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 Outrageous Nov14-12 09:55 AM

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

 Philip Wood Nov14-12 10:02 AM

Re: Work done for irreversible process

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 calculatepΔV, as the change is non-quasi-static and pressure does not change with volume in a defined way.]

 Outrageous Nov14-12 10:32 AM

Re: Work done for irreversible process

Quote:
 Quote by Philip Wood (Post 4158965) 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?

 Philip Wood Nov14-12 11:45 AM

Re: 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.

 Outrageous Nov14-12 06:29 PM

Re: Work done for irreversible process

Quote:
 Quote by Philip Wood (Post 4159091) 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?

 Darwin123 Nov14-12 06:46 PM

Re: Work done for irreversible process

Quote:
 Quote by Outrageous (Post 4158953) 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:
 Quote by Outrageous (Post 4158953) 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.

 Outrageous Nov14-12 10:08 PM

Re: Work done for irreversible process

Quote:
 Quote by Darwin123 (Post 4159659) 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

 Philip Wood Nov15-12 03:16 AM

Re: Work done 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."

 Studiot Nov15-12 04:41 AM

Re: Work done for irreversible process

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?

 DrDu Nov15-12 05:42 AM

Re: Work done for irreversible process

Quote:
 Quote by Darwin123 (Post 4159659) 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.

 Outrageous Nov15-12 12:26 PM

Re: Work done for irreversible process

Quote:
 Quote by Philip Wood (Post 4160260) 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?

 Philip Wood Nov15-12 01:33 PM

Re: Work done for irreversible process

"How can get you make conclusion to say that free expansion is irreversible from ∫pdV] is a positive quantity?"

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!

 Darwin123 Nov15-12 01:55 PM

Re: Work done for irreversible process

Quote:
 Quote by Outrageous (Post 4159941) 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.

 Chestermiller Nov15-12 04:50 PM

Re: Work done for irreversible process

An irreversible process is slightly problematic because, in an irreversible process, the pressure and/or the temperature are non-uniform, and vary with spatial position within the system. But that doesn't mean that we cannot apply the first law to a system that has undergone a reversible process. However, unlike a reversible process where we can calculate the pressure and temperature variation throughout the entire process, in the case of an irreversible process, the system is at thermodynamic equilibrium only at the start of the process and at the end, and it is at these points that the pressure and temperature again become uniform (and can be determined).

It is helpful for the present purposes to restate the first law: The change in internal energy between equilibrium states of a closed system is equal to the heat absorbed from the surroundings (path dependent) minus the work done on the surroundings (path dependent). But, if the pressure varies with spatial position within the system during an irreversible process, what pressure should one use to calculate the work done on the surroundings during the process? For both reversible and irreversible processes, the answer to this question is the pressure at the interface between the system and the surroundings. Since, even in an irreversible process, the pressure is continuous at the interface, the pressure to use in calculating the work is always the pressure of the surroundings Psurr. In a reversible process, the pressure within the system becomes uniform, and thus, becomes equal to the pressure of the surroundings. So, for both reversible and irreversible processes, the mathematical statement of the first law becomes:

ΔU = Qsurr - Wsurr = Qsurr - ∫PsurrdV

In this equation, the change in internal energy ΔU depends only on the initial and final equilibrium states of the system, but is independent of the path between the initial and final states. But both the heat absorbed from the surroundings Qsurr and the word done on the surroundings Wsurr is dependent on the path.

As an aside, we talked about how the work done on the surroundings is calculated. But how is the heat absorbed from the surroundings calculated? What you do is take the heat flux vector at the boundary and evaluate its component normal to the boundary (by dotting the heat flux vector with a unit normal). This then gives the local rate of heat absorbed from the surroundings per unit area of interface. You then integrate over the entire area of the interface to get the total rate of heat absorbed from the surroundings. You then integrate this with respect to time to get the total amount of heat absorbed from the surroundings during the entire process.

I'd like to illustrate how this all works with a specific example. Suppose you have an ideal gas contained within a cylinder, and, in the initial equilibrium state, the pressure, temperature, and volume per mole are Pi, Ti, and Vi, respectively. We will be looking at an isothermal and then an adiabatic expansion of the gas in an irreversible process for which the imposed pressure of the surroundings is suddenly changed at time zero from Pi to Psurr, and subsequently held constant. Of course, the final state for the isothermal process will be different from the final state for the adiabatic process.

First the Isothermal Process:

Wsurr = Psurr(Vf - Vi)

where Vf is the final volume per mole. Now, in the final equilibrium state, the pressure will be Psurr, so that, from the ideal gas law

Vf = RTi/Psurr

where, for the isothermal process, the final temperature is equal to the initial temperature. Also, applying the ideal gas law to the initial state, we get:

Vi = RTi/Pi

If we substitute these results from the ideal gas law into the equation for the work Wsurr, we get

Wsurr = RTi (1 - Psurr/Pi)

I've frequently heard it said that, for an irreversible expansion, in the limit of very low pressures for the surroundings, the amount of work done on the surroundings approaches zero. But that is not what this equation is telling us. It is saying that, for an irreversible isothermal expansion, as the pressure of the surroundings gets very low, the amount of work done on the surroundings approaches RTi. 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.

In this isothermal case, the change in internal energy for an ideal gas is zero, so the amount of heat absorbed from the surroundings is equal to the work done on the surroundings.

In this case, the equation for the work done on the surroundings is again given by:

Wsurr = Psurr(Vf - Vi)

If we again apply the ideal gas law tot this equation for an adiabatic expansion, we obtain:

Wsurr = RTf - RTiPsurr/Pi

For an adiabatic process, the heat absorbed from the surroundings Qsurr is zero, but the change in internal energy is equal to minus the work done on the surroundings:

ΔU = Cv(Tf- Ti) = RTiPsurr/Pi - RTf

This equation can be used to solve for the final equilibrium state temperature Tf in terms of the initial temperature:

Tf=Ti(Cv+RPsurr/Pi)/(Cv+R)

Substitution of this into the equation for the work done by the surroundings then gives:

Wsurr = RTi (1 - Psurr/Pi)Cv/(Cv+R)

Note that this is the same result as for the isothermal case, except for the additional factor of Cv/(Cv+R). As expected, in the adibatic case, less work is done on the surroundings.

Finally, in the limit as the pressure of the surroundings Psurr becomes small, the adiabatic work for this irreversible process is not zero, but becomes:

Wsurr = RTiCv/(Cv+R)

 Studiot Nov15-12 06:43 PM

Re: Work done for irreversible process

Quote:
 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.