Explanation Write the expression for the Gibbs-Duhem equation. ∑ i = 1 j n i d μ i = V d p − S d T (I) Here, volume is V and entropy is S , the number of moles is n i and the chemical potential is μ i . As there is no change in the pressure or the temperature, the right hand side of the equation (I) is zero that is ∑ i = 1 j n i d μ i = 0 . Write the expression for both the solvent and solute. n 1 d μ 1 + n 2 d μ 2 = 0 (II) The sum of the mole fraction of the solvent and solute must be unity. y 1 + y 2 = 1 (III) Here, mole fraction of the solute is y 2 and the mole fraction of the solvent is y 1 . Differentiate Equation (II) with respect to y 2 . n 1 d μ 1 d y 2 + n 2 d μ 2 d y 2 = 0 (IV) Write the expression for fugacity of a component in a mixture. μ i = R ¯ T ln f ¯ i + C i ( T ) Write the expression for the fugacity of the solute. μ 1 = R ¯ T ln f ¯ 1 + C 1 ( T ) Write the expression for the fugacity of the solvent. μ 2 = R ¯ T ln f ¯ 2 + C 2 ( T ) Substitute R ¯ T ln f ¯ 1 + C 1 ( T ) for μ 1 and R ¯ T ln f ¯ 2 + C 2 ( T ) for μ 2 in Equation (IV). n 1 d ( R ¯ T ln f ¯ 1 + C 1 ( T ) ) d y 2 + n 2 d ( R ¯ T ln f ¯ 2 + C 2 ( T ) ) d y 2 = 0 n 1 [ d ( R ¯ T ln f ¯ 1 ) d y 2 + d ( C 1 ( T ) ) d y 2 ] + n 2 [ d ( R ¯ T ln f ¯ 2 ) d y 2 + d ( C 2 ( T ) ) d y 2 ] = 0 (V) Remove the derivatives of constants. n 1 R ¯ T [ d ( ln f ¯ 1 ) d y 2 ] + n 2 R ¯ T [ d ( ln f ¯ 2 ) d y 2 ] = 0 n 1 d ( ln f ¯ 1 ) d y 2 + n 2 d ( ln f ¯ 2 ) d y 2 = 0 (VI) The number of moles is directly proportional to mole fraction. Substitute y 1 for n 1 and y 2 for n 2 in Equation (VI). y 1 d ( ln f ¯ 1 ) d y 2 + y 2 d ( ln f ¯ 2 ) d y 2 = 0 (VII) Write the expression for Henry’s rule

Fundamentals of Engineering Thermodynamics

8th Edition

ISBN: 9781118412930

Author: Michael J. Moran, Howard N. Shapiro, Daisie D. Boettner, Margaret B. Bailey

Publisher: WILEY

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Chapter 11.9, Problem 122P

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The expression for the fugacity of the solvent.

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Chapter 11.9, Problem 121P

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