Question

For the same increase in volume, why work done is more if the gas is allowed to expand reversibly at higher temperature?

Answer 1

"Frictional losses are not considered and no energy is lost during the reversible process. So, the work done becomes maximum in the reversible process.

W.k.t, In reversible process, Work done = P.dv

Here, Pressure = function of volume

dv = Change in volume

If the process is not reversible, then, we will consider "work done" by the surrounding particles on the system. Hence, the "work done" by the surrounding is removed from the work done by system.

In irreversible process, work done = P.dv - "work done" by surrounding.

Thus, the "reversible work done" is always "more than irreversible" work done. "