I was doing a question on thermodynamics and came across this one stating

During isothermal expansion of gas of an ideal gas :

a) Internal energy decreases

b) Enthalpy increases

c) Enthalpy reduces to zero

d) Enthalpy remains unchanged

I cannot agree to any of these as $\Delta H = \Delta U + W$, $\Delta U = 0$ as $\Delta T = 0$ but $W \neq 0$ so $\Delta H = W$ which is $\neq 0$ and as $W \lt 0$ the enthalpy should decrease but how much how do I know and to add to the confusion the answer given is enthalpy remains unchanged in the light of the equation $\Delta H = \Delta U +\Delta n_\mathrm gRT$ and $\Delta U$ and $\Delta n = 0$ so $\Delta H = 0$ is it right.

  • $\begingroup$ @ChetMiller dPV or nRdT also defines work and nRTln(v2/v1) also defines work....then how work has zero and non zero values at the same time in the same process(isothermal expansion of ideal gas) $\endgroup$ Dec 8, 2023 at 7:35

1 Answer 1


You have already mentioned that the change of internal energy is zero $(\Delta U=0)$ since, for an ideal gas, the internal energy $U$ only depends on amount of substance $n$ and temperature $T$, and in a closed system $n$ is constant $(\Delta n=0)$ and during an isothermal process also $T$ remains constant $(\Delta T=0)$.

Enthalpy $H$ is defined as


and the ideal gas law states that




Since $n$ (closed system) and $T$ (isothermal process) are constant, the product $nRT$ is constant, and therefore, according to the ideal gas law, also the product $pV$ is constant.

Furthermore, since $U$ is constant during the given process, the sum $H=U+nRT$ as well as $H=U+pV$ has to remain unchanged (answer d).

Note that your assumption

$$\Delta H = \Delta U + W$$

(where $W=p\Delta V$ is the pressure-volume work) only applies to processes under constant pressure $p$, since enthalpy is defined as


and thus

$$\begin{align} \mathrm dH&=\mathrm dU+\mathrm d(pV)\\ &=\mathrm dU+V\mathrm dp+p\mathrm dV \end{align}$$

which simplifies to

$$\mathrm dH=\mathrm dU+p\mathrm dV$$

at constant pressure $(\mathrm dp=0)$.

However, the pressure does not remain constant during the process given in the question.

  • $\begingroup$ Yes. The enthalpy of an ideal gas is a function only of temperature, no matter what kind of process is imposed. $\endgroup$ Aug 17, 2016 at 13:17
  • $\begingroup$ @chester Miller if it were so then all isothermal process would be isoenthalpic which is not the case $\endgroup$ Aug 18, 2016 at 3:33
  • $\begingroup$ @PrateekChauhan For a single phase ideal gas, of course it is. $\endgroup$ Aug 18, 2016 at 13:14
  • $\begingroup$ @ChesterMiller If I have an ideal gas, then will the change in enthalpy be zero, for an isothermal path? Does it matter whether the process is carried out reversibly or not? $\endgroup$ Sep 28, 2018 at 6:39
  • $\begingroup$ If the initial and final equilibrium temperatures are the same (irrespective of what happens along the path), the change in enthalpy of an ideal gas is zero. The enthalpy and internal energy of an ideal gas are functions only of temperature. $\endgroup$ Sep 29, 2018 at 1:23

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.