Question

For a pure substance, the following data at saturated conditions are given: 
\[ \begin{array}{c c} \ln P^{sat} \, (\text{bar}) & T \,(\text{K})\\ 0.693 & 350\\ 1.386 & 370 \end{array} \] Assume the vapor behaves ideally, liquid molar volume is negligible, and latent heat of vaporization is constant over this range. The universal gas constant is $R=8.314$ J mol$^{-1}$ K$^{-1}$. From the above data, the estimated latent heat of vaporization at 360 K is \(\underline{\hspace{2cm}}\) kJ/mol (rounded to one decimal place).

Hint:

The Clausius–Clapeyron equation is best used when vapor behaves ideally and latent heat is nearly constant.

Answer 1

Correct Answer: 36.3 - 38.3

For phase equilibrium, Clausius–Clapeyron relation applies: \[ \frac{d(\ln P^{sat})}{dT} = \frac{\Delta H_{vap}}{RT^2} \] Using two-point approximation: \[ \Delta H_{vap} \approx R \, \frac{\ln(P_2/P_1)}{(1/T_1)-(1/T_2)} \] Substitute values: \[ P_1 = e^{0.693} = 2.00 \text{ bar}, P_2 = e^{1.386} = 4.00 \text{ bar} \] \[ \ln(P_2/P_1) = \ln 2 = 0.693 \] \[ (1/T_1)-(1/T_2) = \frac{1}{350} - \frac{1}{370} \] Compute denominator: \[ \frac{1}{350} - \frac{1}{370} = 0.0001539 \] Thus, \[ \Delta H_{vap} = 8.314 \times \frac{0.693}{0.0001539} \approx 3.63\times10^{4} \text{ J/mol} \] \[ \Delta H_{vap} \approx 36.3 \text{ to } 38.3 \text{ kJ/mol} \]