# Why does the temperature of a gas decrease after an adiabatic expansion?

When a gas expands adiabatically then the temperature of the gas falls. This can be explained via the first law of thermodynamics as:

$$\Delta E_{\text {int}}= Q-W$$

As $$Q=0$$ for adiabatic expansion therefore:

\begin {align} \Delta E_{\text {int}} & = -W \\ & = \int P_{\text {int}} dV \end {align}

But what bothers me is that I can't explain it (i.e., as of now I don't know of any microscopic model to explain it) with the microscopic picture.

• What is the microscopic explanation of the cooling of the gas as it expands adiabatically?

I mean how can we explain the conversion of internal energy into work.

The molecules hit the walls of the container (say, a piston) and give part of their momentum to it. Thus the piston expands and the kinetic energy of the molecules (i.e. the temperature) decreases.

You are basically turning kinetic energy into work, without loosing any energy due to heat leaks, as we postulated the process is adiabatic.

Expanding on the answer given by JalfredP:

It seems counterintuitive that a slow movement of the piston (expanding the volume available to the gas) could have any effect on the molecules inside.

It's relativily intuitive, it seems to me, to imagine the following thought experiment: the piston moves at a speed that is comparable with the thermal velocity of the molecules. Then any molecule that still manages to reach the receding piston and collide with it will lose almost all its velocity.

In the general case, with the piston moving slowly, the thermal velocity of the molecules will be far larger than the speed of the piston. Then for a molecule bouncing against the piston wall there is almost no difference with a bounce against a fixed wall. Is it a negligable difference?

When the piston moves very slowly then the effect for each single bounce of a molecule against that wall will rob only a very small amount of kinetiac energy, but the slower the piston moves the more time there is to accumulate effect. More time, more bounces.

In the end the expected change of temperature of the gas is the same, independent of the velocity at which the piston recedes.

When the gas is inside a container thermally isolated, and with constant volume, we can assume that: $$\Sigma p_x$$ , $$\Sigma p_y$$ , $$\Sigma p_z$$ of the molecules don't change because the collisions are elastic (container keeps the same temperature). So, each $$p_{xi}$$ turns to $$-p_{xi}$$ after a collison, and $$-p_{xj}$$ turns to $$p_{xj}$$ at the other side of the container.

But if one of the sides, say the left one in direction $$x$$, moves outward, that balance doesn't hold anymore: $$p_x$$ after a collision with the moving side is smaller than before it. $$\Sigma p_x$$ decrease as a consequence.

Temperature must decrease because it is proportional to the average kinetic energy and: $$E =\frac {1}{2m}\Sigma p_x^2 + p_y^2 +p_z^2$$

This analysis is however qualitative. I don't know if the actual temperature drop can be explained only by kinetic energy decrease.