Estimate ϕ1 and ϕ2 by Eqs. (10.63) for an equimolar mixture of methyl ethyl ketone(1)/ toluene(2) at 50°C and 25 kPa. Set all kij = 0.

Question

Estimate [latex] \hat{\phi}_1 ext { and } \hat{\phi}_2 [/latex] by Eqs. (10.63) for an equimolar mixture of methyl ethyl ketone(1)/ toluene(2) at 50°C and 25 kPa. Set all [latex] k_{ij} = 0[/latex] .

[latex] \ln \hat{\phi}_1=\frac{P}{R T}\left(B_{11}+y_2^2 \delta_{12} ight) [/latex] (10.63a)

[latex] \ln \hat{\phi}_2=\frac{P}{R T}\left(B_{22}+y_1^2 \delta_{12} ight) [/latex] (10.63b)

Accepted Answer

The required data are as follows:

ij	[latex] T_{c i j} \mathrm{~K} [/latex]	[latex] P_{c i j} ext { bar } [/latex]	[latex] V_{c i j} \mathrm{~cm}^3 \cdot \mathrm{mol}^{-1} [/latex]	[latex] Z_{c i j} [/latex]	[latex] \omega_{i j} [/latex]
11	535.5	41.5	267.	0.249	0.323
22	591.8	41.1	316.	0.264	0.262
12	563.0	41.3	291.	0.256	0.293

where values in the last row have been calculated by Eqs. (10.70) through (10.74). The values of [latex] T_{rij} [/latex] , together with [latex] B^0, B^1, and B_{ij} [/latex] calculated for each ij pair by Eqs. (3.65), (3.66), and (10.69), are as follows:

[latex] \omega_{i j}=\frac{\omega_i+\omega_j}{2} [/latex] (10.70)

[latex] V_{c i j}=\left(\frac{V_{c i}^{1 / 3}+V_{c j}^{1 / 3}}{2} ight)^3 [/latex] (10.74)

[latex] \hat{C}=C^0+\omega C^1 [/latex] (3.65)

[latex] C^0=0.01407+\frac{0.02432}{T_r}-\frac{0.00313}{T_r^{10.5}} [/latex] (3.66)

ij	[latex] T_{r i j} [/latex]	[latex]B^0 [/latex]	[latex]B^1 [/latex]	[latex] B_{i j} \mathrm{~cm}^3 \cdot \mathrm{mol}^{-1} [/latex]
11	0.603	–0.865	–1.300	–1387.
22	0.546	–1.028	–2.045	–1860.
12	0.574	–0.943	–1.632	–1611.

Calculating [latex]δ_{12}[/latex] according to its definition gives:

[latex] \delta_{12}=2 B_{12}-B_{11}-B_{22}=(2)(-1611)+1387+1860=25 \mathrm{~cm}^3 \cdot \mathrm{mol}^{-1} [/latex]

Equation (10.63) then yields:

[latex] \ln \hat{\phi}_1=\frac{P}{R T}\left(B_{11}+y_2^2 \delta_{12} ight)=\frac{25}{(8314)(323.15)}\left[-1387+(0.5)^2(25) ight]=-0.0128 [/latex]

[latex] \ln \hat{\phi}_2=\frac{P}{R T}\left(B_{22}+y_1^2 \delta_{12} ight)=\frac{25}{(8314)(323.15)}\left[-1860+(0.5)^2(25) ight]=-0.0172 [/latex]

Thus,

[latex] \hat{\phi}_1=0.987 \quad ext { and } \quad \hat{\phi}_2=0.983 [/latex]

These results are representative of values obtained for vapor phases at typical conditions of low-pressure vapor/liquid equilibrium.

Question 10.10: Estimate ϕ1 and ϕ2 by Eqs. (10.63) for an equimolar mixture ...

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