If a gas expands in a container with constant external pressure then:

$W=-P_{ext}\Delta V$

Instead of $P_{ext}$ if we use $P_{int}$ then we will not get the same value because $P_{int}$ is constantly decreasing from a certain value until equilibrium is attained, while the change in volume remains the same.

This is the reason why we use $P_{ext}$ here:

$\Delta H= \Delta U +P_{ext} \Delta V$

But then, $\Delta H$ for a reaction is:

$\Delta H= \Delta U +\Delta n_{g}RT$ ($n_{g}$ is for change in number of gaseous moles as the solids and liquids are ignored)

where $PV=nRT$ is used. However, the $P$ in the ideal gas equation is $P_{int}$.

So my question is how can $P_{int}$ be substituted in the place of $P_{ext}$ if the values of work done change when we do so?

  • $\begingroup$ Who says the change in enthalpy is $\Delta U+P_{ext}\Delta V$? Please provide a reference. This is not correct. $\endgroup$ Mar 20, 2021 at 12:30
  • $\begingroup$ I've seen $\Delta H=\Delta U +P \Delta V$ in text books and when showing that $\Delta H=q$ under constant pressure the work from $\Delta U = q+w$ gets canceled out with $P \Delta V$ from $\Delta H$ so I thought that for it to cancel out with work the $P$ had to be $P_{ext}$.Hence, the $\Delta H=\Delta U +P_{ext} \Delta V$ $\endgroup$
    – whae
    Mar 20, 2021 at 12:58
  • $\begingroup$ How would that work for an adiabatic irreversible expansion at constant external pressure (for which q = 0)? $\endgroup$ Mar 20, 2021 at 13:17
  • $\begingroup$ @ChetMiller I don't see why it woudn't work. Wouldn't $\Delta H=0$ for adiabatic irreversible expansion. $\endgroup$
    – whae
    Mar 20, 2021 at 14:13
  • $\begingroup$ Does the temperature change? Is the enthalpy of an ideal gas a function of temperature? $\endgroup$ Mar 20, 2021 at 14:26

1 Answer 1


Your claim that $\Delta H = \Delta U + P_{ext} \Delta V$ is incorrect. It is only correct when the system is always in mechanical equilibrium with the external pressure, or in other words irreversible expansion against constant pressure. The more general statement is $\Delta H = \Delta U + \Delta (PV) $. Here as you can see even when you compare it with $\Delta H = \Delta U + \Delta n_gRT$ there is no problem as we don't have to replace $P_{in} $ with $P_{ext} $.


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