# How can enthalpy of an ideal gas be independent of pressure?

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., (dH/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?

• 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). Nov 19 at 7:37