# Maximum work done in isothermal or isobaric process? [closed]

In an isothermal process all heat is converted into work while in an isobaric process heat is converted into work as well as internal energy. However, the PV-graph shows that the maximum work is done in an isobaric process. How do we resolve this discrepancy?

• Do you think that the amount of heat received by the system is the same in both cases? – Chet Miller May 6 at 17:32
• You need to specify further constraints. Isothermal processes can do more than, less than, or the same amount of work as an isobaric process depending on the states you are moving between for your system – Aaron Stevens May 6 at 17:48

I am assuming you are looking at the idea gas and wondering about comparing two processes, one isothermal and one isobaric, where the initial state is the same for each process, but the final state is different, but at the same larger volume (of course both the initial and final states cannot be the same). Which, if you then look at the PV-diagrams of these process you will see that you are correct: there is more area under the curve of the isobaric process compared to the isothermal process Of course, if we started at different states but ended at the same state the opposite would be true: the isothermal process would have done the larger work. And we can actually take two systems that start and end at the same volume with these processes and have the same amount of work done for both: So really, this question is not answerable. There is no "maximum work process" if we just specify that one is isothermal and one is isobaric.

We can still look at the energies associated with the first image (since it seems like this is what you had in mind when asking this question).

The work done on the gas during the isothermal expansion is given by $$W_T=-NkT\ln\left(\frac{V_f}{V_i}\right)$$ And, as you said, since there is no change in internal energy, so the heat that enters the system is given by $$Q_T=-W_T=NkT\ln\left(\frac{V_f}{V_i}\right)$$

The work done on the gas during the isobaric process of simply $$W_P=-P(V_f-V_i)$$ and the change in internal energy is given by (assuming a monatomic gas) $$\Delta U=\frac32Nk\Delta T=\frac32P(V_f-V_i)$$ so the heat that enters the system is given by $$Q_P=\Delta U-W_P=\frac52P(V_f-V_i)=-\frac52W_P$$

So we see that for the isobaric process, the heat that enters the system is $$5/2$$ larger than the work required for the process, whereas for the isothermal process the heat and work are equal. Since we know that the isobaric process in this example already requires much more work than the isothermal process, we see that the isobaric process also needs more heat than the isothermal process.

Therefore, even though the heat of the isobaric process is divided between work and changing internal energy, there is still more heat involved, so there is no contradiction even though all of the heat goes into work for the isothermal process.

There’s no “discrepancy “. I don't mean to be presumptuous, but I think you may be confusing maximum work with maximum efficiency.

Efficiency is gross heat added divided by net work done. A reversible isothermal process converts all the heat added to work. As a process it is 100 % efficient. The isobaric process does not.

So although an isobaric process may do more work, it does it less efficiently.

This is one reason why the Carnot cycle, the most efficient cycle possible, has two reversible isothermal processes,

Hope this helps