Explanation Given: Within the solution the solid phase is pure solvent, no solute. Formula used: Write the expression for the equilibrium condition for solid and liquid phases within the solution. μ A L = μ A S ..... (1) Here, μ A L is the chemical potential of the liquid phase of solvent A and μ A S is the chemical potential of the solid phase of solvent A . Write the expression for the chemical potential in terms of standard chemical potential. μ A L = μ 0 ( T , P ) − N B k T 0 N A ..... (2) Here, μ 0 ( T , P ) is the standard chemical potential, N B is the number of solute molecules , T 0 is the freezing temperature and k is Boltzmann constant. Write the expression of chemical potential for pure liquid in terms of the temperature T 0 . μ 0 ( T , P ) = μ 0 ( T 0 , P ) + ( T − T 0 ) ∂ μ 0 ∂ T ..... (3) Write the expression of chemical potential for the solid in terms of the temperature T 0 . μ A S ( T , P ) = μ A S ( T 0 , P ) + ( T − T 0 ) ∂ μ A S ∂ T ..... (4) Write the expression for Gibbs energy for the pure solvent A . G = N A μ 0 ..... (5) Calculation: Substitute μ 0 ( T , P ) − N B k T 0 N A for μ A L in expression (1). μ 0 ( T , P ) − N B k T 0 N A = μ A S ..... (6) Substitute μ 0 ( T 0 , P ) + ( T − T 0 ) ∂ μ 0 ∂ T for μ 0 ( T , P ) and μ A S ( T 0 , P ) + ( T − T 0 ) ∂ μ A S ∂ T for μ A S in expression (6). μ 0 ( T 0 , P ) + ( T − T 0 ) ∂ μ 0 ∂ T − N B k T 0 N A = μ A S ( T 0 , P ) + ( T − T 0 ) ∂ μ A S ∂ T

1st Edition

ISBN: 9780201380279

Author: Daniel V. Schroeder

Publisher: Addison Wesley

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Chapter 5.5, Problem 81P

To determine

The shift in the freezing temperature of a dilute solution.

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Chapter 5.5, Problem 80P

Chapter 5.5, Problem 82P

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Chapter 5 Solutions

An Introduction to Thermal Physics

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