Question

In Problems 3.30 and 3.31 you calculated the entropies of diamond and graphite at 500 K. Use these values to predict the slope of the graphite- diamond phase boundary at 500 K, and compare to Figure 5.17. Why is the slope almost constant at still higher temperatures? Why is the slope zero at T = 0?

Short answer

At T=0 K the slope will be zero.

Step-by-step solution

Given information

$$We have the information of entropies, so

$$\begin{gathered} \text{Entropy of diamond at}500\mathrm{K}\text{is}{\mathrm{S}}_{\mathrm{a}}=7·59\mathrm{J}/\mathrm{K} \\ \text{Entropy of diamond at}500\mathrm{K}\text{is}{\mathrm{S}}_{\mathrm{g}}=12·33\mathrm{J}/\mathrm{K} \\ \text{Molar Volumes of diamond is}{\mathrm{V}}_{\mathrm{d}}=3·42{\mathrm{cm}}^3 \\ =3.42\times {10}^{-6}{\mathrm{m}}^3 \\ \text{Molar Volumes of graphite is}{\mathrm{V}}_{\mathrm{g}}=6·30{\mathrm{cm}}^3 \\ =5·30\times {10}^{-6}{\mathrm{m}}^3 \end{gathered}$$

Now, the change in entropies and molar volume will be:

$$\begin{gathered} \text{Change in entropies}\Delta \mathrm{S}={\mathrm{S}}_{\mathrm{g}}-{\mathrm{S}}_{\mathrm{d}} \\ =12·33\mathrm{J}/\mathrm{K}-7·59\mathrm{J}/\mathrm{K} \\ =4.74\mathrm{J}/\mathrm{K} \end{gathered}$$

role="math" localid="1646931515944" $$\begin{gathered} \mathrm{Change}\mathrm{in}\mathrm{Molar}\mathrm{volumes}\Delta \mathrm{V}={\mathrm{V}}_{\mathrm{g}}-{\mathrm{V}}_{\mathrm{d}} \\ =\left(5·30\times {10}^{-6}{\mathrm{m}}^3\right)-\left(3·42\times {10}^{-6}{\mathrm{m}}^3\right) \\ =1.88\times {10}^{-6}{\mathrm{m}}^3 \end{gathered}$$

Calculation

The slope of phase boundary will be:

$$\begin{gathered} \frac{dP}{dT}=\frac{\Delta S}{\Delta V} \\ =\frac{4.74\mathrm{J}/\mathrm{K}}{1·88\times {10}^{-6}{\mathrm{m}}^3} \\ =2·5213\times {10}^6\frac{\mathrm{N}/{\mathrm{m}}^2}{\mathrm{K}} \\ \therefore \frac{\mathrm{dP}}{\mathrm{dT}}=2·52\times {10}^6\frac{\mathrm{Pa}}{\mathrm{K}} \end{gathered}$$

The graphite-diamond phase boundary slope at 500 K is this.

Calculations

$\Delta S\to 0$at high temperatures because, according to the Dulong-Petits law, both heat capacities approach 3R at high temperatures.

$\therefore \Delta S\to 0$

$\text{As}T\to 0\text{, both entropies approaches zero, so}\Delta S\to 0$

$\therefore \frac{dP}{dT}=\frac{\Delta S}{\Delta V}=0$

At T=0 K the slope will be zero.