1
$\begingroup$

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?

Improve this question
$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Karsten Theis Apr 30 '19 at 22:02
  • $\begingroup$ 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. $\endgroup$ – Poutnik May 1 '19 at 5:59
  • 1
    $\begingroup$ dG=Cp.dT - S.dT - TdS=Cp.dT - S.dT - dQ= - S.dT // dG/dT= -S $\endgroup$ – Poutnik May 1 '19 at 7:08
  • $\begingroup$ I edited the first term in the derivation to something that seems sensible, starting from a "mauled" form that doesn't make sense. $\endgroup$ – Buck Thorn May 1 '19 at 9:53
  • $\begingroup$ 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. $\endgroup$ – Buck Thorn May 1 '19 at 10:00
1
$\begingroup$

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.

Improve this answer
$\endgroup$
  • $\begingroup$ Thank your very much! Could you also explain how we get to $C_pdT=TdS$? $\endgroup$ – Astronguem May 1 '19 at 20:42
  • $\begingroup$ $C_p = (dH/dT)_p$ and at constant p $dQ_p= dH$. Also for a reversible process $dQ= TdS$. $\endgroup$ – Buck Thorn May 2 '19 at 15:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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