Since its isothermal, $\delta U$ is zero. But enthalpy change $\delta H$ is $\delta U + \delta n(\text{gaseous})RT$ or $\delta H = \delta U + \delta (pV)$.

For reversible isothermal expansion of ideal gas, what is the $\delta H$, is it positive, negative or zero.

Ref: I've checked my study material and it says positive, which makes sense but my exam worksheets state it to be zero.


Recall that, for an ideal monatomic gas, $$U = \frac{3}{2}nRT.$$ (Diatomic and polyatomic gases have a larger coefficient.) For an isothermal expansion, reversible or otherwise, we therefore have $$\Delta H = \Delta(U+PV) = \Delta(U+nRT) = \frac{5}{2}nR\Delta T = 0,$$ since $\Delta T = 0$.

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  • $\begingroup$ So for an ideal gas its zero. Is it a any different for non-ideal gas? $\endgroup$ – Cinnamaldehyde Oct 29 '17 at 5:33
  • $\begingroup$ @Cinnamaldehyde, yes, certainly. Exactly how it changes will depend on your equation of state; I don't recall any further detail off the top of my head. $\endgroup$ – a-cyclohexane-molecule Oct 29 '17 at 5:59
  • $\begingroup$ For a non-ideal gas, enthalpy is a function of pressure (in addition to temperature). $\endgroup$ – Chet Miller Nov 13 '17 at 15:54

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