# In an adiabatic process, if there is no heat flow between system and surroundings, then why don't we consider temperature to be constant always?

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.

• 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. Commented Jan 5 at 11:54

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.

• 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? Commented Jan 5 at 17:22
• @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. Commented Jan 5 at 18:20