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Irreversible Process vs. Reversible Process

Posted: Thu Jan 14, 2016 7:54 pm
by Gina Grainda 1G
I understand that an irreversible process isn't at equilibrium because a certain direction reaction is favored, but why does that make it less efficient than a reversible process? Why is a reversible process more efficient?

Re: Irreversible Process vs. Reversible Process

Posted: Fri Jan 15, 2016 12:09 am
by Ronald Yang 2F
Hm...I'm not sure if how I reasoned it makes sense or is correct, but here:

Imagine a gas inside a piston. Let's say initially it is at equilibrium, with external pressure being 2 atm and the pressure of the gas being 2 atm. For expansion to take place, there must be a difference in pressure between that of the gas and external pressure, with Pext<Pgas.

To expand a gas irreversibly, we have to reduce the external pressure very fast initially, let's say to 1 atm. When the gas expands, it does work on the surroundings by moving the piston out against the external pressure, which is constant at 1 atm.

On the other hand, to expand a gas reversibly, we have to reduce the external pressure very slowly by small increments. When the gas expands, it also does work on the surroundings by moving the piston out, but this time, against a varying external pressure, which slowly decreases from the original 2 atm. Thus, as the gas expands here, the gas expands against the maximum possible external pressure at each increment. This is shown when you happen to increase the external pressure by just a little bit, an action that results in compression. At each increment, the gas expands against a pressure greater than 1 atm, as compared to the constant 1 atm that the gas expands against for a irreversible process. This is why it takes more work for a reversible process, given that each process experiences the same change in volume.