# Need help understanding why isobaric heat capacity: $\left(\frac{\partial H}{\partial T}\right)_P=\,{T\left(\frac{\partial S}{\partial T}\right)_P}$

If the state function Enthalpy $H$ is defined by $$H=U+PV\tag{1}$$ where $U$ is the internal energy of the system, $P$ is the systems pressure and $V$ is its volume.

It follows that $$dH=TdS + VdP\tag{2}$$ by use of the product rule and the first law of thermodynamics: $$dU=TdS - PdV\tag{3}$$ where $T$ is the thermodynamic (absolute) temperature and $S$ is the entropy of the system defined by $$dS=\frac{\delta Q_R}{T}\tag{4}$$ where $Q_R$ is the heat energy supplied to a reversible thermodynamic process.

One way of describing enthalpy is to say that the change in enthalpy is the heat flow in an isobaric, reversible process. From equation $(2)$ for such a process it is easily seen that $$𝑑𝐻=TdS=\delta 𝑄_R\tag{5}$$ and this then gives expressions for constant pressure heat capacity: $$C_P=\left(\frac{\partial H}{\partial T}\right)_P=\,\color{red}{T\left(\frac{\partial S}{\partial T}\right)_P}\tag{6}$$

I don't understand how the part marked red in equation $(6)$ was obtained, since heat capacity in general is defined by $$C=\frac{\delta Q}{dT}\tag{7}$$ and from equation $(4)$ $$\delta Q_R=TdS\tag{8}$$ so taking the partial derivative of equation $(8)$ with respect to $T$ means $$\frac{\partial (TdS)}{\partial T}=dS+T\partial (dS)\ne \color{red}{T\left(\frac{\partial S}{\partial T}\right)_P}$$

• Why aren't you substituting eqn. 8 in 7, given that pressure is constant for both expressions? – Deep Sep 20 '16 at 5:12
• @Zero Yes, very good point. If I did so I will get $C=\dfrac{\delta Q}{dT}=T\left(\dfrac{dS}{dT}\right)$. Which is similar to $T\left(\dfrac{\partial S}{\partial T}\right)_P$. How can I justify $d \longrightarrow \partial$? Sorry, it's still not quite clear to me. – BLAZE Sep 20 '16 at 7:05
• It is incorrect to write $\frac{dS}{dt}$, because in your expressions $p$ is held constant. Work with small quantities: $\delta Q=T \delta S, C=\frac{\delta Q}{\delta T}\bigr|_p$, substitute and take the limit $\delta T\rightarrow 0$, holding $p$ constant. – Deep Sep 20 '16 at 8:52

Since Enthalpy $(H)$ is a state function, it can be expressed as a function of either of the two variables of $P,\,V,\,T\,.$

So, consider enthalpy as a function of $T$ and $P\,.$

So, its differential can be written as

$$\mathrm dH = \left(\frac{\partial H}{\partial T}\right)_P ~\mathrm dT + \left(\frac{\partial H}{\partial P}\right)_T~\mathrm dP\tag I$$

From the First Law and assuming non-compression work is zero, we get

$$\mathrm dH= đq + V~\mathrm dP\tag{II}$$

Now,

\begin{align}\mathrm dS &= \frac{đq}{T}\\ \implies \mathrm dS &= \frac{(\mathrm dH- V~\mathrm dP)}{T}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\mathrm{Using~~ II}]\\ \implies \mathrm dS &= \frac1T\left(\frac{\partial H}{\partial T}\right)_P~\mathrm dT + \frac1T\left[\left(\frac{\partial H}{\partial P}\right)_T- V\right]~\mathrm dP~~~~~~~[\mathrm{Using~~ I}]\tag{III}\end{align}

Also, $$\mathrm dS= \left(\frac{\partial S}{\partial T}\right)_P~\mathrm dT + \left(\frac{\partial S}{\partial P}\right)_T~\mathrm dP\tag{IV}$$

Comparing $\rm (III)$ and $\mathrm{ (IV)}$ we get

\begin{align}\left(\frac{\partial S}{\partial T}\right)_P&=\frac1T\left(\frac{\partial H}{\partial T}\right)_P\\\implies \left(\frac{\partial S}{\partial T}\right)_P&= \frac{C_p}T\,. \tag{V} \end{align}

• Thank you very much for your answer, I just fixed a minor typo in the partial derivatives of $(\mathrm{I})$. – BLAZE Sep 20 '16 at 12:17

In freshman physics, we learned that, when heat is added to a constant volume system, we can write Q = CΔT, where C is called the heat capacity. However, when we got more deeply into the basics and learned thermodynamics, we found that this elementary approach is no longer adequate (or precise). We found that Q depends on process path and that, if work W is occurring, this substantially changes things. However, we still wanted C to continue to represent a physical property of the material being processed, and not to depend on process path or whether work is occurring. This is dealt with in thermodynamics by changing the definition of C a little. Rather than associating C with the path dependent heat Q, in thermodynamics, we associate C with parameters relating to the state of the material being processed, in particular internal energy U and enthalpy H. We define the heat capacity at constant volume $C_v$ as the derivative of the internal energy U with respect to temperature at constant volume: $$C_v=\left(\frac{\partial U}{\partial T}\right)_v\tag{1}$$ We also found that we could define a heat capacity at constant pressure $C_p$ as the derivative of the enthalpy H with resepct to temperature at constant pressure:$$C_p=\left(\frac{\partial H}{\partial T}\right)_p\tag{2}$$ The question is, "do either of these definitions reduce to the more elementary version from freshman physics under any circumstances." The answer is "yes." From the first law of thermodynamics, we find that, for a closed system of constant volume (no work being done), $Q=\Delta U=C_v\Delta T$, and, for a closed system experiencing a constant pressure change (with $W=p\Delta v$), $Q=\Delta H=C_p\Delta T$. Of course, Eqns. 1 and 2 are much more generally applicable than this.

• Is it also correct to say that a general definition would be : $C_X = \left(\frac{\delta Q^{rev}}{dT} \right)_X$ ? – StarBucK Feb 25 at 21:42
• Not in my judgment. – Chet Miller Feb 25 at 23:08
• Why ? Isnt this definition path independant as I force to take a reversible path. And it is consistent with the case where the transformation is truly reversible ? – StarBucK Feb 25 at 23:11
• This is not general enough. The equations in my answer apply even if the path is not constant pressure or constant volume. Thus, in the case of an ideal gas for example, dU=CvdT and dH=CpdT irrespective of whether the path is constant volume or constant pressure. – Chet Miller Feb 25 at 23:15
• I see what you mean. Then I dont fully understand your answer. How do we know that even if the volume is not constant we will have $dU=Cv dT$ that will still be true. Because mathematically it is only the partial derivative at fixed constant volume so I dont get why it would still be true. – StarBucK Feb 25 at 23:21