# Entropy of diamond and graphite

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problems 3.30 – 3.31.

In a quasistatic process, the relation between entropy, temperature and the heat flow is

$\displaystyle dS=\frac{Q}{T} \ \ \ \ \ (1)$

where ${Q}$ is the (infinitesimal) amount of heat flowing into or out of the system at temperature ${T}$. For a process at constant pressure but changing temperature, ${Q}$ can be written in terms of the heat capacity ${C_{P}}$ at constant pressure, since the amount of heat required to change the temperature by ${dT}$ is ${C_{P}dT}$. In that case, the entropy change between temperatures ${T_{i}}$ and ${T_{f}}$ is

$\displaystyle \Delta S=\int_{T_{i}}^{T_{f}}\frac{C_{P}\left(T\right)}{T}dT \ \ \ \ \ (2)$

Example 1 From Schroeder’s Figure 1.14, we can estimate a linear relation for ${C_{P}}$ for a mole of diamond between ${T=300\mbox{ K}}$ and ${T=400\mbox{ K}}$. Reading off the graph we get

 $\displaystyle C_{P}\left(300\right)$ $\displaystyle =$ $\displaystyle 6.5\mbox{ J K}^{-1}\ \ \ \ \ (3)$ $\displaystyle C_{P}\left(400\right)$ $\displaystyle =$ $\displaystyle 11\mbox{ J K}^{-1} \ \ \ \ \ (4)$

Between these temperatures, a formula for ${C_{P}\left(T\right)}$ is therefore a straight line:

 $\displaystyle \frac{C_{P}\left(T\right)-6.5}{T-300}$ $\displaystyle =$ $\displaystyle \frac{11-6.5}{400-300}=0.045\ \ \ \ \ (5)$ $\displaystyle C_{P}\left(T\right)$ $\displaystyle =$ $\displaystyle 0.045T-7 \ \ \ \ \ (6)$

If we assume this is valid over the range of temperature from 298 K up to 500 K, we can get the entropy change over that range for a mole of diamond (incidentally, if you want to try this experiment, you’ll need a very big diamond. A mole of diamond (carbon) is around 12 grams, and there are 5 carats per gram, so you’re looking for a 60 carat diamond). The entropy change is

 $\displaystyle \Delta S$ $\displaystyle =$ $\displaystyle \int_{298}^{500}\frac{0.045T-7}{T}dT\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[0.045T-7\ln T\right]_{298}^{500}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 5.47\mbox{ J K}^{-1} \ \ \ \ \ (9)$

The entropy of a mole of diamond at ${T=298\mbox{ K}}$ is given in Schroeder’s appendix as ${2.38\mbox{ J K}^{-1}}$ so the total entropy at ${T=500\mbox{ K}}$ is

$\displaystyle S\left(500\right)=2.38+5.47=7.85\mbox{ J K}^{-1} \ \ \ \ \ (10)$

Example 2 An empirical formula obtained by fitting to measured data for ${C_{P}}$ for one mole of graphite is

$\displaystyle C_{P}=a+bT-\frac{c}{T^{2}} \ \ \ \ \ (11)$

where the constants are

 $\displaystyle a$ $\displaystyle =$ $\displaystyle 16.86\mbox{ J K}^{-1}\ \ \ \ \ (12)$ $\displaystyle b$ $\displaystyle =$ $\displaystyle 4.77\times10^{-3}\mbox{ J K}^{-2}\ \ \ \ \ (13)$ $\displaystyle c$ $\displaystyle =$ $\displaystyle 8.54\times10^{5}\mbox{ J K} \ \ \ \ \ (14)$

The entropy change of a mole of graphite over the range of temperature from 298 K up to 500 K is therefore

 $\displaystyle \Delta S$ $\displaystyle =$ $\displaystyle \int_{298}^{500}\frac{a+bT-\frac{c}{T^{2}}}{T}dT\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[a\ln T+bT+\frac{c}{2T^{2}}\right]_{298}^{500}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 6.59\mbox{ J K}^{-1} \ \ \ \ \ (17)$

Adding on the tabulated value for ${T=298\mbox{ K}}$ we get

$\displaystyle S\left(500\right)=5.74+6.59=12.33\mbox{ J K}^{-1} \ \ \ \ \ (18)$

The entropy of graphite is larger than that of diamond which we’d expect since diamond’s crystal structure is more ordered than that of graphite.