Expansion of an ideal gas not always represent an increase in entropy of the system. Statement is True or False?
What I thought is that whenever an ideal gas expands the randomness of gas molecules increases. So, entropy must increase.
$\begingroup$
$\endgroup$
-
3$\begingroup$ What if, during the expansion, the temperature decreases? $\endgroup$ – Chet Miller May 7 '18 at 16:40
$\begingroup$
$\endgroup$
Entropy does not depend on just the volume the gas has available. It also depends on the energy the gas has to exploit that available volume.
Suppose you expand the gas by pushing against a piston. The gas is doing work to "earn" its extra volume, so even as volume increases the gas is expending energy. Thus unless heat flow into the gas is allowed the gas will cool. This loss in energy tends to decrease random motion, and thus tends to decrease entropy, even as having more volume would have the opposite effect. We have a "tug of war" here.
In an idealized version of such an expansion, where we aallow no heat flow into or out of the gas (adiabatic) and we maintain equilibrium with no "losses" due to friction ("reversible") then the loss of energy to doing work must effectively balance the gain in volume, leaving zero entropy change. With no heat flows through the insulation and no heat generation due to friction in this idealized scenario, there must be zero entropy change in both the gas and the surroundings.
The lesson: Any process (such as expansion of an ideal gas) that is reversible (passing through equilibrium states) and adiabatic (no heat flow) has zero entropy change.
answered May 8 '18 at 10:19
