Problem 5.14 (Schroeder’s Intro to Thermal Physics)

Problem 5.14

The partial-derivative relations derived in Problems 1.46, 3.33, and 5.12, plus a bit more partial-derivative trickery, can be used to derive a completely general relation between $C_P$ and $C_V.$

(a) With the heat capacity expressions from Problem 3.33 in mind, first consider $S$ to be a function of $T$ and $V.$ Expand $dS$ in terms of the partial derivatives $(\partial S/ \partial T)_V$ and $(\partial S/ \partial V)_T.$ Note that one of these derivatives is related to $C_V.$
(b) To bring in $C_P,$ consider $V$ to be a function of $T$ and $P$ and expand $dV$ in terms of partial derivatives in a similar way. Plug this expression for $dV$ into the result of part (a), then set $dP = 0$ and note that you have derived a nontrivial expression for $(\partial S/\partial T)_P.$ This derivative is related to $C_P,$ so you now have a formula for the difference $C_P - C_V.$
(c) Write the remaining partial derivatives in terms of measurable quantities using a Maxwell relation and the result of Problem 1.46. Your final result should be

$C_P = C_V + \dfrac{TV \beta^2}{\kappa_T}.$

(d) Check that this formula gives the correct value of $C_P - C_V$ for an ideal gas.
(e) Use this formula to argue that $C_P$ cannot be less than $C_V.$
(f) Use the data in Problem 1.46 to evaluate $C_P - C_V$ for water and for mercury at room temperature. By what percentage do the two heat capacities differ?
(g) Figure 1.14 shows measured values of $C_P$ for three elemental solids, compared to predicted values of $C_V.$ It turns out that a graph of $/beta$ vs. $T$ for a solid has same general appearance as a graph of heat capacity. Use this fact to explain why $C_P$ and $C_V$ agree at low temperatures but diverge in the way they do at higher temperatures.