# Deriving an expression for the constant pressure heat capacity

Given the relation $G = H − T S$, show that: $$\left(\frac{\partial S}{\partial T}\right)_p=\frac{C_p}T$$

Now I realise that $T\mathrm dS=\mathrm dQ$ and that would I suppose get the result. However, I have been asked to use the expression $G = H - TS$ to find the result. If I just differentiate implicitly I end up getting:

$$\left(\frac{\partial G}{\partial T}\right)_p+\left(\frac{\partial S}{\partial T}\right)_p=\frac{C_p}T$$

What am I doing wrong?

• Btw, how did you get to your final equation. If I just differentiate $G = H -TS$ with respect to $T$ I get something else. Or what did you mean by "differentiate implicitly"? Apr 7 '16 at 8:50
• I differentiated with respect to T yes. Perhaps I did it wrong; what did you get? I think I assumed that dT/dT was zero but it's actually 1 isn't it? Perhaps that's where I went wrong? Apr 7 '16 at 17:31
• Yes, that's where the differnce between your equation and mine came from. And actually this gets you to your desired result even faster than in my answer. You'd have $\left(\frac{\partial G}{\partial T}\right)_p+\left(\frac{\partial S}{\partial T}\right)_p + S=\frac{C_p}T$. Then just use $\left(\frac{\partial G}{\partial T}\right)_p = -S$ and you're there. I should have tried that way first :) Apr 7 '16 at 20:12

You could start by dividing your whole equation by $T$:

\begin{align} \frac{G}{T} = \frac{H}{T} - S \end{align}

and then take the derivative with respect to $T$ at constant $p$:

\begin{align} \left(\frac{\partial\frac{G}{T}}{\partial T}\right)_{p} = \left(\frac{\partial\frac{H}{T}}{\partial T}\right)_{p} - \left(\frac{\partial S}{\partial T} \right)_{p} \ . \end{align}

That way you've already introduced the desired derivative of the entropy with respect to $T$ at constant $p$. Then you can use that

\begin{align} \frac{G}{T} = \frac{H}{T} - S \qquad &\Rightarrow& \qquad \mathrm{d} \!\left(\frac{G}{T}\right) &= - \frac{H}{T^2} \mathrm{d} T + \frac{1}{T} \underbrace{\mathrm{d} H}_{= T \mathrm{d}S + V\mathrm{d}p} - \mathrm{d} S \\ &&&= - \frac{H}{T^2} \mathrm{d} T + \frac{V}{T} \mathrm{d}p \\ &\Rightarrow& \qquad \left(\frac{\partial\frac{G}{T}}{\partial T}\right)_{p} &= - \frac{H}{T^2} \end{align}

and simply using the product rule

\begin{align} \left(\frac{\partial\frac{H}{T}}{\partial T}\right)_{p} = \frac{1}{T} \underbrace{\left(\frac{H}{\partial T}\right)_{p}}_{= C_{p}} + H \underbrace{\left(\frac{\partial \frac{1}{T}}{\partial T} \right)_{p}}_{=-\frac{1}{T^2}} = \frac{C_{p}}{T} - \frac{H}{T^2} \ . \end{align}

Putting it all together gives you the desired result:

\begin{align} \left(\frac{\partial\frac{G}{T}}{\partial T}\right)_{p} &= \left(\frac{\partial\frac{H}{T}}{\partial T}\right)_{p} - \left(\frac{\partial S}{\partial T} \right)_{p} \\ - \frac{H}{T^2} &= \frac{C_{p}}{T} - \frac{H}{T^2} - \left(\frac{\partial S}{\partial T} \right)_{p} \\ \Rightarrow \left(\frac{\partial S}{\partial T} \right)_{p} &= \frac{C_{p}}{T} \ . \end{align}

Starting with $dG=-SdT+VdP$, we have $$\left(\frac{\partial G}{\partial T}\right)_P=-S\tag{1}$$ But, from the equation $G = H - TS$, we also have:$$\left(\frac{\partial G}{\partial T}\right)_P=\left(\frac{\partial H}{\partial T}\right)_P-S-T\left(\frac{\partial S}{\partial T}\right)_P\tag{2}$$ If we combine Eqns. 1 and 2, we obtain:$$\left(\frac{\partial H}{\partial T}\right)_P=T\left(\frac{\partial S}{\partial T}\right)_P\tag{3}$$The left hand side of Eqn. 3 is the definition of $C_P$. So,$$C_P=T\left(\frac{\partial S}{\partial T}\right)_P\tag{4}$$ QED