# Why does the derivation Gibbs-Helmholtz equation hold when the enthalpy is temperature-dependant

The derivation shown here assumes H and U are independant of temperature, right? But what if $$H(T) = H_{0}+C_V \Delta T$$?

Then $$dG = C_V\, dT - SdT$$, right? So $$\displaystyle \left( \frac{\partial G}{\partial T}\right)_p=C_V -S$$ ?

\begin{align} \left(\frac{\partial (G/T)}{\partial T}\right)_p &= \frac{T(\partial G/\partial T)_p - G(\partial T/\partial T)_p} {T^2} \\[8pt] &= \frac{T(C_V-S) - G(1)}{T^2} \\[8pt] &= \frac{TC_V-TS-G}{T^2} \\[8pt] &= -\frac{H_0}{T^2} \end{align}

as desired (since $$G = H_0+C_V\Delta T - TS$$).

But this way we see that the H in this equation would only be containing the constant term? Of course we insert H(T) to get thee exact G, but where in my thinking process did I go wrong? How would this derivation be made correctly, assuming that H changes with Temperature?

• No, there are no assumptions. You should consider that S is temperature-dependent as well if your plan is to derive the temperature-dependence of G by adding the temperature dependence of H and S. I bet something will cancel out. – Karsten Theis Apr 30 at 22:02
• Should not it to be Cp for G, resp. Cv for A? As H and G are at constant pressure, while U and A at constant volume. – Poutnik May 1 at 5:59
• dG=Cp.dT - S.dT - TdS=Cp.dT - S.dT - dQ= - S.dT // dG/dT= -S – Poutnik May 1 at 7:08
• I edited the first term in the derivation to something that seems sensible, starting from a "mauled" form that doesn't make sense. – Buck Thorn May 1 at 9:53
• Your equation $( \frac{\partial G}{\partial T})_p=C_V -S$ is wrong. It should read $( \frac{\partial G}{\partial T})_p=-S$ as in the answer you link to. – Buck Thorn May 1 at 10:00

The derivation of the van't Hoff equation does not require that you specify the T-dependence of H. It arises from computing the derivative (limiting slope) of $$G/T$$ so you need to consider a polynomial expansion of $$G/T$$ to linear terms in $$T$$ around the temperature of interest (that is, an infinitely small interval dT). At the risk of redundancy:

\begin{align} (d(G/T))_p &= ((1/T)dG-G(dT/T^2))_p \\ &= ((1/T)(VdP-SdT)-(H-TS)(dT/T^2))_p \\ &= (-(S/T)dT-(H/T^2-S/T)dT)_p \\ &= (-(H/T^2)dT)_p \end{align}

The answer is therefore found in differential Calculus.

Another comment explains that, if

$$G(T)=H(T) -TS(T)=H_0 +C_p \Delta T -TS(T)$$

where $$H(T)$$ is expressed as a linear polynomial in T ($$C_p$$ and $$p$$ assumed constant) then

$$dG=C_p dT -SdT-TdS$$

Since for a process at constant pressure, $$dq = C_p dT=TdS$$ (since $$C_p = (dH/dT)_p$$ and $$dq_p= dH$$; also for a reversible process $$dq= TdS$$), it follows that

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

and the van't Hoff expression provided in the linked answer also follows.

• Thank your very much! Could you also explain how we get to $C_pdT=TdS$? – Astronguem May 1 at 20:42
• $C_p = (dH/dT)_p$ and at constant p $dQ_p= dH$. Also for a reversible process $dQ= TdS$. – Buck Thorn May 2 at 15:03