## Main Question or Discussion Point

What is the concept behind 'irreversible adiabatic process' ? Why is the expression for work done in this case different from that when it is reversible?

LeonhardEuler
Gold Member
For instance, an irreversible adiabatic compression of a gas would be one where the pressure force exerted on the gas exceeds the pressure of the gas. The work done is different than in a reversible process because the "P" that appears in the PdV expression for the work is actually the external pressure exerted on the gas and not the pressure of the gas itself, so the work would be greater than in the reversible case since the external pressure would need to be greater than the gas pressure to compress it.

What does the 'P' of a reversible adiabatic process stand for? External or Internal Pressure?

LeonhardEuler
Gold Member
In a reversible process the external and internal pressure are the same at every point in time so there is no need to distinguish.

What about the work done in an irreversible adiabatic expansion.... is it again greater than that in reversible expansion?

LeonhardEuler
Gold Member
Here the external pressure is less than the internal pressure, so the absolute value of the work done on the system is less, but the work done on the system is negative, so the work done on the system is again greater.

DrDu
Volume work is not too good an example to exemplify irreversible work (although possible).
Consider instead e.g. stirring where the deposited mechanical energy leads to an increase in internal energy which could have been achieved also by heat transfer.

Consider instead e.g. stirring where the deposited mechanical energy leads to an increase in internal energy which could have been achieved also by heat transfer.
Is stirring an irreversible process? But the liquid can lose its internal energy on cooling and revert to its original state.

LeonhardEuler
Gold Member
Is stirring an irreversible process? But the liquid can lose its internal energy on cooling and revert to its original state.
An irreversible process is one where net entropy is created in the universe. It is irreversible because that entropy can never be destroyed. Stirring is irreversible because once the fluid settles down, work has been completely converted into heat, and this increases entropy. The fluid can be brought back to it's original state, but it must be always in a way that does not also bring the surroundings back to their original state.

Volume work is not too good an example to exemplify irreversible work (although possible).
Consider instead e.g. stirring where the deposited mechanical energy leads to an increase in internal energy which could have been achieved also by heat transfer.
It's not the clearest example of irreversible work, but I think it is the clearest example to answer the second part of the question about why the expression for the work differs from the reversible case, since there is no reversible stirring except of an inviscid fluid that retains the kinetic energy imparted by the stirrer forever, which is not usually what is thought of by stirring work.

DrDu
Is stirring an irreversible process? But the liquid can lose its internal energy on cooling and revert to its original state.
How can it loose its internal energy if the process is adiabatic?

DrDu
What is the concept behind 'irreversible adiabatic process' ? Why is the expression for work done in this case different from that when it is reversible?
One remark on the original question: It is not possible to connect the same initial and final states by both a reversible and an irreversible adiabatic process. If one of the two is adiabatic, the other one has to be non-adiabatic.

How can it loose its internal energy if the process is adiabatic?
You did not mention that the process is adiabatic For instance, an irreversible adiabatic compression of a gas would be one where the pressure force exerted on the gas exceeds the pressure of the gas. The work done is different than in a reversible process because the "P" that appears in the PdV expression for the work is actually the external pressure exerted on the gas and not the pressure of the gas itself, so the work would be greater than in the reversible case since the external pressure would need to be greater than the gas pressure to compress it.
For a reversible A.P.,
Work done in expansion = -PextΔV

For an irreversible A.P.,
Work done in expansion = -PextΔV

In both the cases Pext is same. So the difference in work done arises due to this Pext or a difference in volume change?

LeonhardEuler
Gold Member
The difference arises because the external pressure is different. In a reversible expansion or compression the external pressure equals the pressure of the system. In an irreversible one it does not.

DrDu
You did not mention that the process is adiabatic 