The solubility parameter and liquid molar volume for n-hexane are given in Table 9.6-1 as \delta_{2}=7.3 \text { and } \underline{V}_{2}^{ L }=132 cc/mol, respectively. Since the liquidmolar volume of naphthalene given in the data is for a temperature 80°C higher than the temperature of interest, and the volume change on melting of naphthalene is small, the molar volume of liquid naphthalene below its melting temperature will be taken to be that of the solid; that is,
\underline{V}_{1}^{ L }=\frac{128.19 g / mol }{1.0253 g / cc }=125 cc / mol
The heat of sublimation of naphthalene is not given. However, we can compute this quantity from the vapor pressure curve of the solid and the Clausius-Clapeyron equation (Eq. 7.7-5a) by taking P to be equal to the sublimation pressure P^{ sub }, \Delta \underline{H} o equal the heat of sublimation, and setting \Delta \underline{V}=\underline{V}^{\overline{ V }}-\underline{V}^{ S }=\Delta_{ sub } \underline{V} \cong R T / P^{ sub }. Thus
\frac{d \ln P^{ vap }}{d T}=\frac{\Delta_{ vap } \underline{H}}{R T^{2}} (7.7-5a)
\frac{\Delta_{\text {sub } \underline{H}}}{R T^{2}}=\frac{d \ln P^{ sub }}{d T}=2.303 \frac{d \log _{10} P^{ sub }}{d T}=+2.303 \frac{(3783)}{T^{2}}
and
\Delta_{\text {sub }} \underline{H}=2.303(3783)(8.314 J / mol )=72434 J / mol
Next,
\Delta_{\text {vap }} \underline{U}=\Delta_{\text {sub }} \underline{H}-\Delta_{\text {fus }} \underline{H}-R T=72434-18804-8.314 \times 293.15=51193 J / mol
so that
\delta_{1}=\left(\frac{51193 J / mol }{125 cc / mol \times 4.184 J / cal }\right)^{1 / 2}=9.9( cal / cc )^{1 / 2}
Now using Eq. 12.1-7 with \Delta C_{ P }=0, and the regular solution expression for the activity coefficient, we obtain
\ln x_{1}=-\ln \gamma_{1}-\left\{\frac{\Delta_{ fus } \underline{H}\left(T_{t}\right)}{R T}\left[1-\frac{T}{T_{t}}\right]+\frac{\Delta C_{ P }}{R}\left[1-\frac{T_{t}}{T}+\ln \left(\frac{T_{t}}{T}\right)\right]\right\} (12.1-7)
\ln x_{1}=-\frac{V_{1}^{ L }\left(\delta_{1}-\delta_{2}\right)^{2} \Phi_{2}^{2}}{R T}-\frac{\Delta_{ fus } H\left(T_{m}\right)}{R T}\left(1-\frac{T}{T_{m}}\right)
As a first guess, assume that x_{1} will be small, so that
\Phi_{2}=\frac{x_{2} \underline{V}_{2}^{ L }}{x_{1} \underline{V}_{1}^{ L }+x_{2} \underline{V}_{2}^{ L }} \approx 1
In this case
\begin{aligned}\ln x_{1}=& \frac{-125 \frac{ cc }{ mol } \times(9.9-7.3)^{2} \frac{ cal }{ cc } \times 4.184 \frac{ J }{ cal }}{8.314 \frac{ J }{ mol K } \times 293.15 K } \\&-\frac{18804 \frac{ J }{ mol }}{8.314 \frac{ J }{ mol K } \times 293.15 K }\left(1-\frac{293.15}{353.35}\right) \\=&-1.451-1.314=-2.765\end{aligned}
x_{1}=0.063
With such a large value for x_{1} we must go back and correct the value of \Phi_{2} for the presence of the solute and repeat the computation. Thus
\Phi_{2}=\frac{0.937 \times 132}{0.937 \times 132+0.063 \times 125}=0.94
and
\begin{aligned}\ln x_{1} &=1.282-1.314=2.596 \\x_{1} &=0.0746\end{aligned}
The results of the next two iterations are x_{1}=0.0768 \text { and } x_{1}=0.0772, respectively. This last prediction is in reasonable agreement with the experimental result of x_{1}=0.09 .^{4}
Comment
Note that had we assumed ideal solution behavior, \gamma_{1}=1 \text { and } \ln \gamma_{1}=0, so that
\begin{aligned}\ln x_{1} &=-1.314 \\x_{1} &=0.269\end{aligned}
which is a factor of 3 too large.
^4G. Scatchard, Chem. Rev., 8, 329 (1931).