# What is the internal energy at constant pressure?

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?

• it is an ideal gas
– Jake
Sep 24 '15 at 19:25

$$\mathrm dU = \delta Q + \delta W$$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, $\Delta U=\pu{9.19 kJ}$