91Ó°ÊÓ

The formula for experimental measurements of the heat capacity of aluminum at low temperatures (below about $50K$):

$C_V=aT+b{T}^3$

Where,

$C_V$is the heat capacity of one mole of aluminum, and the constants $a$and $b$are $a=0.00135\mathrm{J}/{\mathrm{K}}^2$,and $b=2.48×{10}^{-5}\mathrm{J}/{\mathrm{K}}^4$.

The change in entropy is given as:

$\mathrm{Δ}S={∫}_{T_i}^{T_f}\frac{C_V}{T}dT$

By substituting the value of $C_V$,the above equation can be modified as:

role="math" localid="1647250975365" $$\begin{gathered} S\left(T_f\right)-S\left(T_i\right)={∫}_{T_i}^{T_f}\frac{aT+b{T}^3}{T}dT \\ S\left(T_f\right)-S\left(T_i\right)={∫}_{Ti}^{T_f}\frac{{\displaystyle T}\left(a+b{T}^2\right)}{T}dT \\ S\left(T_f\right)-S\left(T_i\right)={∫}_{Ti}^{T_f}a+b{T}^{2}dT \end{gathered}$$

Consider initial temperature as 0 and by integrating, we get,

$$\begin{gathered} S\left(T_f\right)-S\left(0\right)={\left[\left(aT+\frac{b}{3}{T}^3\right)\right]}_0^{T_f} \\ S\left(T_f\right)-S\left(0\right)=a{T}_f+\frac{b}{3}{T}_f^3 \end{gathered}$$

By assuming $S\left(0\right)=0$,

role="math" localid="1647251199556" $S\left(T_f\right)=a{T}_f+\frac{b}{3}{T}_f^3..........\left(1\right)$

For $T=1K$,

$S\left(1\right)=a×1+\frac{b}{3}×1^3$

By substituting the values of $a$and $b$, we get,

role="math" localid="1647251363775" $$\begin{gathered} S\left(1\right)=0.00135×1+\frac{2.48×{10}^{-5}}{3}×1^3 \\ S\left(1\right)=1.35×{10}^{-3}{\mathrm{JK}}^{-1} \end{gathered}$$

For unitless form, divide the equation by $k=1.38×{10}^{-23}J{K}^{-1}$

$$\begin{gathered} \frac{S\left(1\right)}{k}=\frac{1.35×{10}^{-3}}{k} \\ \frac{S\left(1\right)}{k}=\frac{1.35×{10}^{-3}}{1.38×{10}^{-23}} \\ \frac{{\displaystyle S\left(1\right)}}{{\displaystyle k}}=9.79×{10}^{19} \end{gathered}$$

For $T=10K$, equation (1) becomes,

$S\left(10\right)=a×10+\frac{b}{3}×{10}^3$

By substituting the values of $a$and $b$, we get,

$$\begin{gathered} S\left(10\right)=0.00135×10+\frac{2.48×{10}^{-5}}{3}×{10}^3 \\ S\left(10\right)=2.176×{10}^{-2}{\mathrm{JK}}^{-1} \end{gathered}$$

For unitless form, divide the equation by $k=1.38×{10}^{-23}J{K}^{-1}$

$$\begin{gathered} \frac{S\left(10\right)}{k}=\frac{2.176×{10}^{-2}}{k} \\ \frac{S\left(10\right)}{k}=\frac{2.176×{10}^{-2}}{1.38×{10}^{-23}} \\ \frac{S\left(10\right)}{k}=1.576×{10}^{21} \end{gathered}$$

Hence, the required formula for entropy is $S\left(T_f\right)=a{T}_f+\frac{b}{3}{T}_f^3$.

Entropies at $T=1\mathrm{K}$and $T=10\mathrm{K}$are $S\left(1\right)=1.35×{10}^{-3}{\mathrm{JK}}^{-1}$and $S\left(10\right)=2.176×{10}^{-2}{\mathrm{JK}}^{-1}$respectively.

In dimensionless form, entropy will be $\frac{S\left(1\right)}{k}=9.79×{10}^{19}$and $\frac{S\left(10\right)}{k}=1.576×{10}^{21}$.