# Understanding efficiency

This is a repost of a previous question, following the advice of an experienced user.

I was told that reversible processes are more efficient than irreversible ones, and that this fact is expressed in the inequality $\delta Q_{rev}\ge \delta Q_{ir}$ or $\delta W_{rev}\le \delta W_{ir}$, where $W$ stands for work. I'll write $\delta W$ instead of $\delta W_{rev}$.

My question is how do I compare a reversible process with an irreversible one? Let me explain what I mean. Textbooks say that expansion of a gas against zero pressure is zero, hence $\delta W\le \delta W_{ir}$ implies by integration that the work done by a reversible process should be negative. But between to states there are infinitely many processes, what are the processes that do negative work. Say the initial state is $(p_i,V_i)$ and the final $(p_f,V_f)$, where $V_f>V_i$, then the process depicted below does positive work.

Basically, you must expand the gas at very low temperature and compres it at high temperature. Even more, given any number $A$, there is a path exchanging $A$ Joules of work.

In general, given an irreversible process, how can I say what are the more efficient reversible processes?

I'd appreciate any help.

## 1 Answer

It sounds to me like you have two confusions you need to watch out for. The first is the difference between the "work on" a system, and the "work by" a system, which differ in sign. So you should not speak of the "work done by a process", because it's not clear which sign you refer to, so you will have to essentially equip the sign manually (expansion of a system does "work by" the system, so that's negative "work on" the system). The second confusion to watch out for is that when you do an integral to get, say, the area under a P-V curve, if you go from right to left, rather than left to right, the area needs a negative sign to be manually attached to it. So the area between the curves in your figure is actually a negative "work by" the system, so that's a positive "work on" the system. (It's often best just to figure out the logically reasonable sign, rather than relying on formulae to tell you, though such formulae do exist.)

So the upshot is, the process you showed does negative work on its surroundings, so it's not a heat engine. Run it in the opposite direction and you have a heat engine that you could talk about the "efficiency" of. That should also resolve your issues with why reversible processes are more efficient-- they have a more negative work done on the system, so a larger positive work done by the system on its environment. That also means that the inequalities you wrote must treat dW as the work done on the system, they should all be reversed in direction if you mean the work done by the system on its surroundings (which is what you want if efficiency is your issue). Or if you are fixing the work done by the system to be a given value, and varying the heat that crosses from hot to cool to run that engine, then the amount of heat is smaller for the reversible process (that's what higher efficiency means), so your inequalities would only be true if dQ is regarded as negative (a strange interpretation of its sign, by the way). I'd check those again, they seem an odd way to talk about efficiency.