Explanation Write the expression for the volume of the saturated liquid. R T ( ∂ ln f ∂ p ) T = v v R T = ( ∂ ln f ∂ p ) T (I) Here, the gas constant is R , the temperature of the liquid is T . The state is a saturated liquid state hence v = v f . Substitute v f for v in Equation (I). R T ( ∂ ln f ∂ p ) T ≈ v f R T (II) Integrate the Equation (II). ∫ f f s a t L ∂ ln f = v f ( T ) R T ∫ p p s a t ∂ p [ ln f − ln f s a t L ] = v f ( T ) R T [ p s a t − p ] ln f f s a t L = v f ( T ) R T [ p s a t − p ] f ( T , p ) ≈ f s a t L ( T ) exp { v f ( T ) R T [ p − p s a t ] } (III) Here, the fugacity at saturated liquid is f s a t L . Substitute v f ( T ) R T [ p − p s a t ] for x in Equation (III). f ( T , p ) ≈ f s a t L ( T ) exp { x } (IV) Write the expression for expansion of equation (IV). f ( T , p ) f s a t L ( T ) ≈ 1 + x + x 2 2 + x 3 3 + ..... Substitute x for ln f f s a t in Equation (III). x = v f ( T ) R T [ p − p s a t ] (V) Conclusion: Substitute f s a t L ( T ) for f ( T , p ) when the terms in power k are small enough to ignore the higher order in Equation (III). Refer to the Table A-2, “Properties of saturated water (Liquid-Vapor):Temperature Table”, to obtain the saturated pressure as 0.03567 bar and the specific volume at saturated liquid is 1.0035 × 10 − 3 m 3 / kg at a temperature of 300.15 K . Substitute 1.0035 × 10 − 3 m 3 / kg for v f , 0

Fundamentals of Engineering Thermodynamics

8th Edition

ISBN: 9781118832301

Author: SHAPIRO

Publisher: JOHN WILEY+SONS,INC.-CONSIGNMENT

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Chapter 11.9, Problem 113P

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The expression for the fugacity of a liquid.

The range of pressure for the expression.

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The expression for the fugacity of a liquid. The range of pressure for the expression.