2
$\begingroup$

Does change in internal energy $\Delta U =0$ implies that the process is always isothermal while considering a ideal or real gas ?

If not, then what are the other process where $\Delta U =0$ and $\Delta T≠0$ while considering a ideal or real gas ?

$\endgroup$
  • $\begingroup$ Please edit your question by precising whether you are considering an ideal gas, a real gas or something else. $\endgroup$ – thermomagnetic condensed boson Jan 16 '16 at 14:39
3
$\begingroup$

$U$ is a function of temperature only for an ideal gas. For a real gas, it is also a function of pressure (or specific volume). Unlike the case of an ideal gas, if you have an insulated chamber divided in half, with a gas in one half and vacuum in the other half, and you remove the divider (allowing the system to re-equilibrate), the temperature of the gas will change (even though the internal energy does not change).

$\endgroup$
  • $\begingroup$ I'm looking at en.wikipedia.org/wiki/Free_expansion but can't see why ΔU=0. It's non-equilibrium so you can't use the differential expression of the first law. Maybe you just say ΔU = ΔQ-ΔW = 0? $\endgroup$ – ignacio Jan 16 '16 at 13:13
  • $\begingroup$ Yes, the latter. $\endgroup$ – Chet Miller Jan 16 '16 at 14:38
1
$\begingroup$

Considering an ideal gas,ΔU=0 for isotermal processes. However there could be other processes where ΔU=0 even if process is not isothermal. If the process is represented in pv graph and finally the gas is anywhere on its original isotherm then ΔU=0. For example, all cyclic processes have ΔU=0. This is because in ideal gas there is no potential enegry and internal energy is only from its kinetic energy. Translation kinetic energy is 0.5mv^2 where v would be rms velocity. Rms velocity only depends on temperature for a particular gas. Other kinetic energy would be rotational and vibrational. These are dependent on temperature as well. You can find more details by looking up "degrees of freedom".

$\endgroup$
1
$\begingroup$

$U = cnT$ gives the relation of internal energy of a system and it's temperature, for an ideal gas. As you can see, the temperature is directly proportional to the internal energy of a gas. Hence, if $\Delta U= 0$, then $\Delta T=0$.

One possible exception is the case of liquefaction or condensation of a gas. This is, of course, true only for non-ideal gases as ideal gases do not liquify. In these cases, the internal energy of the gas reduces, without the temperature being changed. I'm not sure if this would answer your question as after the process, the substance is no longer a gas.

$\endgroup$
1
$\begingroup$

It's not clear whether you are taking real gases into account or not.

  • If it is a ideal gas, then internal energy is dependent of temperature only (among P, V & T) and hence $\Delta U =0$ implies that $\Delta T =0$ but not that the process is isothermal as isothermal process implies the temperature is constant throughout, which is not necessary if $\Delta T =0$.

  • If it is a real gas, then U is a dependent on P & V apart from T and hence $\Delta U =0$ does not imply $\Delta T =0$. An example of such a process is 'Joule Thomson Effect'

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.