91Ó°ÊÓ

$$We have the information of entropies, so

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

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

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

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

The slope of phase boundary will be:

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

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

$\mathrm{Δ}S→0$at high temperatures because, according to the Dulong-Petits law, both heat capacities approach 3R at high temperatures.

$∴\mathrm{Δ}S→0$

$\text{As}T→0\text{, both entropies approaches zero, so}\mathrm{Δ}S→0$

$∴\frac{dP}{dT}=\frac{\mathrm{Δ}S}{\mathrm{Δ}V}=0$

At T=0 K the slope will be zero.