# Heat Capacity

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

Let $$\alpha\in\{p,V\}$$. So, $$S(T,\alpha,N)$$ with $$N$$ implied. $$dS = \left(\frac{\partial S}{\partial T}\right)_\alpha\:dT + \left(\frac{\partial S}{\partial \alpha}\right)_T\:d\alpha$$. So, $$\left(\frac{\partial S}{\partial T}\right)_\alpha=\frac{C_\alpha}{T}$$, $$C_\alpha=T\left(\frac{\partial S}{\partial T}\right)_\alpha$$.

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, $$dU=\left(\frac{\partial U}{\partial T}\right)_{V,N}\:dT = C_V\:dT$$. $$C_P\neq\left(\frac{\partial U}{\partial T}\right)_P$$. $$C_P=\left(\frac{\partial H}{\partial T}\right)_P=T\left(\frac{\partial S}{\partial T}\right)_P$$

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

Abstractly, f(g,h) <> g(h,f) or h(f,g). $$df=\left(\frac{\partial f}{\partial g}\right)_h\:dg + \left(\frac{\partial f}{\partial h}\right)_g\:dh$$. $$dg=\left(\frac{\partial g}{\partial f}\right)_h\:df + \left(\frac{\partial g}{\partial h}\right)_f\:dh$$. So, $$df=\left[\left(\frac{\partial f}{\partial g}\right)\left(\frac{\partial g}{\partial h}\right)_f+\left(\frac{\partial f}{\partial h}\right)_g\right]dh + \left[\left(\frac{\partial f}{\partial g}\right)_h\left(\frac{\partial g}{\partial f}\right)_h\right]df$$. Thus, the second coefficient is 1 so the first coefficient must be zero. Then, $$\left(\frac{\partial f}{\partial g}\right)_h\left(\frac{\partial g}{\partial h}\right)_f\left(\frac{\partial h}{\partial f}\right)_g=1$$. And, $$\left(\frac{\partial f}{\partial g}\right)_h = \frac{1}{\left(\frac{\partial g}{\partial f}\right)_h}$$.

We get the Maxwell relations: $$\left(\frac{\partial}{\partial h}\left(\frac{\partial f}{\partial g}\right)_h\right)_g=\left(\frac{\partial}{\partial g}\left(\frac{\partial f}{\partial h}\right)_g\right)_h$$. Note that $$F(f,g,h)=0$$. $$dF=dU-TdS-SdT=-SdT-pdV+\mu dN = -\left(\frac{\partial}{\partial V}\left(\frac{\partial F}{\partial T}\right)_{VN}\right)_{TN} = - \left(\frac{\partial}{\partial T}\left(\frac{\partial F}{\partial V}\right)_{TN}\right)_{VN}$$. So, $$\left(\frac{\partial S}{\partial V}\right)_{TN} = \left(\frac{\partial P}{\partial T}\right)_{VN}$$.

Created: 2024-05-30 Thu 21:19

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