Heat Capacity

$C_{V or P} = \frac{d Q}{d T}$ This should be related to $\frac{\partial S}{\partial T}$ at constant volume or poressure, N is implied. Remark: $d S := \frac{d Q}{d T}$ .

Let $α \in {p, V}$ . So, $S (T, α, N)$ with $N$ implied. $d S = {(\frac{\partial S}{\partial T})}_{α} d T + {(\frac{\partial S}{\partial α})}_{T} d α$ . So, ${(\frac{\partial S}{\partial T})}_{α} = \frac{C_{α}}{T}$ , $C_{α} = T {(\frac{\partial S}{\partial T})}_{α}$ .

We could also get this from checking dimensions of the original formulation and seeing that entropy has units of Energy per temperature and checking extrinsic and intrinsic.

dU=dQ+dW. dW is constant for constant volume. So, dU=dQ. So, $d U = {(\frac{\partial U}{\partial T})}_{V, N} d T = C_{V} d T$ . $C_{P} \neq {(\frac{\partial U}{\partial T})}_{P}$ . $C_{P} = {(\frac{\partial H}{\partial T})}_{P} = T {(\frac{\partial S}{\partial T})}_{P}$

Recall: dU=TdS-pdV and dS=(dS/dT)dT+(dS/dP)dP dH=T(dS/dT)dT=C_PdT.

Abstractly, f(g,h) <> g(h,f) or h(f,g). $d f = {(\frac{\partial f}{\partial g})}_{h} d g + {(\frac{\partial f}{\partial h})}_{g} d h$ . $d g = {(\frac{\partial g}{\partial f})}_{h} d f + {(\frac{\partial g}{\partial h})}_{f} d h$ . So, $d f = [(\frac{\partial f}{\partial g}) {(\frac{\partial g}{\partial h})}_{f} + {(\frac{\partial f}{\partial h})}_{g}] d h + [{(\frac{\partial f}{\partial g})}_{h} {(\frac{\partial g}{\partial f})}_{h}] d f$ . Thus, the second coefficient is 1 so the first coefficient must be zero. Then, ${(\frac{\partial f}{\partial g})}_{h} {(\frac{\partial g}{\partial h})}_{f} {(\frac{\partial h}{\partial f})}_{g} = 1$ . And, ${(\frac{\partial f}{\partial g})}_{h} = \frac{1}{{(\frac{\partial g}{\partial f})}_{h}}$ .

We get the Maxwell relations: ${(\frac{\partial}{\partial h} {(\frac{\partial f}{\partial g})}_{h})}_{g} = {(\frac{\partial}{\partial g} {(\frac{\partial f}{\partial h})}_{g})}_{h}$ . Note that $F (f, g, h) = 0$ . $d F = d U - T d S - S d T = - S d T - p d V + μ d N = - {(\frac{\partial}{\partial V} {(\frac{\partial F}{\partial T})}_{V N})}_{T N} = - {(\frac{\partial}{\partial T} {(\frac{\partial F}{\partial V})}_{T N})}_{V N}$ . So, ${(\frac{\partial S}{\partial V})}_{T N} = {(\frac{\partial P}{\partial T})}_{V N}$ .