## Work done for irreversible process

For reversible, work done =∫P dV
Then for irreversible ,we can't use the above equation, because we have to consider the dissipative work. Correct?

Thank you
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 Yes, but there are other cases where the 'dissipative' label might be misleading. An example would be a gas expanding into a small extra evacuated volume, ΔV. Since the extra volume is small the gas pressure, p, is hardly affected, so pΔV has a definite, non-zero value. Yet the expanding gas does no external work, as no force resists its expansion. [If the expansion is into a large extra evacuated volume, we can't even calculate ∫pΔV, as the change is non-quasi-static and pressure does not change with volume in a defined way.]

 Quote by Philip Wood Yet the expanding gas does no external work, as no force resists its expansion. [If the expansion is into a large extra evacuated volume, we can't even calculate ∫pΔV, as the change is non-quasi-static and pressure does not change with volume in a defined way.]
Since we can calculate the work done ,that means there must be external pressure, so how you say no force?

So do you mean we can't calculate the work done for irreversible process?

## Work done for irreversible process

I can't see the logic of your assertions. In the cases I quoted the external work done is zero. This is because there is no external pressure and no external force, as the gas is expanding into a vacuum (empty space). You can calculate pΔV in which p is the gas pressure, if the gas expands into a very small (initially empty) volume, ΔV. The result is non-zero. So pΔV ≠ work in this case.

It is an irreversible change, though I'd hesitate to say that dissipation is involved. That's the point I was trying to make.

 Quote by Philip Wood I can't see the logic of your assertions. In the cases I quoted the external work done is zero. This is because there is no external pressure and no external force, as the gas is expanding into a vacuum (empty space). You can calculate pΔV in which p is the gas pressure, if the gas expands into a very small (initially empty) volume, ΔV. The result is non-zero. So pΔV ≠ work in this case. It is an irreversible change, though I'd hesitate to say that dissipation is involved. That's the point I was trying to make.
I know in free expansion , there will be no external pressure so no work done, I don't understand why do you want to use gas pressure times ΔV. So do you mean that PΔV is the dissipative work?

 Quote by Outrageous For reversible, work done =∫P dV
Yes. That is the work performed by the system whether or not the process is reversible.

I think that your intuition is better stated this way. If the process is reversible and there is no heat conduction, then the work is equal to the change in internal energy.

 Quote by Outrageous Then for irreversible ,we can't use the above equation, because we have to consider the dissipative work. Correct?
No. That is still the work performed by the system. The irreversibility of the process does not change the definition of work.

Pressure is normal force divided by area. "Normal" refers to the direction of the force, not the type of force. The force can be dissipative. For instance, one can have a normal force that includes a component of friction. The same equation for work is valid even if frictional forces are included in the pressure.

From a thermodynamics point of view, dissipation refers to the generation of entropy. Any process that creates entropy is dissipative. A force is dissipative if the process that generates the force creates entropy.

If the process is irreversible and there is no heat conductivity, then the work does not equal the change in internal energy.

 Quote by Darwin123 If the process is irreversible and there is no heat conductivity, then the work does not equal the change in internal energy.
Then the change in internal energy for irreversible process equal to ?
Or we can't calculate that for irreversible process
 The main point I was trying to make was that expansion into a vacuum is irreversible, but not, in the usual sense, dissipative. Darwin: "[∫pdV] is the work done whether or not the process is reversible." No. Take the case of (irreversible) expansion into a vacuum. Suppose we have a container of 1000 litre, partitioned into a 999 litre chamber containing gas at a pressure of 100000 Pa, and a 1 litre evacuated container. Suppose the partition starts to leak, and gas flows into the vacuum, until the pressures equalise in the two chambers. The gas pressure will now be slightly less than 100000 Pa, but to a very good approximation, ∫pdV] for the expansion will equal 100000 x 1 x 10-3 J = 100 J. Note that p is the gas pressure - the state variable. But no external work has been done by the gas! The work term in the first law of thermod., as applied to the gas, is zero! The system is isolated from the work point of view. The gas hasn't done work ON anything.Yet ∫pdV] is a positive quantity. To quote Pippard (Classical Thermodynamics) "w is zero, even though pdV does not vanish."
 Good example, Philip. @Outrageous As an adjunct consider the implications of the first law. What does this mean for the other properties of the gas?

Recognitions:
 Quote by Darwin123 Yes. That is the work performed by the system whether or not the process is reversible.
In an irreversible process p need not even be defined. If it is, it is mostly not homogeneous and equal to the equilibrium pressure of a system at equal volume. Even if it were homogeneous, the force on the boundary is not given by p alone but has contributions from longitudinal friction.
So the OP is correct that there is a difference between int -pdV and the work done in an irreversible process.

 Quote by Philip Wood Yet ∫pdV] is a positive quantity.
My teacher used to say that should be external pressure, I still don't understand what is the point to use gas pressure.
How can get you make conclusion to say that free expansion is irreversible from ∫pdV] is a positive quantity?

From my understanding so far is that we can calculate the work done for irreversible but not using a formula ,instead we have to see the condition( free expansion is zero ,if have external pressure then use formula) , and it has nothing to do with dissipative work.
Am I correct?
 "How can get you make conclusion to say that free expansion is irreversible from ∫pdV] is a positive quantity?" That wasn't my argument at all. Please re-read post(s). In fact I didn't make any argument for free expansion being irreversible - I just took that as uncontroversial. If I did want to show that free expansion was irreversible, I'd say that the gas won't spontaneously go back to its original chamber, leaving a vacuum in the other chamber!

 Quote by Outrageous Then the change in internal energy for irreversible process equal to ? Or we can't calculate that for irreversible process
My mistake.
If the process is irreversible and there is no conductivity, the work is not equal to the change in free energy.
If the process is reversible and there is no conductivity, the work is equal to the change in free energy.
Irreversible processes create entropy, resulting in a decrease of free energy.

 The reason for this result is that even though the surroundings pressure is very low, the final volume of the system becomes very large, and so a finite amount of work gets done.
Let us examine this statement in the light of Philip Wood's example (post#8) where the final volume is a negligable increase on the original.

Work is still done but, as Dr Du pointed out, the pressure of expansion is undefined.

It starts out at zero but increases rapidly to the pressure of the system.

So actual work is done against an inceasing pressure, as the system volume expands, or the empty chamber fills.

Further if we try to sum the PdV we have to ask what by what value does the volume increase as the gas leaks into the empty chamber? The whole empty chamber volume or part of it?

If we can define the intial and final states the work can be calculated as referred by ChesterMiller or by other paths not yet stated.

@ Outrageous.