Calculation of the Fugacity of Naphthalene from Density Data The saturation pressure of solid naphthalene can be represented by log10 P^sat(bar) = 8.722 - 3783/T (T in K) The density of the solid is 1.025 g/cm^3, and the molecular weight of naphthalene is 128.19.Estimate the fugacity of solid

Question

Calculation of the Fugacity of Naphthalene from Density Data

The saturation pressure of solid naphthalene can be represented by

[latex]\log _{10} P^{ ext {sat }}(\mathrm{bar})=8.722-\frac{3783}{T}[/latex] (T in K)

The density of the solid is 1.025 g/cm[latex]^3[/latex], and the molecular weight of naphthalene is 128.19.Estimate the fugacity of solid naphthalene at 35°C at its vapor pressure, and also at 1, 20,30, 40, 50, and 60 bar. (This information is used in Sec. 12.1 in a study of supercritical phase behavior.)

Accepted Answer

We start with Eq. 7.4-23 for a single solid phase with the assumption that the solid is incompressible:

[latex]f^{\mathrm{S}}(T, P)=P^{\mathrm{sat}}(T)\left(\frac{f}{P} ight)_{\mathrm{sat}, T} \exp \left[\frac{1}{R T} \sum_{J=1} \int_{P^{\mathrm{J}}}^{P^{\mathrm{J}+1}} \underline{V}^{\mathrm{J}} d P ight][/latex] (7.4.23)

[latex]f^{\mathrm{S}}(T, P)=P^{\mathrm{sat}}(T)\left(\frac{f}{P} ight)_{\mathrm{sat}, T} \exp \left[\frac{\underline{V}\left(P-P^{ ext {sat }}(T) ight)}{R T} ight][/latex]

Using the data in the problem statement,

[latex]P^{ ext {sat }}\left(35^{\circ} \mathrm{C} ight)=10^{[8.722-(3783 /(273.15+35))]}=2.789 imes 10^{-4} \mathrm{bar}[/latex]

Since the sublimation pressure is so low, we can use (see Fig. 7.4-1)

[latex]\left(\frac{f}{P} ight)_{\mathrm{sat}}=1[/latex]

Therefore,

[latex]f^{\mathrm{S}}\left(T=35^{\circ} \mathrm{C}, P^{\mathrm{sat}} ight)=2.789 imes 10^{-4} \mathrm{bar}[/latex]

At any higher pressure, we have

[latex]\begin{aligned}&f^{\mathrm{S}}\left(T=35^{\circ} \mathrm{C}, P ight)=2.789 imes 10^{-4} ext { (bar) }\\& imes \exp \left[\frac{\frac{128.19}{1.025} \frac{\mathrm{cm}^{3}}{\mathrm{~mol}} imes\left(P-2.879 imes 10^{-4} ight) ext { bar } imes 10^{-6} \frac{\mathrm{m}^{3}}{\mathrm{~cm}^{3}}}{8.314 imes 10^{-5} \frac{\mathrm{bar} \mathrm{m}}{\mathrm{mol} \mathrm{K}} imes 308.15 \mathrm{~K}} ight]\end{aligned}[/latex]

The result is

P (bar)	[latex]f^s[/latex](3°C, P) (bar)
2.789 × [latex]10^{−4}[/latex]	2.789 × [latex]10^{−4}[/latex]
1	2.803 × [latex]10^{−4}[/latex]
10	2.933 × [latex]10^{−4}[/latex]
20	3.083 × [latex]10^{−4}[/latex]
30	3.241 × [latex]10^{−4}[/latex]
40	3.408 × [latex]10^{−4}[/latex]
50	3.582 × [latex]10^{−4}[/latex]
60	3.766 × [latex]10^{−4}[/latex]

Question 7.4.9: Calculation of the Fugacity of Naphthalene from Density Data...

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