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, \(\left(\frac{\partial S}{\partial V}\right)_{TN} = \left(\frac{\partial P}{\partial T}\right)_{VN}\).

Author: Christian Cunningham

Created: 2023-06-25 Sun 02:30

Validate