 5136\) K and \(T_{1} = 281.15\) K:
\[ \begin{aligned} \frac{1}{T_{2}} &= \frac{1}{281.15} - \frac{1.609}{5136} \\ &= 3.557\times10^{-3} - 3.13\times10^{-4} \\ &= 3.243\times10^{-3}\ \text{K}^{-1} \end{aligned} \]
so \(T_{2} = 308.3\) K = 35.2 °C. The other values, 44.3, 51.4 and 58.7 °C, would correspond to vapour pressures well above 13.5 kPa on this curve.
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