Calculation of the Fugacity of a Gas Using the Virial Equation Compute the fugacities of pure ethane and pure butane at 373.15 K and 1, 10, and 15 bar, assuming the virial equation of state can describe these gases at these conditions.

Question

Calculation of the Fugacity of a Gas Using the Virial Equation
Compute the fugacities of pure ethane and pure butane at 373.15 K and 1, 10, and 15 bar, assuming the virial equation of state can describe these gases at these conditions.

Data:

[latex]\begin{aligned}&B_{\mathrm{ET}}(373.15 \mathrm{~K})=-1.15 imes 10^{-4} \mathrm{~m}^{3} / \mathrm{mol} \\&B_{\mathrm{BU}}(373.15 \mathrm{~K})=-4.22 imes 10^{-4} \mathrm{~m}^{3} / \mathrm{mol}\end{aligned}[/latex]

[Source: E. M. Dontzler, C. M. Knobler, and M. L. Windsor, J. Phys. Chem. 72, 676 (1968).]

Accepted Answer

Using Eq. 7.4-12, we find

[latex]\ln \frac{f^{\mathrm{V}}(T, P)}{P}=\frac{2 B(T)}{\underline{V}}-\ln Z=\frac{2 P B(T)}{Z R T}-\ln Z[/latex] (7.4.12)

	Ethane		Butane
P (bar)	Z	f (bar)	Z	f (bar)
1	0.996	0.996	0.986	0.986
10	0.961	9.629	0.838	8.628
15	0.941	14.16	0.714	11.86

Since the pressures are not very high, these results should be reasonably accurate. However, the virial equation with only the second virial coefficient will be less accurate as the pressure increases. In fact, at slightly above 15 bar and 373.15 K, n-butane condenses to form a liquid.

In this case the virial equation description is inappropriate, as it does not show a phase change or describe liquids.

Question 7.4.4: Calculation of the Fugacity of a Gas Using the Virial Equati...

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