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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.

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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 $Q=W$.

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  • $\begingroup$ 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? $\endgroup$ – Aaryan Dewan Dec 26 '16 at 5:18
  • $\begingroup$ @AaryanDewan no, because when two objects collide, energy is conserved, it doesn't create more energy $\endgroup$ – DavePhD Dec 26 '16 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? $\endgroup$ – Aaryan Dewan Dec 26 '16 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" ? $\endgroup$ – onurcanbektas Apr 11 at 4:55
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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 '14 at 15:29
  • $\begingroup$ gj255 please turn your comment into answer , as it precisely answers the given question $\endgroup$ – Chemist Sep 16 at 19:51
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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.

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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.

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    $\begingroup$ 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 '14 at 16:44

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