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?



2 Answers 2


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.


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,

$$\md S = \frac{\md q_\mathrm{rev}}{T} = \frac{C_p\,\md T}{T} \Longrightarrow \pdiff{S}{T}{p} = \frac{C_p}{T} \endgroup$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.