In isothermal process $\Delta U =0$. But I am having trouble understanding it.

Say we have an ideal gas, and say my temperature is constant but I move the pressure, volume from $(P, V) \to (P-dP, V+dV) $. So the volume has expanded and system has done some work to the surrounding. So my work is non-zero.

So how come $\Delta U=0$? I am really confused here.


5 Answers 5


It is not generally true that $\Delta U = 0$ in an isothermal process.

An ideal gas by definition has no interactions between particles, no intermolecular forces, so pressre change at constant temperature does not change internal energy.

Real gases have intermolecular interactions, attractions between molecules at low pressure and repulsion at high pressure. Their internal energy changes with change in pressure, even if temperature is constant.

For an ideal gas, in an isothermal process, $\Delta U = 0 = Q-W$, so $Q=W$.

  • $\begingroup$ Great answer! So if i have an ideal gas and then I increase the pressure, then wont the particles bump into each other more frequently and thus that would increase the avg KE of the particles and that in turn would increase the internal energy of the system? $\endgroup$ Dec 26, 2016 at 5:18
  • 2
    $\begingroup$ @AaryanDewan no, because when two objects collide, energy is conserved, it doesn't create more energy $\endgroup$
    – DavePhD
    Dec 26, 2016 at 10:45
  • $\begingroup$ I think that it should be that "because when two objects collide *in the absence of an external force *, energy is conserved". If we apply external pressure then won't the internal energy increase? $\endgroup$ Dec 26, 2016 at 14:09
  • $\begingroup$ However, you didn't derive that $\Delta U =0$ for ideal gases, instead you assumed it and derive some conclusion on top of it. What is the mathematical representation of having "no intermolecular forces" ? $\endgroup$
    – Our
    Apr 11, 2019 at 4:55
  • $\begingroup$ @Our In statistical mechanics, it is represented as the absence of correlation between particle states. $\endgroup$
    – Noldorin
    Dec 1, 2021 at 16:34

In Isothermal process the temperature is constant.

The internal energy is a state function dependent on temperature. Hence, the internal energy change is zero.

For the process you are describing the work is done by the system, but had you not supplied heat, then the temperature would have dropped. That is a adiabtic cooling process. If no heat is supplied and internal energy is not maintained at the same level, then the process wont be a isothermal process.

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    $\begingroup$ In general the internal energy will depend on more variables than just temperature, and so it's mistaken to conclude that just because the temperature stays constant, so will the internal energy. In addition we need the relation, true for an ideal gas but not true in general, that $U = 3Nk_B T/2$. $\endgroup$
    – gj255
    Sep 18, 2014 at 15:29
  • $\begingroup$ gj255 please turn your comment into answer , as it precisely answers the given question $\endgroup$
    – Chemist
    Sep 16, 2019 at 19:51

Internal energy is due to motion of particles in a system. As internal energy depends on temperature. As we know temperature in isothermal process is constant so the internal energy will also be constant thus the change in internal energy will be zero.


You might have got the answer, but this answer is for new visitors ... When you say isothermally it means the system is somehow allowed to exchange thermal or mechanical energy. In an isothermal compression, the system is allowed to release heat otherwise (adiabatic process) change in temperature will change the internal energy. Similarly in isothermal expansion, the system does work on the expense of its internal energy which is compensated by influx of heat otherwise the temperature will decrease. For your question, the system must have absorbed some heat and the expansion work consumes it. This satisfies the equation: the inward heat is negative and the work is done by the system is positive both are equal so no net change in internal energy, yet the volume increases.


See consider a cylinder with piston fixed, (i.e. it doesn't move) and the system is provided with a source with temperature $T$. Now as the piston does not move the volume is constant, so no work is done and internal energy is also constant and no heat is added since system and source are at the same temperature.

Now let us release the piston. Then the system does work using internal energy, but the change in internal energy is spontaneously overcome since system is copped to maintain at constant temperature. In other words we can say that system never uses internal energy since it is supplied with constant heat from a source.

  • 9
    $\begingroup$ Hi rahul, and welcome to Physics.SE! Your post is quite difficult to parse due to lack of paragraphs and spaces after periods - you might consider to improve this. Also, I'm not certain you are actually answering the question - it seems to me as if you are just putting your thoughs on some thermodynamical cycle out here, which this is not the place for. $\endgroup$
    – ACuriousMind
    Jul 21, 2014 at 16:44

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