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Unformatted text preview: us expression.
(a) K i = (e) K i = Pi s
is not rigorous. It assumes ideal solutions and an ideal gas,
such that φiV , γ iL , and γ iV are 1.0. It also assumes a low
pressure for the liquid phase, such that φiL =1.0. (f) K i =
(g) K i = γ iLφiL
is a rigorous expression
γ iV φiV
γ iL Pi s
P is not rigorous. It assumes low pressure for the liquid
and the ideal gas law, such that φiL = Pi s
and φiV =1.0.
P Exercise 2.6
Subject: Comparison of experimental K-values to Raoult's law predictions.
Given: For the propane-isopentane system at 167oF and 147 psia, propane mole fractions
of 0.2900 in the liquid phase and 0.6650 in the vapor phase. Vapor pressures at 167oF of
409.6 psia for propane and 58.6 psia for isopentane.
Find: (a) Experimental K-values.
(b) Raoult's law K-values.
Analysis: (a) Mole fraction of isopentane in the liquid phase = 1 - 0.2900 = 0.7100
Mole fraction of isopentane in the vapor phase = 1 - 0.6650 = 0.3350
From Eq. (2-9), Ki = yi /xi
= 0.6650/0.2900 = 2.293 for propane
= 0.3350/0.7100 = 0.472 for isopentane (b) From Eq. (3), Table 2.3, K i = Pi s / P
= 409.6/147 = 2.786 for propane
= 58.6/147 = 0.399 for isopentane
This is rather poor agreement. Note that the experimental values give a
relative volatility of 2.293/0.472 = 4.86, while Raoult's law predicts 2.786/0.399 = 6.98.
A modified Raoult’s law should be used to incorporate a liquid-phase activity
coefficient. Also a fugacity correction for the gas phase might improve the agreement. Exercise 2.7
Subject: Liquid-liquid phase equilibrium data.
Given: Experimental solubility data at 25oC for the isooctane (1)-furfural (2) system.
In the furfural-rich liquid phase, I, x1 = 0.0431
In the isooctane-rich liquid phase, II, x1 = 0.9461
Assumptions: The furfural activity coefficient in phase I = 1.0
The isooctane activity coefficient in phase II = 1.0
Find: (a) The distribution coefficients for isooctane and furfural.
(b) The relative selectivity for isooctane relative to furfural.
(c) The activity coefficients.
From summation of mole fractions in each liquid phase,
x2 in phase I = 1 - 0.0431 = 0.9569
x2 in phase II = 1 - 0.9461 = 0.0.0539
(a) From Eq. (2-20), K Di = xi( I ) / xi( II ) K D1 = 0.0431
= 0.456 and K D2 =
βij = K Di / K D j (b) From Eq. (2-22), β12 =
they K D1
K D2 = 0.0456
17.75 Note that the I and II designations for the two liquid phases are arbitrary. If
were interchanged, the relative selectivity would be 1/0.00257 = 389.
(c) From rearrangements of Eq. (2-30), γ 1( I) = γ 1(II) x1(II)
0.0431 γ (I)
2 =γ (I)
2 Exercise 2.8
Subject: Activity coefficients of solids dissolved in solvents.
Given: Solubility at 25oC of naphthalene in 5 solvents. Vapor pressure equations for
solid and liquid naphthalene
Find: Activity coefficient of naphthalene in each solvent.
Analysis: From a rearrangement of Eq. (2-34) for solid-liquid phase equilibrium of
γL = s
x L Pliquid (1) From the given naphthalene vapor pressure equations at 25oC = 298.15 K,
Psolid = ex...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
- Spring '11
- The Land