
Stanton D. answered 06/25/21
Tutor to Pique Your Sciences Interest
Hi Eve K.,
Let's assume that the two are connected by a tube and that a valve in the tube is opened. The pressurized air will inflate the balloon. If the balloon is under normal conditions (with exterior pressure), the air will cool slightly as a result of doing work inflating the balloon (think this is the Joule-Thomson effect?). So there is a slight decrease of enthalpy (heat == work). But there is an increase in entropy -- the air now occupies a larger space, and this makes the positions of individual molecules less predictable. That was "stage 1".
If you want to squeeze the last of the air into the balloon ("stage 2"), that requires an input of work to deflate the ball completely, so a little enthalpy has been added back (but not as much as was lost in the first stage). Since the pressure has increased only negligibly doing this, the entropy decreased only negligibly (not as much as it decreased in stage 1).
You can figure each of these things out yourself from first principles, once you master the concepts of enthalpy and entropy.
--Cheers, --Mr. d.