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I made a little derivation for the Enthalpydifference $\Delta H$. I got as a solution:

$\Delta U+\underbrace{\Delta p\cdot V_{1}+p_{2}\cdot\Delta V}_{a}$

Now I do want to compute the transferred heat of the process by using the 1st law of thermodynamics:

$Q=\Delta U+\Delta E_{kin}+\Delta E_{pot}+W_{rest}+W_{vol}$

Where I divided Work $W$ in to $W_{rest}$ and $W_{vol}=\int_{1}^{2}p\left(V\right)\cdot dV$

So for the case that $\Delta V=0$ I can easily calculate $Q$ with the a table for $U$, if I know $\Delta E_{kin}$, $\Delta E_{pot}$ and $W_{rest}$:

$Q=\Delta U+\Delta E_{kin}+\Delta E_{pot}+W_{rest}$

because $W_{vol}=0$

It is similar for the case $\Delta p=0$:

Although $W_{vol}=p \cdot \Delta V \neq 0$, $a$ does simplify to $a=p \cdot \Delta V$ and therefore I can write $\Delta H = \Delta U + W_{vol}$

$Q=\Delta H+\Delta E_{kin}+\Delta E_{pot}+W_{rest}$.

But what about all the other cases? Am I right that there is no simple way to calculate $Q$ by using $\Delta H$?

B. Preiss
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1 Answers1

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According to Denbigh, "The enthalpy change of a closed system is thus equal to the heat absorbed under two restrictive conditions, namely, that there is constancy of the pressure and that the only form of work is due to the volume change of the system. In a galvanic cell, where electrical work is done, the heat absorbed is not equal to the change in enthalpy."

Chet Miller
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