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Q.5.81

Expert-verified
Found in: Page 208

### An Introduction to Thermal Physics

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

# 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.

See the 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.

${\mu }_{A.liq}={\mu }_{A.solid}$

Here, ${\mu }_{A.solid}$ is the chemical potential of $A$ in the solid phase and ${\mu }_{A.liq}$ 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.

${\mu }_{A.liq}={\mu }_{0}\left(T,P\right)-\frac{{N}_{B}kT}{{N}_{A}}$

Here, ${\mu }_{0}$ is the chemical potential of the pure solvent and $k$ is the Boltzmann's constant.

Substitute ${\mu }_{0}\left(T,P\right)-\frac{{N}_{B}kT}{{N}_{A}}={\mu }_{A.solid}\left(T,P\right)......\left(1\right)$

## 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.

${\mu }_{0}\left(T,P\right)={\mu }_{0}\left({T}_{0},P\right)+\left(T-{T}_{0}\right)\frac{\partial {\mu }_{0}}{\partial T}\phantom{\rule{0ex}{0ex}}{\mu }_{A.solid}\left(T,P\right)={\mu }_{A.solid}\left({T}_{0},P\right)+\left(T-{T}_{0}\right)\frac{\partial {\mu }_{A.solid}}{\partial T}\phantom{\rule{0ex}{0ex}}$

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

${\mu }_{0}\left({T}_{0},P\right)+\left(T-{T}_{0}\right)\frac{\partial {\mu }_{0}}{\partial T}-\frac{{N}_{B}kT}{{N}_{A}}={\mu }_{A.solid}\left({T}_{0},P\right)+\left(T-{T}_{0}\right)\frac{\partial {\mu }_{A.solid}}{\partial T}.....\left(2\right)$

At the temperature ${T}_{0}$, pure liquid phase is equilibrium with the solid phase. Hence,

${\mu }_{0}\left({T}_{0},P\right)={\mu }_{A.solid}\left({T}_{0},P\right)$

The Gibbs free energy for pure solvent A is,

$G={N}_{A}{\mu }_{0}$

Here, ${N}_{A}$ is the number of particle in that phase.

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

$\frac{\partial G}{\partial T}={N}_{A}\frac{\partial {\mu }_{0}}{\partial T}\phantom{\rule{0ex}{0ex}}-S={N}_{A}\frac{\partial {\mu }_{0}}{\partial T}\left(Since\frac{\partial G}{\partial T}=-S\right)\phantom{\rule{0ex}{0ex}}\frac{\partial {\mu }_{0}}{\partial T}=-\frac{S}{{N}_{A}}$

Substitute ${\mu }_{0}\left({T}_{0},P\right)$ for ${\mu }_{A.solid}\left({T}_{0},P\right)$ and $-\frac{S}{{N}_{A}}$ for $\frac{\partial {\mu }_{0}}{\partial T}$ in the equation (2) and simplify.

${\mu }_{0}\left({T}_{0},P\right)+\left(T-{T}_{0}\right)\left(-\frac{S}{{N}_{A}}\right)-\frac{{N}_{B}kT}{{N}_{A}}={\mu }_{0}\left({T}_{0},P\right)+\left(T-{T}_{0}\right){\left(-\frac{S}{{N}_{A}}\right)}_{solid}\phantom{\rule{0ex}{0ex}}\left(T-{T}_{0}\right){\left(-\frac{S}{{N}_{A}}\right)}_{liq}-\frac{{N}_{B}kT}{{N}_{A}}=\left(T-{T}_{0}\right){\left(-\frac{S}{{N}_{A}}\right)}_{solid}\phantom{\rule{0ex}{0ex}}\left(T-{T}_{0}\right)\left({S}_{liq}-{S}_{solid}\right)=-{N}_{B}kT\phantom{\rule{0ex}{0ex}}T-{T}_{0}=\frac{-{N}_{B}kT}{{S}_{liq}-{S}_{solid}}$

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

${S}_{liq}-{S}_{solid}=\frac{L}{{T}_{0}}$

Here, $L$ is the latent heat of condensation and $T\approx {T}_{0}$.

Substitute $\frac{L}{T}$ for ${S}_{liq}-{S}_{solid}$ in the equation $T-{T}_{0}=\frac{-{N}_{B}kT}{{S}_{liq}-{S}_{solid}}$ and solve for $T-{T}_{0}$

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