# Complete transfer of Heat into Work

During an isothermal reversible expansion of a gas, all heat absorbed by the gas from the reserviour is being used up to push the piston up. But doesn't this contradict something like the second law? When I push a brick on a rough table, the motion will eventually stop and some heat will be generated due to friction, but If I get a bunsen burner and supply the same amount of heat, it brick won't move back to where it started.

• I'll have to think if exactly this it allowed at all, but generally such a thermodynamical process can only be reversible if it is done infinitely slow, in which case you never get any work out of it. (?) – Karl Jul 4 '16 at 14:03
• And that's why there's no such thing as fully reversible process. – Mithoron Jul 4 '16 at 23:07
• only in a cycle a complete conversion of absorbed heat to work is prohibited. – hyportnex Jul 5 '16 at 0:22
• Yes, the above comment by hyportnex is correct. Levine's Physical Chemistry 6th ed. writes: i.stack.imgur.com/i3NzD.png – orthocresol Sep 4 '16 at 15:02

In an isothermal reversible process the total entropy change of piston and energy reservoirs (surroundings) must be zero, $\Delta S_T = 0$. Thus the entropy of the expanding gas may increase but only if it is compensated for by a decrease in entropy of the reservoirs. So no contradiction of second law. (The heat absorbed from the surroundings is $T\Delta S$ and if this is positive more work can be done by a chemical reaction than the heat of reaction.)