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What is the internal energy when a reaction is run at constant volume and $9.19\ \mathrm{kJ}$ of heat is absorbed, when run at constant pressure $8.62\ \mathrm{kJ}$ is absorbed? (for an ideal gas)

I know that internal energy for constant volume is equal to the heat then $\Delta U=9.19\ \mathrm{kJ}$, but what is the internal energy for a reaction at constant pressure? I know that $\Delta H=\Delta U+p\Delta V$ and from this I know that $\Delta H=8.62\ \mathrm{kJ}$ but how does it relate to the internal energy?

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  • $\begingroup$ it is an ideal gas $\endgroup$
    – Jake
    Sep 24, 2015 at 19:25

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If you are working with an ideal gas mixture, for which the internal energy of the reactants and products is not a function of pressure, the change in internal energy in going from reactants to products at constant temperature and volume is the same as the change in internal energy in going from reactants to products at constant temperature and pressure. With regard to the changes in enthalpy, the same is true. If the number of moles of reactants and products is the same, then the change in enthalpy for each of these processes will be the same as the change in internal energy. However, if the number of moles changes, the change in enthalpy will differ from the change in internal energy in proportion to the change in the number of moles.

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$$\mathrm dU = \delta Q + \delta W$$For an ideal gas expanding against an external pressure $p$: $$\mathrm dU = \delta Q -p\,\mathrm dV$$

So at constant volume, $\mathrm dU = \delta Q$. Therefore $\mathrm dU = \pu{9.19kJ}$

If the reaction is identical, then $\mathrm dU = \pu{9.19kJ}$ for the ideal gas at constant pressure. From this we can then work out $\delta w$ at constant pressure as $\pu{9.19kJ} - \pu{8.62kJ} = \pu{0.57kJ}$.

Asking for the 'internal energy' and not the change in internal energy is meaningless here, because internal energy, $U$, is a state-function without a scale (unlike $S$) to define absolute values against.

So really you answered your own question in your question, $\Delta U=\pu{9.19 kJ}$

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