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.

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