Understanding Wrev > Wirrev in Thermodynamics

In summary: I suppose in principle they could be equal. 2. I would say they must not be equal or it wouldn't move, but I'm sure I'm getting corrected pretty quickly.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...so I suppose in principle they could be equal.
  • #1
Ale_Rodo
32
6
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 Wrev is always greater than Wirrev , 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 Pext = Pint - dP which means Pext ≅ Pint and therefore Wrev = ∫Pext dV = ∫Pint dV = ∫(RT/V) dV = RTlog(Vf /Vi ) . I suppose T didn't change, but as I said, it's not clear.
Since we considered an expansion, Wrev > 0 .
2) In the second case, Pext = Pint - 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 Wirrev = ∫Pext dV = PΔV .

The professor then just states Wrev is always > Wirrev 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.
 
Science news on Phys.org
  • #2
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?
 
  • Like
Likes vanhees71
  • #3
Chestermiller said:
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.
 
  • #4
Ale_Rodo said:
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?
Ale_Rodo said:
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?
 
Last edited:
  • #5
Chestermiller said:
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.

Chestermiller said:
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.
 
  • #6
Ale_Rodo said:
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?
 
  • #7
Chestermiller said:
Are you okay with these conclusions so far?
Yes, everything clearer so far.
 
  • #8
Ale_Rodo said:
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?
 
  • #9
Chestermiller said:
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:
Wrev = ∫PdV = nRT∫dV/V = nRTlog(V2/V1) = nRTlog(V1P12/V2P22) .

Chestermiller said:
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 Ptot has P1 as a contribute which varies but cancels itself out, we will use Ptot as a constant since after doing the cancels we remain with P2 which was held constant.
So Wirrev = ∫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.
 
  • #10
Ale_Rodo said:
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:
Wrev = ∫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, $$W_{rev}=P_1V_1\ln{(V_2/V_1)}$$
Ale_Rodo said:
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 Ptot has P1 as a contribute which varies but cancels itself out, we will use Ptot as a constant since after doing the cancels we remain with P2 which was held constant.
So Wirrev = ∫PdV = P2∫dV = P2(V2-V1) = (P1V1/V2)(V2 - V1) .
Carrying the algebra one step further gives you $$W_{irrev}=P_1V_1\left(1-\frac{V_1}{V_2}\right)$$
Ale_Rodo said:
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?
 
  • #11
Chestermiller said:
This is correct, but I asked for the answer only in terms of P1, V1, and V2.
Oh yes, my bad.

Chestermiller said:
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.
 
  • #12
Ale_Rodo said:
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).
 
Last edited:
  • Like
Likes vanhees71
  • #13
Chestermiller said:
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 Wrev to be larger than Wirrev.
 
  • #14
Ale_Rodo said:
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 Wrev to be larger than Wirrev.
Do you have Excel? If not, do some quick calculations with a selection of representative values for V2/V1
 
  • #15
Chestermiller said:
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.
 
  • #16
Ale_Rodo said:
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 Wrev is always greater than Wirrev , 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.
 
  • Like
Likes vanhees71
  • #17
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.
 
  • #18
Chestermiller said:
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.
 

Related to Understanding Wrev > Wirrev in Thermodynamics

1. What is the difference between Wrev and Wirrev in thermodynamics?

Wrev and Wirrev are two terms used to describe different types of work in thermodynamics. Wrev, or reversible work, refers to work done in a system where there are no energy losses and the system can be returned to its initial state. Wirrev, or irreversible work, refers to work done in a system where there are energy losses and the system cannot be returned to its initial state.

2. How do Wrev and Wirrev affect the efficiency of a thermodynamic process?

The efficiency of a thermodynamic process is affected by the ratio of Wrev to the total work done, which includes both Wrev and Wirrev. The higher the ratio of Wrev, the more efficient the process is considered to be. This is because Wrev represents the ideal amount of work that can be obtained from a system, while Wirrev represents the amount of work that is lost due to energy losses.

3. Can you give an example of a process where Wrev and Wirrev are both present?

A common example of a process where both Wrev and Wirrev are present is the compression of a gas in a piston-cylinder system. While the gas is being compressed, some work is done on the gas (Wrev) to change its volume. However, there are also energy losses due to friction and heat transfer (Wirrev) which result in a decrease in the efficiency of the process.

4. How does the concept of entropy relate to Wrev and Wirrev?

The concept of entropy, which is a measure of the disorder or randomness in a system, is closely related to Wrev and Wirrev. In a reversible process, there is no increase in entropy as the system can be returned to its initial state. However, in an irreversible process, there is an increase in entropy due to energy losses, making it more difficult to return the system to its initial state.

5. Is it possible to have a process with only Wrev or only Wirrev?

In theory, it is possible to have a process with only Wrev or only Wirrev. However, in practical applications, both types of work are usually present to some degree. This is because it is difficult to completely eliminate energy losses in a system, and therefore, there will always be some amount of Wirrev present in a process.

Similar threads

Replies
22
Views
2K
Replies
4
Views
867
Replies
56
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
857
Replies
5
Views
1K
Replies
1
Views
731
Replies
13
Views
2K
Replies
4
Views
1K
Replies
8
Views
1K
  • Thermodynamics
Replies
8
Views
657
Back
Top