Question

As shown in Figure 1.14, the heat capacity of diamond near room temperature is approximately linear in T. Extrapolate this function up to $500\mathrm{K}$, and estimate the change in entropy of a mole of diamond as its temperature is raised from$298\mathrm{K}$ to $500\mathrm{K}$. Add on the tabulated value at$298\mathrm{K}$ (from the back of this book) to obtain $S(500\mathrm{K})$.

Short answer

The required entropies are:

$$\begin{gathered} ∆S=5.46J{K}^{-1} \\ S\left(500\right)=7.85J{K}^{-1} \end{gathered}$$

Step-by-step solution

Given Information

From the graph:

At $T=300\mathrm{K}$, role="math" localid="1647289402840" ${\mathrm{C}}_P(300\mathrm{K})=6.5{\mathrm{JK}}^{-1}$.

At $T=400\mathrm{K}$, role="math" localid="1647289412982" ${\mathrm{C}}_P(400\mathrm{K})=11{\mathrm{JK}}^{-1}$

From the table at the back,

At $T=298\mathrm{K}$, ${\mathrm{C}}_P(298\mathrm{K})=2.38{\mathrm{JK}}^{-1}$

Calculation

The slope of a straight line is given as:

$m=\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$

The heat capacity can be found by using the above equation as:

$$\begin{gathered} \frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1} \\ \frac{C_P\left(T\right)-6.5}{T-300}=\frac{11-6.5}{400-300} \\ \frac{C_P\left(T\right)-6.5}{T-300}=0.045 \\ {C}_P\left(T\right)=0.045T-7 \end{gathered}$$

Now, the change in entropy is given as:

$\Delta S={\int }_{T_i}^{T_f}\frac{C_P\left(T\right)}{T}dT$

The change in entropy when the temperature changes from $298K$to $500K$can be given by:

$$\begin{gathered} \Delta S={\int }_{298}^{500}\frac{0.045T-7}{T}dT \\ \Delta S=[0.045T-7\ln T{]}_{298}^{500} \\ \Delta S=0.045(500)-7\times \ln 500-0.045(298)+7\times \ln 298 \\ \Delta S=5.46J{K}^{-1} \end{gathered}$$

The entropy at $500K$can be given as:

role="math" localid="1647290666617" $$\begin{gathered} S(500K)=2.38+5.46 \\ S(500K)=7.85J{K}^{-1} \end{gathered}$$

Final answer

The required change in entropy is $5.46J{K}^{-1}$and $S(500K)$is calculated to be$7.85J{K}^{-1}$.