I need to understand a concept, which is Adiabatic expansion. I know that this process is where no heat exchange happens. Now, can i say that when an ideal gas undergoes an adiabatic expansion,then the temperature of the gas has to decrease? Is this a valid statement to make ? and if so , is it always true or are there cases in where the temperature in constant or increases if the ideal gas undergoes an adiabatic expansion?

• Just for clarification: is it important for you that the fas is ideal? i think for non-ideal gases the temperature can actually increase. Apr 12, 2017 at 15:35

The first law of thermodynamics states the change in internal energy of a system equal the heat input to the system minus the work done by the system.
Chemists tend to use "plus the work done on the system".

$\Delta U = Q-W$

For an ideal gas $\Delta U =\frac 32R\Delta T$ and for an adiabatic process $Q=0$.

$\Rightarrow \Delta U =\frac 32R\Delta T=-W$

If the ideal gas expands then $W$ is positive and hence $\Delta T$ is negative i.e. the temperature decreases.

Update as a result of some comments

If the expansion is into a vacuum then no work is done by the gas and the temperature of the gas stays constant.

• This is not generally true. You could remove a wall to expand the gas (so no mechanical work done); this would be ab example of an irreversible process. Since the internal energy doesnt change in the process the ideal gas temperature stays constant. Apr 12, 2017 at 5:49
• @lalala Do you mean expand into a vacuum or just remove the walls? Apr 12, 2017 at 6:17
• I mean there is the gas on one side and a vacuum container on the otherside, and then I remove the wall. The gas will violently expand by itself into this void until it reaches equilibrium. Apr 12, 2017 at 8:31
• @lalala I assumed a standard container with a piston type arrangement not expansion into a vacuum.. Apr 12, 2017 at 8:38
• ok, I understand now. The question seems to be though:"can i say that when an ideal gas undergoes an adiabatic expansion,then the temperature of the gas has to decrease?" Which doesnt seem to be limited to a piston type arrangement. Apr 12, 2017 at 9:11

Alternative way to look at this is to use polytropic process where, for adiabatic process, the exponent is the specific heat ratio $\gamma$. For a polytropic process $PV^n=const$. If it is ideal gas, you then get $TV^{n-1}=const$. When volume increases (expands), you can conclude the temperature will not be constant and will not increase.

• again, this only applies if the expansion/compression is also reversible. Apr 13, 2017 at 15:05