Let's assume the system is undergoing isobaric and isothermal process, simultaneously.

The gas of the system is assumed ideal, and its volume is changed from $V_1$ to $V_2$.

Internal energy of the system does not change, since the gas is ideal. ($\Delta U=0$)

Work done by/to system will be $w=-p_{const}(V_2-V_1)$, and the heat will be $q=-w$.

But how about enthalpy change $\Delta H$?

I know that by the definition of enthalpy, $H=U+pV$ changes to

$$\Delta H=\Delta U+\Delta(pV)$$

and since pressure is constant, the enthalpy is just

$$\Delta H=0+p_{const}\Delta V=p_{const}(V_2-V_1)$$

But other though is, using ideal gas law,

$$\Delta H=\Delta U+\Delta(pV)=\Delta U+nR\Delta T$$

since the process is also isothermal, $\Delta T=0$

Then, enthalpy is also

$$\Delta H=0$$

Which one is correct?

-Edit- I might have to specify that this question came out when solving textbook problem, Physical Chemistry by Atkins, 9th Edition, Chapter 2, excercise 2.3

"A sample consisting of 1.00 mol Ar is expanded isothermally at 273.15K from 22.4$dm^3$ to 44.8$dm^3$ (a) reversibly, (b) against a constant external pressure equal to the final pressure of the gas and (c) freely. Calculate $q, w, \Delta U, \Delta H$."

I got above question when solving for part (b)

  • 2
    $\begingroup$ Unless the process is just species changes without any heat release or consumption, how is the process both constant pressure and constant temperature while still doing something other than sitting there? If both $P$ and $T$ are constant, than all other thermodynamic properties are also constant unless the composition changes (while still keeping pressure and temperature fixed). $\endgroup$
    – tpg2114
    Commented Apr 6, 2020 at 21:51
  • $\begingroup$ Or the system isn't isolated and something is being injected or removed... $\endgroup$
    – tpg2114
    Commented Apr 6, 2020 at 21:52
  • $\begingroup$ If you suddenly drop that pressure from its initial value to a lower value so that the gas can expand at the lower constant external pressure, this is not an isobaric process. There is a change in pressure. For a process to be isobaric, the final pressure, the intermediate pressures, and the initial pressure must all be equal. $\endgroup$ Commented Apr 6, 2020 at 23:04
  • $\begingroup$ @ChetMiller I see... $\endgroup$
    – user65452
    Commented Apr 7, 2020 at 17:33

1 Answer 1


The processes described in part (b) and part (c) are irreversible, which you are not taking into account. More specifically, the expansion in part (b) is not isobaric - the pressure of the gas is not constant, it is only the external pressure which is fixed.

You know $(P,V,T)$ both before and after the expansion (via the given parameters and the ideal gas law), which means you can easily calculate the changes in the state variables $U$ and $H$. Calculating $q$ and $w$ is more subtle. Note that while the internal pressure of the gas is not constant, the external pressure is, which allows you to calculate the work that the environment does on the gas.

This, along with the first law of thermodynamics, should be all that you need.


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