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Q.5.81

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An Introduction to Thermal Physics
Found in: Page 208
An Introduction to Thermal Physics

An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279

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Short Answer

Derive a formula, similar to equation 5.90, for the shift in the freezing temperature of a dilute solution. Assume that the solid phase is pure solvent, no solute. You should find that the shift is negative: The freezing temperature of a solution is less than that of the pure solvent. Explain in general terms why the shift should be negative.

The negative sign in the above result shows that adding a solute will lower the temperature of the freezing point of the liquid.

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Step by Step Solution

Let us consider a pure solid A. If a solution of solute B in solvent A is equilibrium with solid A, then the chemical potentials of A for the two phases must be equal.

μA.liq=μA.solid

Here, μA.solid is the chemical potential of A in the solid phase and μA.liq is the chemical potential of A in liquid phase.

Use the equation 5.69, the chemical potential of A in liquid phase is as follows.

μA.liq=μ0(T,P)-NBkTNA

Here, μ0 is the chemical potential of the pure solvent and k is the Boltzmann's constant.

Substitute μ0(T,P)-NBkTNA=μA.solid(T,P) ......(1)

At a fixed pressure P, the pure liquid is equilibrium with the solid at temperature T0. Then the chemical potential can be written in terms of T0 as follows.

μ0(T,P)=μ0(T0,P)+(T-T0)μ0TμA.solid(T,P)=μA.solid(T0,P)+(T-T0)μA.solidT

Substitute the above two equations in the equation (1) and simplify.

μ0(T0,P)+(T-T0)μ0T-NBkTNA=μA.solid T0,P+(T-T0)μA.solidT .....(2)

At the temperature T0, pure liquid phase is equilibrium with the solid phase. Hence,

μ0(T0,P)=μA.solid(T0,P)

The Gibbs free energy for pure solvent A is,

G=NAμ0

Here, NA is the number of particle in that phase.

Partial differentiate the equation on both sides with respect to the temperature.

GT=NAμ0T-S=NAμ0T Since GT=-Sμ0T=-SNA

Substitute μ0(T0,P) for μA.solid(T0,P) and -SNA for μ0T in the equation (2) and simplify.

μ0(T0,P)+(T-T0)-SNA-NBkTNA=μ0(T0,P)+(T-T0)-SNAsolid(T-T0)-SNAliq-NBkTNA=(T-T0)-SNAsolid(T-T0) Sliq-Ssolid=-NBkTT-T0=-NBkTSliq-Ssolid

The difference in the entropy between the liquid and solid is :

Sliq-Ssolid=LT0

Here, L is the latent heat of condensation and TT0.

Substitute LT for Sliq-Ssolid in the equation T-T0=-NBkTSliq-Ssolid and solve for T-T0

T-T0=-NBkTLT=NBkT2L

Hence, the shift in the freezing temperature of the dilute solution is negative.

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