91Ó°ÊÓ

${\mathrm{μ}}_{A.liq}={\mathrm{μ}}_{A.solid}$

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

${\mathrm{μ}}_{A.liq}={\mathrm{μ}}_0(T,P)-\frac{N_{B}kT}{N_A}$

Here, ${\mathrm{μ}}_0$is the chemical potential of the pure solvent and $k$is the Boltzmann's constant.

Substitute${\mathrm{μ}}_0(T,P)-\frac{N_{B}kT}{N_A}={\mathrm{μ}}_{A.solid}(T,P)......\left(1\right)$

$$\begin{gathered} {\mathrm{μ}}_0(T,P)={\mathrm{μ}}_0(T_0,P)+(T-T_0)\frac{∂{\mathrm{μ}}_0}{∂T} \\ {\mathrm{μ}}_{A.solid}(T,P)={\mathrm{μ}}_{A.solid}(T_0,P)+(T-T_0)\frac{∂{\mathrm{μ}}_{A.solid}}{∂T} \end{gathered}$$

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

${\mathrm{μ}}_0(T_0,P)+(T-T_0)\frac{∂{\mathrm{μ}}_0}{∂T}-\frac{N_{B}kT}{N_A}={\mathrm{μ}}_{A.solid}\left(T_0,P\right)+(T-T_0)\frac{∂{\mathrm{μ}}_{A.solid}}{∂T}.....\left(2\right)$

At the temperature $T_0$, pure liquid phase is equilibrium with the solid phase. Hence,

${\mathrm{μ}}_0(T_0,P)={\mathrm{μ}}_{A.solid}(T_0,P)$

The Gibbs free energy for pure solvent A is,

$G=N_A{\mathrm{μ}}_0$

Here, $N_A$is the number of particle in that phase.

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

$$\begin{gathered} \frac{∂G}{∂T}=N_A\frac{∂{\mathrm{μ}}_0}{∂T} \\ -S=N_A\frac{∂{\mathrm{μ}}_0}{∂T}\left(Since\frac{∂G}{∂T}=-S\right) \\ \frac{∂{\mathrm{μ}}_0}{∂T}=-\frac{S}{N_A} \end{gathered}$$

Substitute ${\mathrm{μ}}_0(T_0,P)$for ${\mathrm{μ}}_{A.solid}(T_0,P)$and $-\frac{S}{N_A}$for $\frac{∂{\mathrm{μ}}_0}{∂T}$in the equation (2) and simplify.

$$\begin{gathered} {\mathrm{μ}}_0(T_0,P)+(T-T_0)\left(-\frac{S}{N_A}\right)-\frac{N_{B}kT}{N_A}={\mathrm{μ}}_0(T_0,P)+(T-T_0){\left(-\frac{S}{N_A}\right)}_{solid} \\ (T-T_0){\left(-\frac{S}{N_A}\right)}_{liq}-\frac{N_{B}kT}{N_A}=(T-T_0){\left(-\frac{S}{N_A}\right)}_{solid} \\ (T-T_0)\left(S_{liq}-S_{solid}\right)=-N_{B}kT \\ T-T_0=\frac{-N_{B}kT}{S_{liq}-S_{solid}} \end{gathered}$$

$S_{liq}-S_{solid}=\frac{L}{T_0}$

Here, $L$is the latent heat of condensation and $T≈T_0$.

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 $T-T_0$

$$\begin{gathered} T-T_0=\frac{-N_{B}kT}{\left({\displaystyle \frac{L}{T}}\right)} \\ =\frac{N_{B}k{T}^2}{L} \end{gathered}$$

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