0
$\begingroup$

I know that when one applies a manipulation of the Ideal Gas Law to the Maxwell Relations the result that enthalpy is independent of pressure tumbles out of it, i.e., $(\mathrm dH/\mathrm dp)=0$.

I don't understand what this actually means though, because how could it be? If $H=U+pV$ where $U$ is internal energy and $pV$ is the pressure-volume work, then the pressure clearly must play a role. Imagine some other planet where their atmosphere is 10 times the pressure of the Earth atmosphere. It takes more energy to do the $pV$ work to create the space for the system to exist. So for an equal volume is there not a greater enthalpy as a result of increased pressure? More heat at constant pressure, i.e. enthalpy, would need to flow into the system to reach that final state under that larger pressure regime. What is meant that enthalpy is in fact independent of pressure?

$\endgroup$
1
  • 1
    $\begingroup$ PV is not work. It is just multiple of P and V. Ať constant T, PV is constant for ideal gas, in spite of nonzero volume work (and heat compensating this work). $\endgroup$
    – Poutnik
    Nov 19, 2023 at 7:37

1 Answer 1

3
$\begingroup$

The statement means that enthalpy is independent of pressure, for an ideal gas, if temperature and the amount of matter are both held constant.

For example, if you have 1 mol of idealium on Earth's surface at 273 K, it has a volume of 22.4 L. Take that same 1 mol of gas to your exotic planet, where the temperature is still 273 K but the pressure is 10 atm, and the volume will now be just 2.24 L. Although P and V have both changed, PV is the same. The internal energy U is also unchanged because the temperature is unchanged. Therefore the enthalpy is unchanged.

I think you are imagining a system that has a pressure of 10 atm but takes up the same 22.4 L on the exotic planet. But to achieve that, either the temperature would have to be higher, or you would have to have more moles of idealium, or some combination of those. Either way, one of the assumptions is violated.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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