# A The exact formula for PV-work in an irreversible process

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1. Mar 9, 2018

### Philip Koeck

I've written a short text adapted from a previous post by Count Iblis. I'll append it below. It shows that irreversible PV-work is always smaller than reversible, which fits very nicely with W = Pext ΔV. I'd be interested if there's a way to show that W is exactly equal to Pext ΔV for all expansions and compressions where the content of the cylinder qualifies as an ideal gas (pressure not too high and not too low). Another interesting thing would be experiments where this formula for work has been checked, maybe via the temperature change during a sudden expansion or compression.
Here comes the text:

PV-work for irreversible processes

The pV-work done by the system on the surroundings in an irreversible expansion is always smaller than the pV-work done in a reversible process between the same initial and final state.

In an irreversible compression the work done by the surroundings on the system is larger than in a reversible compression between the same states.

To see why this is the case, consider the first law of thermodynamics.

dU = dQ – dW (1)

In case of a reversible change of the system, we know that

dQ = T dS (2)

and

dW = P dV (3)

Inserting (2) and (3) in (1) gives:

dU = T dS - P dV (4)

which is known as the fundamental thermodynamic relation. Now, even though this was derived assuming the change in the system was reversible, it is also valid for irreversible changes. The reason is that the inner energy U is a thermodynamic state variable which can be uniquely specified by the entropy and volume. Then, if you change the volume and entropy, the internal energy change is some fixed quantity, independent of how that change is realized (reversible or not).
The fundamental thermodynamic relation (4) is just the first law of thermodynamics (1) written only with state variables and functions.

For irreversible changes we know that (2) is not valid.
In general we have

T dS ≥ dQ (5)

Inserting (4) on the left and (1) on the right of (5) and cancelling dU gives

P dV ≥ dW (6)

So, the work done in irreversible processes will, in general, be less than the integral of P dV.
Of course, if both dW and P dV are negative, as for a compression, this means that the absolute value of dW will be larger than the absolute value of P dV for compressions.

??Without proof we state that the PV-work in an irreversible expansion or compression at constant external pressure is given by
W = Pext ΔV, which fulfills (6)??
Notice that for Pext = 0 the PV-work becomes 0, just as it should for a free expansion.

Last edited: Mar 9, 2018
2. Mar 9, 2018

### Staff: Mentor

For any expansion, whether reversible or irreversible, the work done by the gas on its surroundings is given by: $$W=\int{P_{ext}dV}$$where $P_{ext}$ is the force per unit area exerted by the surroundings on the gas at the interface where the boundary movement is occurring and, by Newton's 3rd Law, it is also equal to the force per unit area exerted by the gas on the surroundings at the interface. For example, in the case of a gas contained in a cylinder by a piston, $P_{ext}$ is the force per unit area exerted by the gas on the inner face of the piston (and is also the force per unit area exerted by the inner face of the piston on the gas). For a reversible process, the force per unit area exerted by the gas on the inner face of the piston is equal to the pressure calculated from the ideal gas law; for an irreversible process, it is not.

The origin of the P...V work equation is the relationship for W we learned in mechanics: $$W=\int{Fds}=\int{\frac{F}{A}Ads}$$where A is the area of the piston. But, $$F=P_{ext}A$$ and $$dV=Ads$$So, if we believe what we learned in freshman physics, then $W=\int{P_{ext}dV}$. Beyond this, the validity of using this equation to calculate the work in the first law have been observationally confirmed millions of times over the past couple of hundred years for both reversible and irreversible processes.

I agree with all this, although the fundamental thermodynamic relation also includes a contribution from the 2nd law of thermodynamics (in addition to the first law). I also like to think of this equation as a relationship between the changes dU, dS, and dV for two very closely neighboring thermodynamic equilibrium states of a material at (U,S,V) and (U+dU, S+dS, V+dV).

The next section of discussion, I'm not so comfortable with.
I think that this equation should more properly read $Q=T_B\Delta S+\delta$. This equation assumes that all the heat transfer takes place at the boundary temperature $T_B$ (typically an ideal constant temperature reservoir) and $\delta$ is the entropy generated within the system as a result of irreversibility. In the current form that Eqn. 5 is expressed above, T should represent the temperature of the gas at the heat transfer interface, and not some average temperature of the gas. In an irreversible process, the gas temperature is generally not spatially uniform. The use of $T_B$ in this form of the 2nd law equation for the heat is very much analogous to the use of $P_{ext}$ in the 1st law equation for the work.
Call it a personal preference, but I am uncomfortable with using differentials when analyzing irreversible processes. So, using finite changes, we would have: $$\Delta U=T_B\Delta S+\delta - W$$From this, it seems more difficult to make the case for less irreversible work in an expansion. Still, for an adiabatic irreversible process, we know that $\Delta S$ is greater than zero, so $Q=T_B\Delta S+\delta$ would have to be greater than zero. That would mean that W would have to be less for the irreversible path.

A while ago, I wrote a Physics Forums Insights article that provides a more intuitive comparison between reversible and irreversible gas expansions and compressions. This relates closely to what we have discussing in the present thread.
https://www.physicsforums.com/insights/reversible-vs-irreversible-gas-compressionexpansion-work/

Last edited: Mar 10, 2018
3. Mar 10, 2018

### Philip Koeck

I have a problem with this. Assume a cylinder with an ideal gas and surroundings with an ideal gas. The pressure inside is larger than outside. The cylinder has a piston with a large mass. Initially the piston is held in position by the experimenter. That means the balance of forces is Fint = Fext + Fman, where Fman is the force the experimenter needs to apply. If the experimenter lets go the piston is accelerated, but the accelaration is very small, since the mass of the piston is large. Now the balance of forces is Fint = Fext + m a, where a is the acceleration of the piston with mass m. The gas inside is always homogenous since the piston moves so slowly and accelerates so little. I don't see that Fint has to be equal to Fext.

4. Mar 10, 2018

### mfig

I believe that first equation is not quite right. The usual way I have seen this written is

$\Delta S= \int \frac{\delta Q}{T_b} + \sigma$

Thus sigma is always positive for real processes, has the same units as entropy, and is a measure of the "strength" of the irreversibilities present in the process. Just a minor adjustment.

Last edited: Mar 10, 2018
5. Mar 10, 2018

### Staff: Mentor

Thanks. You're right. My algebra screwed up. Even for my assumed $T_B$ constant, the equation should read: $Q=T_B(\Delta S-\delta)$. And, of course, for an adiabatic irreversible process, Q=0, so that $\Delta S=\delta$.

Now, back to adiabatic reversible and irreversible expansions and contractions. If we assume for the reversible process that all the heat transfer can be carried out by contacting the system with a constant temperature reservoir at $T_B$ (which I assert, we can), then:

$$W=-\Delta U$$
$$\Delta S=\delta$$

$$W=-\Delta U+T_B\Delta S$$

where $\Delta S$ is the same for both paths (since they have the same end points). In both compression and expansion, entropy is generated within the system during the irreversible path. So $\Delta S > 0$ for both paths. This means that, for expansion, the work done by the gas on the surroundings is greater for the reversible path than for the irreversible path. It also means that, for compression, the work done by the surroundings on the system (-W) is greater for the irreversible path than the reversible path.

Last edited: Mar 10, 2018
6. Mar 10, 2018

### Staff: Mentor

Did you not notice where I said (irrespective of what the piston is doing) that $P_{ext}$ is defined as the force per unit area exerted by the gas acting on the inner face of the piston (and also the force of the inner face of the piston acting on the gas).

If there is also a force acting on the outside of the piston, say $P_{out}A$, then the force balance on the piston is given by:
$$P_{ext}A=P_{out}A+m\frac{dv}{dt}$$
If we multiply this equation by $v=dx/dt$ and integrate from 0 to t, we obtain (for the work done by the gas on its surroundings):
$$W=\int{P_{ext}dV}=\int{P_{out}dV}+m\frac{v^2}{2}\tag{1}$$
where v is the velocity of the piston at time t after releasing the piston. This equation applies irrespective of how massive the piston is.

Eqn. 1 describes the system behavior before it attains final thermodynamic equilibrium (since the piston is still moving). However, even if the piston is frictionless, its motion will eventually be damped to a stop by viscous stresses within the gas that affect the behavior of $P_{ext}$ as a function time at the inner piston face. Therefore, at final thermodynamic equilibrium of the system, we will have $$W=\int{P_{ext}}dV=\int{P_{out}dV}$$irrespective of how massive the piston is. If the force acting on the outside of the piston is held constant during the expansion, this equation becomes:
$$W=\int{P_{ext}dV}=P_{out}\Delta V$$
Even if the piston is extremely massive, this same equation will still apply. This means that the viscous stresses within the gas and the small deviation of the gas from being homogeneous, when acting over the large amount of time required to damp the piston motion, are sufficient to render this deformation highly irreversible, and to cause the work done to deviate substantially from the reversible work (that could only be achieved by controlling the outside force in such a way that the applied forces are always nearly in balance and the piston is not allowed to gain any significant kinetic energy. during the deformation).

Last edited: Mar 10, 2018
7. Mar 12, 2018

### Philip Koeck

I'm loosing track of what all the forces and pressures mean now. Could we go back to my thought experiment?
Let's say the pressure from the outside is Pext and this is either a pressure that's the same everywhere outside the cylinder or it's a local pressure acting on the piston. In the same way the pressure acting on the inside of the piston is Pint.
Fext is the force from the surrounding gas and Fint is the force from the gas inside the cylinder.
That means the balance of forces is Fint = Fext + Fman, where Fman is the force the experimenter needs to apply to keep the piston from moving. If the experimenter lets go the piston is accelerated. Now the balance of forces is Fint = Fext + m a, where a is the acceleration of the piston with mass m.
Do you agree with this?

8. Mar 12, 2018

### Philip Koeck

I've had a look at your text (mainly the analysis of the ideal gas).
You don't actually show that W = Pext ΔV, I would say.
In equation 9b you introduce the formula and then you get results similar to what you got for the damped spring. You write that this analogy is compelling evidence for the correctness of the above formula for PV-work. I don't think an analogy can ever be compelling evidence, although it can be interesting. I still haven't seen a derivation that clearly shows that W = Pext ΔV in general.
There's also an interesting paper ("Expansion Work without the External Pressure and Thermodynamics in Terms of Quasistatic Irreversible Processes", Klaus Schmidt-Rohr, J. Chem. Educ. 2014, 91, 402−409). The author is very critical of using this equation in general, but does accept it for special cases:
"...the concept of surroundings-based work enables analysis of some processes without well-defined P(V) or with a fast-moving frictionless piston. Here, the requirement is that at the end of the process, no macroscopic object moves as a result of the process; then, surroundings-based work is given by eq 2."
Eq. 2 in this paper is just W = Pext ΔV. Unfortunately I can't see where he proves this equation in this paper either.

Last edited: Mar 12, 2018
9. Mar 12, 2018

### Staff: Mentor

If you want to do it this way and want to treat the gas as your thermodynamic system, then $F_{int}$ takes the place of what we had been calling $P_{ext}A$, where $F_{int}$ is the force exerted by the gas on the piston face. So, we have changed the notation. It is important to note that, even if the piston is very massive, $F_{int}$ includes a contribution from viscous stresses (over and above what one would calculate from the ideal gas law). Furthermore, this deviation from the ideal gas pressure will be small for a very massive piston, but the time required for the motion of the piston to be damped (i.e., for final thermodynamic equilibrium to be achieved) will be very large, so the viscous stresses will still contribute significantly to the difference between the irreversible work done by the gas on the piston in this spontaneous process, and the reversible work that could be achieved (if the force on the outside face of the piston were varied with time in such a way that there was essentially never any net force on the piston). So, in terms of the work done by the gas on the piston, there is a significant difference between the piston always having a net force on it in the spontaneous irreversible deformation, and the piston never having a net force on it in the reversible deformation.

The work done by the gas on its surroundings is then $W=\int{\frac{F_{int}}{A}dV}=\int{P_{int}dV}$, where, even though we are using the symbol P, it really isn't solely a "pressure," and includes an added contribution from viscous stresses. So, the force balance equation on the piston is: $$P_{int}A=P_{ext}A+m\frac{dv}{dt}$$. This equation applies irrespective of the mass of the piston. If we multiply this equation by the piston velocity v = dx/dt and then integrate from time zero (the time at which the man releases the piston) to time t after release, we obtain: $$W(t)=\int{P_{int}dV}=\int{P_{ext}dV}+\frac{1}{2}mv^2$$ Even if the piston movement oscillates about the final equilibrium position, eventually its movement will be damped out by the viscous stresses in the gas, and, at final thermodynamic equilibrium, the piston will come to rest (so that v = 0). At this final equilibrium, we will have $$W(\infty)=\int{P_{int}dV}=\int{P_{ext}dV}$$ If the external force per unit area on the outside face of the piston were constant during the deformation of the gas to its final equilibrium volume, then this would then reduce to:
$$W(\infty)=\int{P_{int}dV}=P_{ext}\Delta V$$

10. Mar 12, 2018

### Staff: Mentor

I don't have access to this paper (since I am retired) but the arguments he seems to be making appear to be consistent with I have been saying (although he probably doesn't specifically refer to viscous stresses in the gas as the source of the piston motion being damped). And note that the analysis I presented (and, presumably the analysis he presented) applies irrespective of the mass of the piston.

11. Mar 12, 2018

### Philip Koeck

I think I'm finally catching on. This looks like just the kind of derivation I was looking for. Thanks.

12. Mar 12, 2018

### Staff: Mentor

To say I'm thrilled would be an understatement. This was not easy to explain.

Last edited: Mar 12, 2018
13. Mar 14, 2018

### Philip Koeck

Just a bit more discussion of this derivation:
It looks like the formula W = Pext ΔV should be very accurate for a heavy piston since it will move very slowly and the gas both inside and outside the cylinder will have plenty of time to spread out evenly. Obviously the piston has to be allowed to finish oscillating. For the massless piston the pressure inside and outside will probably not be so uniform and the pressure close to the piston won't be constant either so that should mean that the simple formula is only approximately correct. Do you agree? Another question is how friction between piston and cylinder plays in. I've found quite a few papers on the topic of work in various irreversible processes, many of them in the Journal of Chemical Education. Schmidt-Rohr and Gilason are among the authors. It's definitely a topic that requires discussion.

14. Mar 14, 2018

### Staff: Mentor

What a great question!!! Your observation is very perceptive. I do agree that, if the piston has low mass and the deformation of the air outside the cylinder is also very rapid, then viscous stresses and pressure non-uniformities will also contribute to the force per unit area history on the outside face of the piston. And the equation for the work that the outside face of the piston does on the outside air Pext ΔV may not be totally accurate. I do think however, that, because the outside air is not constrained (unlike the gas inside the cylinder), it is likely that the outside irreversibility will be less of a factor. But, this is far from a certainty. I would like to see some computational gas dynamics calculations done to quantify the magnitude of the difference. Meanwhile, I can say that, in problems encountered in thermodynamics books, they expect you to be unaware of these considerations, and either to assume that the equation is accurate or to assume that, somehow, the force per unit area is automatically controlled so that the force per unit area on the outside piston face does not change.
Before I give my thoughts on this, I would like to hear yours.

15. Mar 15, 2018

### Philip Koeck

Okay, I'll try something. If we describe friction as a velocity-dependent force, the force balance equation on the piston is:
Pint A = Pext A + m dv/dt + C v
Multiplying with v = dx/dt and integrating from 0 to t gives:
W(t) = ∫Pint dV = ∫Pext dV + ½ m v2 + ∫C v(x) dx
The last integral is a bit tricky for oscillations, but one could imagine that the damping due to the friction is so big that the piston never oscillates, but rather comes to a gradual stop when the pressure inside and outside is the same. Otherwise one can regard x as the distance traveled by the piston rather than a co-ordinate. When the piston has stopped the kinetic energy term is zero as in the friction-less case. It seems that the work done is a bit bigger than in the friction-less case (C=0). In the limit of a quasistatic expansion v(x) goes to zero and so does the friction.
The heat from the friction has to go somewhere, ouside, inside or both, and it will produce entropy by doing so. The details of the process, such as what the final state is, should depend on where the heat can go and how well the cylinder is insulated.

16. Mar 15, 2018

### Staff: Mentor

I agree with all of this. In the case of the frictional work, I would write it in the form $C\int{v^2}dt$ to emphasize its being positive definite.

17. Mar 15, 2018

### Staff: Mentor

Hi again Philip,

I was just thinking that it might be interesting to do an analysis of a very crude model of the frictional case roughly approximate the frictional work in more detail. Here is the crude model I had in mind:

1. Massless piston
2. Significant friction so that the gas deformation is not exceedingly rapid (and so we can neglect gas inertia)
3. Constant force on outer face of the piston, equal to $P_{atm}A$
4. Rapid heat transfer in the cylinder gas, with the cylinder held in contact with a constant temperature reservoir at the initial gas temperature, so that the expansion is isothermal at the initial gas temperature.

For this model, the force balance on the piston is $$F_{int}=F_{ext}\tag{1}$$ where $F_{int}$ represents the force exerted by the gas on the inside face of the piston, and $F_{ext}$ represents the combined force of the atmosphere on the outside face of the piston plus the friction force $F_R$. $$F_{ext}=P_{atm}A+F_R\tag{2}$$In this development, I propose to use a functional form for the kinetic friction force $F_R$ which (1) guarantees it will have the proper direction and (2) can be readily integrated as part of the final differential force balance equation:
$$F_R=\frac{CA}{V}\frac{dV}{dt}\tag{3}$$where V is the gas volume. As with the velocity-dependent approximation you proposed, this functional form depends on the rate of change of volume, although here, the "equivalent frictional coefficient" decreases with increasing volume.

For the irreversible expansion behavior of the gas, we use a crude approximation that combines its equilibrium (isothermal) PV behavior with its viscous character: $$F_{int}=\frac{P_0V_0}{V}A-\frac{\eta A}{V}\frac{dV}{dt}\tag{4}$$where $\eta$ is proportional to the gas viscosity. Finally, if we combine Eqns. 1-4, we obtain: $$\frac{P_0V_0}{V}-\frac{\eta }{V}\frac{dV}{dt}=P_{atm}+\frac{C}{V}\frac{dV}{dt}\tag{5}$$

Any interest in pursuing further?

18. Mar 16, 2018 at 2:56 PM

### Philip Koeck

Yes, that's interesting.
I've solved the differential equation (I think) for V(0) = V0 and get $$V(t)=\frac{P_0V_0}{V_{atm}}+V_0 (1-\frac{P_0}{P_{atm}}) exp(-\frac{P_{atm}}{C+η}t)$$ I apologize for my LaTex. I have to work on that.
For t → ∞ this becomes $$V(∞) = \frac{P_0V_0}{V_{atm}}$$
Do you get the same?
I guess one should look at the work done as a function of time next.

Why do you write the friction term as you do? It's inversely proportional to V, just like the viscous force, and why is the latter inversely proportional to V?

19. Mar 16, 2018 at 9:49 PM

### Staff: Mentor

I think so, but I think you made a couple of typos. I get
$$V(t)=\frac{P_0V_0}{P_{atm}}+V_0 (1-\frac{P_0}{P_{atm}}) exp(-\frac{P_{atm}}{C+η}t)$$and$$V(∞) = \frac{P_0V_0}{P_{atm}}$$From this point on, I think it would be better to work solely in terms of volumes:
$$V(t)=V_{\infty}+(V_0-V_{\infty}) \exp{\left(-\frac{P_{atm}}{(C+η)}t\right)}$$
Yes. What do you get for the work done by the gas on the piston?
I chose this functional form for the frictional force because it can be integrated analytically for this particular problem. It isn't quantitative, but, since we don't know what the exact relation will be anyway, I decided that, qualitatively, it will be OK.

As far as the viscous stress is concerned, the relation I used with the V in the denominator gives us the extensional rate of strain. For a homogeneous gas deformation, the extensional rate of strain is the same as dv/dx, where v is the local axial gas velocity in the cylinder, equal to 0 at the dead end of the cylinder (x = 0) and $\frac{1}{A}\frac{dV}{dt}$ at the piston face x = V/A.

20. Mar 17, 2018 at 4:07 AM

### Philip Koeck

Yes, that's what it should be. I got it wrong when I wrote it in LaTex.
I'll look at the work a bit later.