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How can the temperature of a body change unless heat flows in or out of it? So shouldn't all adiabatic process be considered isothermal too? but that is not true as both the processes have different equations.

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  • $\begingroup$ The first law of thermodynamics tells us that $\Delta U=Q-W$, which reduces to $\Delta U=-W$ for adiabatic conditions. So if the gas does work, or you do work on the gas, you are also changing the internal energy of the gas, and thus its temperature. $\endgroup$ Commented Jan 5 at 11:54

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Only the heat exchange with the surroundings is zero. We never said anything about the body using up its own, internal energy. In adiabatic processes, the body increases/decreases its own internal energy when it does work, or when work is done on it.
If you want to look at it with the F.L.O.T.: $\Delta U=q-W$. In adiabatic processes,$q=0$. So $\Delta U=-W$., and because $\Delta U=nC_v\Delta T$, there will be change in temperature when there work is done on/by the gas.

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  • $\begingroup$ This is incorrect and should not be the accepted answer. If a system is isolated (which means that no heat flow occurs to or from the surroundings and now work is done on or by the system), then the energy of the system cannot change, and so the temperature $T$ cannot change either (for systems such as the ideal gas). In introductory thermodynamics, $T$ changes in an adiabatic process because work is done on or by the system, leading to the thermal energy to change and thereby $T$ to change. What does "the body uses its own internal energy to change temperature" even mean? $\endgroup$
    – march
    Commented Jan 5 at 17:22
  • $\begingroup$ @march I wasn't seeing what I was writing. I'm sorry. What I meant was: A change in internal energy ( and thus a change in temperature) is caused by/ causes work to be done on/by the gas. $\endgroup$
    – Stuti
    Commented Jan 5 at 18:20
  • $\begingroup$ @Ars Please read the (edited) answer again. $\endgroup$
    – Stuti
    Commented Jan 5 at 18:23
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    $\begingroup$ Okay, that's better, although I don't really like the language "the body decreases/increases its own internal energy", because that ascribes some agency to the system in a strange way. I'll reverse my downvote though. $\endgroup$
    – march
    Commented Jan 5 at 18:27
  • $\begingroup$ @march How would you phrase it then? $\endgroup$
    – Stuti
    Commented Jan 5 at 18:36
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Have you ever had to pump air in a tire? It is essentially an adiabatic process, heat conduction losses between the pump/tire and the air is negligible, but if you touch the pump it feels warmer. Or just touch a tire after driving the car a few minutes, it will feel much warmer than when you started out. Both are adiabatic but showing that external work can warm up the body.

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There actually exists the input force, but we will neglect it when calculating, but it doesn't means it actually never exists, it might can do a lot of works.
Here is a very important point is that we suppose all the changes undergo slowly, but you can imagine that without the input force, we can actually not make sure the change will go slowly, then it will not be a reversible process, then it is not able to be calculated, to make it reversible, we choose to put some force on it, to make the process slow, but those force will give out a negative work, so I think generally speaking, we will have the energy of the gas decreases after the change.
Main Idea
It is an idealic process in theory but in fact it needs other unneglectable force participating.

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