Understanding Wrev > Wirrev in Thermodynamics

[Ale_Rodo]

Hi,

I'm an engIneering undergrad and haven't done Thermodynamics just yet. The little I know about it comes from a Chemistry course I have.

Recently, I happened to struggle understanding the professor's and books' demonstration on how W_rev is always greater than W_irrev , which is a statement that was presented to us just after the First Law of Thermodynamics and just before the Second Law.

My problem is that said statement is not generally demonstrated but only displayed with an example, such as a piston pushed by an expanding ideal gas in a reversible way and then doing the same irreversibly. The initial hypothesis aren't defined and clear though and some assumptions made during the demonstration are not consistent with the experiment setup.

I'll try to explain more precisely.

-Firstly, let us consider a system composed by a cylinder containing 1 mol of an ideal gas and closed by a piston.
Suppose we want to conduct two experiments: one where a slow expansion happens (reversible) and the other is a sudden expansion (irreversible).

1) In the first case, the external pressure P_ext = P_int - dP which means P_ext ≅ P_int and therefore W_rev = ∫P_ext dV = ∫P_int dV = ∫(RT/V) dV = RTlog(V_f /V_i ) . I suppose T didn't change, but as I said, it's not clear.
Since we considered an expansion, W_rev > 0 .
2) In the second case, P_ext = P_int - ΔP and it's considered constant when integrated.
How can it be? The expansion takes place and, even if in a sudden, if the pressure remained the same value it was before we wouldn't have any expansion at all.
The demonstration continues then by stating W_irrev = ∫P_ext dV = PΔV .

The professor then just states W_rev is always > W_irrev which of course it's true, but it doesn't necessarily follow from this reasonment!
To actually cite him "This can be applied to all the situations in general"; yeah thanks...but why and how?
The initial condition not only aren't clear, but are not even the same!

So here I am, asking you to please elaborate and make it more rigorous to me in a way I might comprehend.
I thank anyone who will be answering this in advance, your patience is much appreciated.
Thank you.

 

[Chestermiller]

Welcome to the world of crappy thermodynamics textbooks and crappy professors. Don't blame yourself. The reason that you are struggling with this is that the material is presented so poorly, imprecisely, and, yes, even incorrectly. So let's first take a step back and look at things a little more fundamentally. Here are a couple of questions to get our conversation started:

1. In an irreversible expansion or compression, is the internal force per unit area exerted ideal gas on the piston described by the ideal gas law?

2. In an irreversible expansion or compression, Is the internal force per unit area exerted by the ideal gas on the inside face of the piston equal to the external force per unit area exerted by the inside face of the piston on the ideal gas?

 

[Ale_Rodo]

Welcome to the world of crappy thermodynamics textbooks and crappy professors. Don't blame yourself. The reason that you are struggling with this is that the material is presented so poorly, imprecisely, and, yes, even incorrectly. So let's first take a step back and look at things a little more fundamentally. Here are a couple of questions to get our conversation started:

1. In an irreversible expansion or compression, is the internal force per unit area exerted ideal gas on the piston described by the ideal gas law?

2. In an irreversible expansion or compression, Is the internal force per unit area exerted by the ideal gas on the inside face of the piston equal to the external force per unit area exerted by the inside face of the piston on the ideal gas?

Okay, so:

1. I don't have more reasons to say "yes" than I have to say "no" unfortunately, so I don't really know at this point;

2. I would say they must not be equal or it wouldn't move, but I'm sure I'm getting corrected pretty quickly.

 

[Chestermiller]

Okay, so:

1. I don't have more reasons to say "yes" than I have to say "no" unfortunately, so I don't really know at this point;

The ideal gas law is valid only for a gas that is at thermodynamic equilibrium (or for a gas undergoing a reversible expansion or compression, which consists of a continuous sequence of thermodynamic equilibrium states). When you study fluid dynamics, you will learn that that in an irreversible expansion or compression, there are viscous stresses present in the gas over and above the pressure calculated from the ideal gas law which contribute to the overall forces per unit area internally and at surfaces (such as the face of a piston). So, in an irreversible expansion, using the internal pressure from the ideal gas law to calculate the force per unit area on the piston and the work done by the gas on the piston gives the wrong answer.

OK so far?

2. I would say they must not be equal or it wouldn't move, but I'm sure I'm getting corrected pretty quickly.

From freshman physics, when we do a force balance on a body to see if it will move, we include only the forces that other bodies exert on the body under consideration, not the reaction forces that it exerts on other bodies.

From Newton's 3rd law of action-reaction, if a gas exerts a force F on the inside face of a piston, the inside face of the piston exerts an equal and oppositely directed force on the gas: true or false?

 

[Ale_Rodo]

OK so far?

Perfect, I got this part. It was an information I imagined but couldn't assume. It looks like where I'm studying professor tend to avoid such "details". Such a shame.

From Newton's 3rd law of action-reaction, if a gas exerts a force F on the inside face of a piston, the inside face of the piston exerts an equal and oppositely directed force on the gas: true or false?

I'd say it's true.

 

[Chestermiller]

Perfect, I got this part. It was an information I imagined but couldn't assume. It looks like where I'm studying professor tend to avoid such "details". Such a shame.
I'd say it's true.

That is correct. So, from these answers, we conclude that

1. In an irreversible expansion or compression of an ideal gas, we can't use the ideal gas equation to get the "internal pressure" of the gas (force per unit area exerted by the gas on the piston) because it would give the wrong answer (because of the presence of viscous stresses). For more details, see Transport Phenomena by Bird, Stewart, and Lightfoot, Chapter 1.

2. In both reversible and irreversible expansions or compressions of ideal gases, the "internal pressure" of the gas (force per unit area exerted by the gas on the inside face of the piston) is exactly equal to the "external pressure" (force per unit area exerted by the piston face on the gas), so the work done by the gas on its surroundings is always equal to the "external pressure" integrated over the change in volume. In an reversible expansion or compression, we can also get the work done by the gas by calculating the internal pressure from the ideal gas law and integrating over the volume change, but, in an irreversible expansion or compression, we can't do that. So, if we want to get the work for an irreversible expansion or compression, we are stuck with having to impose the external pressure manually (using a pressure transducer with feedback control or adding/removing weights from the piston).

Are you okay with these conclusions so far?

 

[Ale_Rodo]

Are you okay with these conclusions so far?

Yes, everything clearer so far.

 

[Chestermiller]

Yes, everything clearer so far.

OK. Let's analyze a focus problem to see if we can address your original question. I'll try to specify it as precisely as I can.

I have an ideal gas at temperature T, pressure $P_1$, and volume $V_1$ in a cylinder with a massless frictionless piston. The cylinder is held in contact with a constant temperature reservoir, also at temperature T, while it is expanded to final equilibrium volume $V_2$ (and, according to the ideal gas law, pressure $P_2=V_1P_1/V_2$).

In the reversible case, the external pressure on the piston is reduced gradually to $P_2$ as the gas volume increases. In terms of $P_1$, $V_1$, and $V_2$, how much work is done?

In the irreversible case, the external pressure on the piston is suddenly reduced at time zero to the final pressure $P_2$, and then held at that value until the system re-equilibrates at the final volume and temperature T. Initially, when we do this, the internal force per unit area of the gas also suddenly changes from $P_1$ to $P_2$ and the gas begins expanding rapidly. This rapid expansion of the gas is accompanied by viscous stresses within the gas. Since, initially, the volume and temperature have not changed yet, the "ideal gas contribution" to the initial total force per unit area remains momentarily at $P_1$, while the magnitude of the viscous stress contribution to the initial force per unit area of the gas on the piston must be $P_2-P_1\ (<0)$. So the total force per unit area of the gas on the piston is $P_1+(P_2-P_1)=P_2$. This indicates that the viscous stress contribution to the force per unit area on the piston is initially a tensile stress as the gas rapidly begins "stretching" axially. Now, for this irreversible expansion, in terms of $P_1$, $V_1$, and $V_2$, how much work is done?

 

[Ale_Rodo]

In the reversible case, the external pressure on the piston is reduced gradually to P₂ as the gas volume increases. In terms of P₁, V₁, and V₂, how much work is done?

The system's evolution is just a sequence of equilibrium states, so we might just sum all the infinitesimal work done by each and every state with an integral from V₁ to V₂ with the pressure being the internal one so:
W_rev = ∫PdV = nRT∫dV/V = nRTlog(V₂/V₁) = nRTlog(V₁P₁²/V₂P₂²) .

Now, for this irreversible expansion, in terms of P₁, V₁, and V₂, how much work is done?

Similarly to the reversible case, I think we might want to calculate the work done by using an integral from V₁ to V₂ but this time, although P_tot has P₁ as a contribute which varies but cancels itself out, we will use P_tot as a constant since after doing the cancels we remain with P₂ which was held constant.
So W_irrev = ∫PdV = P₂∫dV = P₂(V₂-V₁) = (P₁V₁/V₂)(V₂ - V₁) .

It feels extremely close to the procedure I explained in my initial question, but this is what I came up with so please bear with me if I'm not right.

 

[Chestermiller]

The system's evolution is just a sequence of equilibrium states, so we might just sum all the infinitesimal work done by each and every state with an integral from V₁ to V₂ with the pressure being the internal one so:
W_rev = ∫PdV = nRT∫dV/V = nRTlog(V₂/V₁) = nRTlog(V₁P₁²/V₂P₂²) .

This is correct, but I asked for the answer only in terms of P1, V1, and V2. From the ideal gas equation, $nRT=P_1V_1$. So, $$W_{rev}=P_1V_1\ln{(V_2/V_1)}$$

Similarly to the reversible case, I think we might want to calculate the work done by using an integral from V₁ to V₂ but this time, although P_tot has P₁ as a contribute which varies but cancels itself out, we will use P_tot as a constant since after doing the cancels we remain with P₂ which was held constant.
So W_irrev = ∫PdV = P₂∫dV = P₂(V₂-V₁) = (P₁V₁/V₂)(V₂ - V₁) .

Carrying the algebra one step further gives you $$W_{irrev}=P_1V_1\left(1-\frac{V_1}{V_2}\right)$$

It feels extremely close to the procedure I explained in my initial question, but this is what I came up with so please bear with me if I'm not right.

In your original question, it was not stated that the system is in contact with a constant temperature reservoir at the same temperature as that of the system initially. This is the missing constraint (limitation) you were intuitively looking for.

Do you know how to compare the two amounts of work to see which is larger?

 

[Ale_Rodo]

This is correct, but I asked for the answer only in terms of P1, V1, and V2.

Oh yes, my bad.

Do you know how to compare the two amounts of work to see which is larger?

As an engineering student, if we had some values that wouldn't be a problem of course, but I suppose you meant to convert one work or the other in a formula which can indeed be comparable. Or we could just try to get the ratio between the twos, but I feel we're not getting any useful information from there?
My final guess would be describing those measures each with a curve function of P and V and then looking at the area which will be our integrals.

 

[Chestermiller]

Oh yes, my bad.As an engineering student, if we had some values that wouldn't be a problem of course, but I suppose you meant to convert one work or the other in a formula which can indeed be comparable. Or we could just try to get the ratio between the twos, but I feel we're not getting any useful information from there?
My final guess would be describing those measures each with a curve function of P and V and then looking at the area which will be our integrals.

The leading factor of each of the expressions is $P_1V_1$, so this is the same for both. So to compare them, all you need to do is make a plot of $\ln{(V_2/V_1)}$ vs $V_2/V_1$ and $1-\frac{V_1}{V_2}$ vs $V_2/V_1$ on the same graph and compare them to see which is bigger. (Of course, you have your graphics package use log scales for both axes).

 

[Ale_Rodo]

So to compare them, all you need to do is make a plot of ln⁡(V₂/V₁) vs V₂/V₁ and 1−V₁V₂ vs V₂/V₁ on the same graph and compare them to see which is bigger. (Of course, you have your graphics package use log scales for both axes).

I'm not sure how to plot it on my laptop or which program to use, but I'm going to cheat if you're ok with that and say that I was expecting W_rev to be larger than W_irrev.

 

[Chestermiller]

I'm not sure how to plot it on my laptop or which program to use, but I'm going to cheat if you're ok with that and say that I was expecting W_rev to be larger than W_irrev.

Do you have Excel? If not, do some quick calculations with a selection of representative values for V2/V1

 

[Ale_Rodo]

Do you have Excel? If not, do some quick calculations with a selection of representative values for V2/V1

I did the second option and ln(V₂/V₁) is indeed the larger one.

 

[Philip Koeck]

Hi,

I'm an engIneering undergrad and haven't done Thermodynamics just yet. The little I know about it comes from a Chemistry course I have.

Recently, I happened to struggle understanding the professor's and books' demonstration on how W_rev is always greater than W_irrev , which is a statement that was presented to us just after the First Law of Thermodynamics and just before the Second Law.

My problem is that said statement is not generally demonstrated but only displayed with an example, such as a piston pushed by an expanding ideal gas in a reversible way and then doing the same irreversibly. The initial hypothesis aren't defined and clear though and some assumptions made during the demonstration are not consistent with the experiment setup.

You could look at this text, which I wrote after a long, interesting discussion with (mainly) Chestermiller: https://www.researchgate.net/publication/333728790_PV-work_for_irreversible_processes
Might help a bit. It does build on the 2.law, however, and I'm not sure if everyone agrees with my discussion.

 

[mfig]

Hello,

I think a general demonstration could be given along these lines. Consider a general closed system operating between some heat source (possibly variable temperature $T$) and the environment (temperature $T_0$) as it evolves from state 1 to state 2. For this case we have:

Energy Balance: $\Delta E = \int_1^2\delta Q - W$

Entropy Balance: $\Delta S = \int_1^2\frac{\delta Q}{T} + S_{gen}$

In the above, the energy of the system includes internal energy, kinetic, and potential energies and the work includes pV and shaft work. Also, the $S_{gen}$ term is the entropy generated during the process, a quantity which is always non-negative. Now multiply the entropy balance by $T_0$ and then subtract the result from the energy balance. This gives:

$W = T_0 \Delta S - \Delta E +\int_1^2 \delta Q \left(1- \frac{T_0}{T}\right) - T_0S_{gen}$

The first two terms on the right depend only on the states, and thus would be the same whether any process is reversible or not. The third term is the maximum work that can be extracted from the heat transfer - the Carnot efficiency. The only term that depends on whether the process is reversible or not is the last term, which has a minimum value of 0 only in the case of no irreversibilities. Thus the maximum work is obtained when the last term is zero, which means the maximum work is obtained when the process is reversible.

 

[cianfa72]

From freshman physics, when we do a force balance on a body to see if it will move, we include only the forces that other bodies exert on the body under consideration, not the reaction forces that it exerts on other bodies.

I would add that since we assume negligible the mass of the piston then, to avoid an infinite acceleration of it, the two forces (internal from gas and external) acting on it must be equal and opposite.

Then as said by @Chestermiller from 3rd Newton law the force that the inside face of the piston exerts on the gas is equal to the opposite of the other.