Thermodynamics: Combining the 1st and 2nd Law

I'm confused how combining the first and second laws can derive


$$dS = \left(\frac{C\_p}{T}\right)dT+ \left[ \left( \frac{1}{T} \right) \left\{ \left( \frac{\partial H}{\partial P} \right)\_T -V \right\} \right] dP$$


The 1st Law states $dU = dq + dw$, and the 2nd Law states $dS = \frac{dq\_{rev}} {T}$


I factored out $\frac{1}{T}$ in the given equation to make


$$dS = \frac{1}{T} \left[ \left( C\_p\, dT \right) + { \left( \frac{\partial H}{\partial P} \right)\_TdP - V dP } \right]$$


$C\_p = \frac{dq}{dT}$


Therefore, 
$$dS = \frac{1}{T} \left[ dq + { \left( \frac{\partial H}{\partial P} \right)\_TdP - V dP } \right] $$


Without analyzing $\left\{ \left( \frac{\partial H}{\partial P} \right)\_T - V dP \right\}$, I already have $dS = \frac{dq}{T}$.


Can anyone explain this for me?


Thanks,




Starting with $$dH=TdS+VdP\tag{1}$$H is a function of T and P, so$$dH=\left(\frac{\partial H}{\partial T}\right)\_PdT+\left(\frac{\partial H}{\partial P}\right)\_TdP\tag{2}$$Substituting Eqn. 2 into Eqn. 1,
$$\left(\frac{\partial H}{\partial T}\right)\_PdT+\left(\frac{\partial H}{\partial P}\right)\_TdP=TdS+VdP\tag{3}$$But, from the **definition of $C\_p$**, $$C\_p\equiv \left(\frac{\partial H}{\partial T}\right)\_P \tag{4}$$Substituting Eqn. 4 into Eqn. 3 gives:$$C\_pdT+\left(\frac{\partial H}{\partial P}\right)\_TdP=TdS+VdP\tag{5}$$Solving for dS then gives the desired relationship.


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Firstly note that the total differential of $S(p,T)$ is 


$$\require{begingroup} \begingroup \newcommand{\md}{\mathrm{d}} \newcommand{\pdiff}[3]{\left(\frac{\partial #1}{\partial #2}\right)\_{\!#3}} \md S = \pdiff{S}{p}{T}\,\md p + \pdiff{S}{T}{p}\,\md T$$


so your question amounts to evaluating the two partial derivatives.


For the first partial derivative, I think one simple possibility is to note that $G = H - TS$ and hence $S = (H - G)/T$. Therefore


$$\begin{align}
\pdiff{S}{p}{T} &= \pdiff{}{p}{T}\frac{H - G}{T} \\
&= \frac{1}{T}\left[\pdiff{H}{p}{T} - \pdiff{G}{p}{T}\right] \\
&= \frac{1}{T}\left[\pdiff{H}{p}{T} - V\right]
\end{align}$$


using the equation $\md G = V\,\md p - S\,\md T$ which implies $(\partial G/\partial p)\_T = V$.


In case you're wondering where the First Law comes in: you need the First Law to derive that equation for $\md G$, via $\md U = \md q + \md w = T\,\md S - p\,\md V$.


The second partial derivative is straightforward: under constant pressure,


$$\md S = \frac{\md q\_\mathrm{rev}}{T} = \frac{C\_p\,\md T}{T} \Longrightarrow \pdiff{S}{T}{p} = \frac{C\_p}{T} \endgroup$$


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