# For an isolated, insulated system for which a gas is allowed to expand, why is the change in entropy for the system nonnegative?

Feb 2, 2016

The entropy is the reversible heat flow possible per unit temperature.

The higher the temperature, the more motion a molecule can have, the more microstates it can assume, the more ways it can absorb heat to use for motion, and thus the more heat it needs to maintain that motion.

$\setminus m a t h b f \left(\Delta S = \int \frac{\delta {q}_{\text{rev}}}{T} \ge 0\right)$

The change in entropy, is therefore the differential (change in) reversible heat flow at a specific temperature.

You know that for a closed system isolated and insulated from the surroundings, $\Delta U = 0$, so using the first law of thermodynamics:

$\setminus m a t h b f \left(\Delta U = {q}_{\text{rev" + w_"rev}}\right) = 0$

Therefore...

$\textcolor{b l u e}{{q}_{\text{rev" = -w_"rev}} = - \left(- \int P \mathrm{dV}\right) \ne 0}$

As a result, you really ought to have some nonzero value for entropy (as the third law of thermodynamics should suggest when you're not at absolute zero).

So if you really want a value for $\Delta S$ of an ideal gas, you need a temperature, some heat flow, and you need a change in volume. Then just perform the above integral (with respect to the differential heat flow $\delta q$) to get:

$\textcolor{b l u e}{\Delta S = {q}_{\text{rev}} / T \ge 0}$

Note that $\Delta S > \frac{q}{T}$ when $q$ is not reversible/efficient.