In 1 and 2, you calculated the entropies of diamond and

Chapter 5, Problem 38P

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Problem 38P

In Problems 1 and 2, 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 below Figure 1. Why is the slope almost constant at still higher temperatures? Why is the slope zero at T = 0?

Figure 1: The experimentalphase diagram of carbon.The stability region of the gasphase is not visible on this scale;the graphite-liquid-gas triplepoint is at the bottom of thegraphite-liquid phase boundary,at 110 bars pressure. FromDavid A. Young, Phase Diagramsof the Elements (Universityof California Press, Berkeley,1991).

Problem 1:

As shown in below Figure 2, the heat capacity of diamond near room temperature is approximately linear in T. Extrapolate this function up to 500 K, and estimate the change in entropy of a mole of diamond as its temperature is raised from 298 K to 500 K. Add on the tabulated value at 298 K (from the backof this book) to obtain S(500 K).

Figure 2: Measured heat capacities at constant pressure (data points) forone mole each of three different elemental solids. The solid curves show the heatcapacity at constant volume predicted by the model used in Section 7.5, with thehorizontal scale chosen to best fit the data for each substance. At sufficiently hightemperatures, CV for each material approaches the value 3R predicted by theequipartition theorem. The discrepancies between the data and the solid curvesat high T are mostly due to the differences between CP and CV. At T = 0 alldegrees of freedom are frozen out, so both CP and CV go to zero. Data from Y. S.Touloukian, ed., Thermophysical Properties of Matter (Plenum, New York, 1970).

Problem 2:

Experimental measurements of heat capacities are often represented in reference works as empirical formulas. For graphite, a formula that works well over a fairly wide range of temperatures is (for one mole)

where a = 16.86 J/K, b = 4.77 × 10–3 J/K2, and c = 8.54 × 105 J∙K. Suppose, then, that a mole of graphite is heated at constant pressure from 298 K to 500 K. Calculate the increase in its entropy during this process. Add on the tabulated value of S(298 K) (from the back of this book) to obtain S(500 K).