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In the book, I have seen that it is written that in an isothermal change, $$\Delta H =\Delta U + \Delta n_\mathrm{g}RT$$ I can clearly understand the derivation where $\Delta H = \Delta U +\Delta (PV)$ And as $PV=nRT$, $\Delta (PV)=\Delta(n_\mathrm{g}RT)$

And as $T=\text{const.}$

$\Delta (PV)=\Delta n_\mathrm{g}RT$

We plunge this value in the above equation. But what my question is, if the process is isothermal, internal energy must be constant i.e. $\Delta U=0$.

Then the equation must be $$\Delta H = \Delta n_\mathrm{g}RT$$

I want to know where is my concept wrong?

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    $\begingroup$ CH SE site prefers not to use MathJax in titles. $\endgroup$
    – Poutnik
    Jun 23 at 4:12
  • $\begingroup$ Wikipedia say ∆U=0 for ideal gas in isothermal process. $\endgroup$
    – Proscionexium
    Jun 23 at 4:39
  • $\begingroup$ ∆U=0 reflect the balance between heat (coming/going) and work (done by system/done on system). $\endgroup$
    – Proscionexium
    Jun 23 at 4:42

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But what my question is, if the process is isothermal, internal energy must be constant.

Thermodynamics in this contexts evaluates changes of state variables. The "state" means their change depends only on the initial and final state, not on the path how the final state has been reached.

Therefore, being isothermal means here that the initial and final temperature are the same, while the temperature profile between these states does not matter.

It says nothing about the $\Delta U$ value. The eventual ongoing process may be exothermic, endothermic or thermally neutral. If you take natural gas and air at $\pu{25 ^{\circ}C}$, burn the gas and cool the products back to $\pu{25 ^{\circ}C}$, it is an isothermal thermodynamic process.

For $\Delta n_\mathrm{g} \ne 0$, a chemical reactions must occur. In such a case, $\Delta U \ne 0$ even if $\mathrm{d}T=0$.

Without any reaction, $\Delta n_\mathrm{g} = \Delta U = \Delta H = 0$ unless the system is open.

With a reaction, generally, $\Delta H = \Delta U + \Delta(pV) \ne 0$ as $\Delta U \ne 0$ and possibly $\Delta (pV) \ne 0$.

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  • $\begingroup$ If $\Delta n_g$ = 0, $p\Delta V =0$ from Avogadro's law of gases ($ V \propto n$) which further means $\Delta H = \Delta U = 0$. I think this is what you mean or not? $\endgroup$
    – Proscionexium
    Jun 23 at 8:32
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    $\begingroup$ Yes, it is what I mean by other words. For $\Delta n=0$ is by consequence of the ideal gas definition $\Delta (pV)_T = 0$ $\endgroup$
    – Poutnik
    Jun 23 at 8:34
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Your equation is based on the condition that a chemical reaction is occurring within a closed system. Even for an ideal gas chemical reaction, $\Delta U$ and $\Delta H$ are not zero even if the change in temperature is zero. Only if the process is adiabatic and in a constant volume system is $\Delta U=0$, but the temperature is not constant in such a case.

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