Thermo - PV work and mechanical work

[Chestermiller]

Since (m+M) d^2 x/dt^2 = (m+M)g−PA we need to multiply that by dx to get the work

\delta W = (m+M)g dx−PAdx

but we need to integrate over time so dividing by dt and multiplying by dt (chain rule)

\delta W = \int ((m+M)g dx/dt−PAdx/dt) dt

This is strange... I don't know how to proceed the integration

However, since dE=\delta W-\delta Q
for the case of the mass dropping fast, the system don't have time to exchange heat with the surroundings, so \delta Q = 0

dE=\delta W implies that dE=dW, thus an exact differential. Is that correct?

Good try, but no. What you are calling ΔW is really the change in kinetic energy of the masses. We want to determine the work done by the gas on the bottom face of the piston. This is the work W done by the "system" on the "surroundings" (since we regard the gas as the system, and everything else as the surroundings). That work W is given by:
W=\int{PdV}=-\int{PAdx}=-\int_{t=0}^{t=∞}{PA\left(\frac{dx}{dt}\right)dt}
We can find this work by first multiplying the force balance by dx/dt to obtain:
(m+M)\frac{d^2x}{dt^2}\left(\frac{dx}{dt}\right)=(m+M)g\left(\frac{dx}{dt}\right)-PA\left(\frac{dx}{dt}\right)=-\frac{(m+M)g}{A}\left(\frac{dV}{dt}\right)+P\left(\frac{dV}{dt}\right)
The left hand side of this equation is an exact differential, so we can rewrite the previous equation as:

\frac{d}{dt}\left(\frac{(m+M)}{2}\left(\frac{dx}{dt}\right)^2\right)=-\frac{(m+M)g}{A}\left(\frac{dV}{dt}\right)+P\left(\frac{dV}{dt}\right)
Rearranging this equation gives:
P\left(\frac{dV}{dt}\right)=\frac{d}{dt}\left(\frac{(m+M)}{2}v^2\right)+\frac{(m+M)g}{A}\left(\frac{dV}{dt}\right)
where v is the velocity of the combined masses dx/dt. The left hand side is rate at which the gas is doing work on the two masses, the first term on the right hand side is the rate of change of kinetic energy of the two masses, and the second term on the right hand side is the rate of change of potential energy of the two masses. If we integrate this equation from time t = 0 to arbitrary time t, we obtain:
\int_{V_1}^{V(t)}{PdV}=\left(\frac{(m+M)}{2}v^2\right)+\frac{(m+M)g}{A}(V(t)-V_1)
where we have made use of the initial conditions that v = 0 and V = V₁ at t = 0.

Let's discuss what's happening here. After we release the masses, they will speed up, reach a maximum downward velocity and then slow down until they reach a maximum downward displacement where they come to a stop. But when they get to this bottom location, there is now a net upward force on them, so they start moving back upward again. They speed up, reach a maximum upward velocity, and then slow down until they reach a maximum upward displacement. But, when they get to the upper location, there is now a net downward force on them, so they start moving down again. The net result of all this is that the masses will oscillate up and down, very much as if they were in simple harmonic motion.

However, even with the piston being frictionless, the amplitude of the oscillation will decrease over time as a result of viscous dissipation within the gas. The details of how this happens is unimportant for our purposes, except that, as time progresses, the velocity of the masses will progressively decrease, and at very long times, it will become equal to zero. Therefore, at final steady state of the system, we will have:
W=\int_{V_1}^{V_2}{PdV}=\frac{(m+M)g}{A}(V_2-V_1)
Amazingly, the work from the initial to the final equilibrium state reduces to this very simple equation.

I'm going to stop here an let you digest what I've presented, as well as let you ask any question you might have. When you're ready, we can continue. We will be looking at both adiabatic and isothermal compression resulting from releasing the mass M.

 

[LoopInt]

I'm sorry, I didn't follow how you got from (m+M)d²xdt²(dx/dt) to d/dt((m+M)2(dx/dt)^2).

Also, you said this will be in harmonic motion, so isn't it "strange" to make the conditions for the final equilibrium stage without specifying how the energy is being lost? How do you see that this makes sense?

 

[Chestermiller]

I'm sorry, I didn't follow how you got from (m+M)d²xdt²(dx/dt) to d/dt((m+M)2(dx/dt)^2).

I started out by saying:

"We can find this work by first multiplying the force balance by dx/dt to obtain:

(m+M)\left(\frac{dx}{dt}\right)\frac{d^2x}{dt^2}=-\frac{(m+M)g}{A}\left(\frac{dV}{dt}\right)+P\left(\frac{dV}{dt}\right)"

If we make the substitutions v = dx/dt, and dv/dt = d²x/dt²l on the left hand side, we get
(m+M)\left(\frac{dx}{dt}\right)\frac{d^2x}{dt^2}=(m+M)\frac{vdv}{dt}=(m+M)\frac{d(\frac{v^2}{2})}{dt}

Also, you said this will be in harmonic motion, so isn't it "strange" to make the conditions for the final equilibrium stage without specifying how the energy is being lost? How do you see that this makes sense?

This isn't exactly what I meant. I meant that the motion would be similar to simple harmonic motion, but damped. There is no energy being lost. Part of the mechanical energy (the kinetic energy) is being converted into internal energy and heat. Even with the piston being frictionless, you wouldn't expect the masses to continue bobbing up and down forever, would you? Neither would I. The gas in the cylinder has a physical property called viscosity. Have you heard of this property before? Even though the viscosity of the gas is very low, it is not zero. The viscosity acts to convert mechanical energy to heat when the gas deforms. So, not only does the gas act "elastically" like a spring, it also partially acts like a damper. This is analogous to a car shock absorber where you have a spring and a damper in series. When you try to oscillate the front end of the car up and down, the deforming viscous oil in the damper causes the amplitude of the oscillation to decay with time. The same kind of thing happens in our system until all the kinetic energy is converted. It is possible to formulate and solve the gas dynamics problem describing all the details of what is happening within the cylinder as a function of both spatial position and time, but this is not necessary for our purposes, because we are mainly interested in the initial and final equilibrium states of the system. So all we really need to know for our purposes is that the kinetic energy gets damped out over time (but this energy does not just vanish). I hope this makes better sense.

Chet

 

[LoopInt]

Ok, let me see if I get it. We are not specifying what is this damping factor and how it interacts with the system in detail, but we know that this exists and eventually all the kinetic energy will be converted to internal energy and heat, and so v final will be 0, which is all that matters to calculate the work.

If that's correct, let's move on!

 

[Chestermiller]

Ok, let me see if I get it. We are not specifying what is this damping factor and how it interacts with the system in detail, but we know that this exists and eventually all the kinetic energy will be converted to internal energy and heat, and so v final will be 0, which is all that matters to calculate the work.

If that's correct, let's move on!

Yes. That's right on.

So, the work equation we've derived so far makes no assumptions about whether the compression is isothermal or adiabatic (or anything else for that matter). Which case would you prefer to consider first? Isothermal or adiabatic?

Chet

 

[LoopInt]

I would like to start by the isothermal. If we can calculate the total energy too it would be better!

 

[Chestermiller]

I would like to start by the isothermal.

Isothermal

In the isothermal case, we are going to express the work specifically for the isothermal case, we are going to determine the heat added Q, and we are going to look at the change in entropy of the gas.

When we compress the gas irreversibly like in our problem, the temperature and pressure of the gas within the cylinder are both going t be non-uniform, except in the initial and final equilibrium states. So during the deformation we can't use the ideal gas law, but at the beginning and end we can. We will make use of this.

During the deformation, the gas being compressed irreversibly in the cylinder is going to get hotter than in the initial state. In order to guarantee that the process is considered "isothermal," we are going to need to make sure that the final temperature T2 is equal to the initial temperature T1. This is what the term "isothermal" means for an irreversible process. We can guarantee that this condition is met by holding all or part of the inside surface of the cylinder at T1. It is not necessary to hold the entire surface at T1, and the remaining part of the surface can be insulated. But, in the part where the temperature is T1, all the heat removal -Q will take place through the portion of the surface held at T1. For our purposes, we can assume that the base of the cylinder is held at T1, and the remainder of the cylinder, as well as the piston, are insulated.

Previously, we showed that the work done by the gas on the surroundings (i.e., the two masses) is given by:
W=\int_{V_1}^{V_2}{PdV}=\frac{(m+M)g}{A}(V_2-V_1)
We also showed that:

P1=mg/A

P2=(m+M)g/A

and

P2/P1=(m+M)/m

Please use these these relationships in conjunction with the ideal gas law to express W exclusively in terms of T1, R, n (the number of moles), m, and M.

Chet

 

[LoopInt]

Ok. First off, I have one question. It doesn`t matter if the system increases or decreases temperature in the middle, as far as Tinicial=Tfinal?

Now to the problem.

(m+M)g/A (V2-V1)

At the initial state volume is V1 and we can use ideal gas law, because the system is at equilibrium. So,

V1=nRT/P1 and V2=nRT/P2, since the system is in equilibrium at the final stage. T1=T2=T

nRT(1/P1-1/P2) --> nRT( (P2-P1)/P1P2) ) --> nRT( A/gm - A/mg(m+M)) --> AnRT(1/mg-1/mg(m+M)) = (V2-V1)

W = (m+M)nRT(1/m-1/m(m+M)) --> W = nRT((m+M)-1)/m

 

[Chestermiller]

Ok. First off, I have one question. It doesn`t matter if the system increases or decreases temperature in the middle, as far as Tinicial=Tfinal?

Yes. Apparently, in this problem, because of the constraints imposed by the weights of the masses in determining the amount of work done, the details of the path between the initial and final states is not as important as in other problems. After all, we really haven't made any assumptions in deriving our equation for the amount of work. Our equation tells us that the only parameters really involved in determining the amount of work are the initial and final volumes, and the final pressure.

Now to the problem.

(m+M)g/A (V2-V1)

At the initial state volume is V1 and we can use ideal gas law, because the system is at equilibrium. So,

V1=nRT/P1 and V2=nRT/P2, since the system is in equilibrium at the final stage. T1=T2=T

nRT(1/P1-1/P2) --> nRT( (P2-P1)/P1P2) ) --> nRT( A/gm - A/mg(m+M)) --> AnRT(1/mg-1/mg(m+M)) = (V2-V1)

W = (m+M)nRT(1/m-1/m(m+M)) --> W = nRT((m+M)-1)/m

Excellent. This is what I got, except I had a minus sign, and, I combined terms further to obtain W = -nRT₁M/m.
Now, the next step is to use the first law (we haven't even used it yet) to determine the amount of heat added to the system (gas) from the surroundings Q, and the change in internal energy of the system. What is your assessment of this?

We are also going to show that the Clausius inequality is satisfied by our system. According to the Clausius inequality, ΔS≥Q/T_I, where T_I is the temperature at the interface with the surroundings (in our problem T₁) over the portion of the interfacial area that the heat Q passes. So, first of all, what do you get for Q/T_I? Secondly, have you learned the equation for the change in entropy of an ideal gas from an initial equilibrium state to a final equilibrium state in terms of T₂/T₁ and V₂/V₁? If so, please write it out, and apply it to our specific problem. Then, from these two results, please determine whether the Clausius inequality is satisfied by our irreversible process.

This will complete the analysis of the isothermal case. We can then start on the more interesting adiabatic case, if you still have the energy. In my judgement, you're been doing extremely well so far, and have been covering a lot of interesting material as well as gaining valuable experience.

Chet

 

[LoopInt]

Why the minus sign? The work done by the system shouldn't be positive? (convention)
Well, I don't know much of the first law. What I know is that dE = dQ - dW.
So I will guess that, since E is the total energy of the system, we can ignore some terms of E. The kinetic energy of the system at initial and final states are 0, because the system is in equilibrium. So only the internal energy is significant for our system. So,
U2-U1 = Q - W
Now I guess we should evaluate U2 U1 somehow or calculate Q=mc(T2-T1).

I know S is entropy, and that's all hehe.

Can you guide me through?

And I sure want to continue our adventure hehe. Thank's for this great help!

 

[Chestermiller]

Why the minus sign?

In #38, you made a small algebra error. The factor should have been V2-V1, not V1-V2.

Well, I don't know much of the first law. What I know is that dE = dQ - dW.
So I will guess that, since E is the total energy of the system, we can ignore some terms of E. The kinetic energy of the system at initial and final states are 0, because the system is in equilibrium. So only the internal energy is significant for our system. So,
U2-U1 = Q - W

Nice analysis.

Now I guess we should evaluate U2 U1 somehow or calculate Q=mc(T2-T1).

For an ideal gas, the internal energy is a function only of temperature. Since we are looking at the isothermal case, U2-U1=0. So

Q = W = -nRT₁M/m

Basically, when we compress the gas isothermally, we have to remove heat.

I know S is entropy, and that's all hehe.

Can you guide me through?

To get the change in entropy, you need to first identify a reversible path between the initial and final states, and then you need to calculate ∫dQ/T for that path. Fortunately, we've already done that when we analyzed the problem for the case where we lower the weight gradually (which constitutes a reversible path). In that case you determined that:

W=nRT_1\ln (\frac{L}{L-d})=nRT_1\ln (V_1/V_21)=-nRT_1\ln \left(1+\frac{M}{m}\right)
So, for the change in entropy, we have

ΔS=-nR\ln \left(1+\frac{M}{m}\right)

and for the irreversible path we have:

Q/T = -nR(M/m)

These results clearly satisfy Cauchy's inequality that ΔS ≥ Q/T

Gotta leave not, but I'll be back later to get us started on the adiabatic case.

Chet

 

[Chestermiller]

Adiabatic Irreversible Compression Case

In the adiabatic compression case, the work done on the gas is going to cause its temperature to rise. We don't know the final temperature yet, and we will have to use the first law to calculate it. Let T₂ represent the final temperature of the gas. From our previous analysis, we have equations for P₁ and P₂, and we can use the ideal gas law for the initial and final states. We also have a general equation for the work that we can apply to this case. Please use these equations to express the work W in terms of n, R, T₁, T₂, M, and m. After we have this relationship, we can apply the first law to get T₂.

Chet

 

[LoopInt]

Ok. Let's check some concepts first.

To calculate the entropy I need an expression of Q. To determine how Q changes we looked at the work that we calculated, because

U2-U1=Q-W

We know U2-U1 for the isothermal case, so we know that Q=W

Now at the integration we have

\int dQ/T

We know T is constant so its
1/T \int dQ = Q/T = W/T

Ok, now I don't understand why can we do the same for the irreversible path, since we need to calculate the work from a reversible path.

Now to express the work

I found that

W=(T2m-T1(m+M))nR

And since Q=0 in the adiabatic case

U2-U1=-W

If I knew how the internal energy is calculated for a perfect gas I would continue substituting for U2 and U1.

 

[Chestermiller]

Ok. Let's check some concepts first.

To calculate the entropy I need an expression of Q. To determine how Q changes we looked at the work that we calculated

Not exactly. As I said in post #41, in order to calculate the change in entropy, you need to first dream up a reversible path between the same initial and final states as for your irreversible path. You calculate ΔS=∫dQ/T for that reversible path.

U2-U1=Q-W

We know U2-U1 for the isothermal case, so we know that Q=W

Now at the integration we have

\int dQ/T

We know T is constant so its
1/T \int dQ = Q/T = W/T

Ok, now I don't understand why can we do the same for the irreversible path, since we need to calculate the work from a reversible path.

In general, if the reversible path is not isothermal, we don't calculate the entropy from the work, but we always calculate if from ∫dQ/T for the reversible path.

Now Back to the Adiabatic Irreversible Case.

Now to express the work

I found that

W=(T2m-T1(m+M))nR

Try the algebra again. I get W=(T2-T1(m+M)/m)nR

And since Q=0 in the adiabatic case

U2-U1=-W

If I knew how the internal energy is calculated for a perfect gas I would continue substituting for U2 and U1.

For an ideal gas, the internal energy is a function only of temperature. Assuming that the molar heat capacity at constant volume C_v is constant between temperatures T1 and T2, the change in internal energy is given by:

U2-U1=nC_v(T2-T1).

You now have what you need to calculate T2. Please solve for T2, and then we can continue.

Chet

 

[LoopInt]

Yea I left one m behind.

So we have
U2-U1=-W
U2-U1=Cv(T2-T1)

So, sunstituting the terms and solving for T2, I got

T2= (nR(m+M)-Cvm)T1/(m(nR-Cv))

 

[Chestermiller]

Yea I left one m behind.

So we have
U2-U1=-W
U2-U1=Cv(T2-T1)

So, sunstituting the terms and solving for T2, I got

T2= (nR(m+M)-Cvm)T1/(m(nR-Cv))

You made a couple of algebra errors. You left out a factor of n in your equation for the change in internal energy, and you forgot to change the sign on the W term to get -W. Also, the n's should cancel out.

Please try again.

Chet

 

[LoopInt]

Sorry, my head is somewhere else.

T2=(((m+M)R+Cvm)T1)/(mR+Cvm)

 

[Chestermiller]

Sorry, my head is somewhere else.

T2=(((m+M)R+Cvm)T1)/(mR+Cvm)

That's a lot of parenthesis. I took your result and expressed it in a little different way:
\frac{T_2}{T_1}=1+\frac{R}{(C_v+R)}\frac{M}{m}
In adiabatic compression and expansion problems, a parameter which occurs very often is γ, the ratio of the heat capacity at constant pressure to the heat capacity at constant volume:
γ=\frac{C_p}{C_v}=\frac{C_v+R}{C_v}
where we have made use here of the condition that, for an ideal gas, C_p=C_v+R
See if you can use this to express T₂/T₁ exclusively in terms of γ and M/m.

Then we'll go on to determining the change in entropy, and comparing it with ∫dQ/T for our irreversible adiabatic compression process.

Chet

 

[LoopInt]

Sorry I took so long,

\frac{T_2}{T_1}=\frac{(y-1)M}{ym}

And I need to ask you something. I am trying to understand how viscoelastic materials behave. I'm getting hysteresis on my system and I think that it is mainly due to the latex rubber I'm using. I'm trying to model the dissipated energy of the latex due to streching. Can you suggest a way I can do that? Or the theory I can use?

 

[Chestermiller]

Sorry I took so long,

\frac{T_2}{T_1}=\frac{(y-1)M}{ym}

Almost. You left out the 1

And I need to ask you something. I am trying to understand how viscoelastic materials behave. I'm getting hysteresis on my system and I think that it is mainly due to the latex rubber I'm using. I'm trying to model the dissipated energy of the latex due to streching. Can you suggest a way I can do that? Or the theory I can use?

You are getting into a complicated area. Is it large deformation or small deformation? That is, is it small deformation like in dynamic mechanical analysis? Also, regarding hysteresis, if you do the deformation very slowly, or just deform to a new equilibrium state (i.e., wait until the system equilibrates), is the hysteresis still present?

Chet

 

[Soumalya]

Hello,
If I am not wrong the issue is why displacement work is an inexact differential whereas mechanical work is not always denoted as an exact differential?

Short answer: W=∫Fdx is the generalized equation for mechanical work ,where,

It's most appropriate to write W=∫δw and NOT ∫dw as work is a path dependent process.
So, writing W=∫dw or dw=Fdx is a serious malpractice.

Explanation: You need to specify the properties pressure and volume for a series of quasi static or equilibrium states to specify the path upon which work can be calculated. As work is defined as the area under the curve between the properties P and V in a displacement process one needs to add up all the elemental work for each equilibrium states to get total work done.As there can be more than one quasi static paths possible between two end states of a thermodynamic process different work transfers are possible.Thus work is essentially a path dependent process.

Even when considering mechanical extension or compression of a spring you need intermediate equilibrium positions to be achieved after each oscillation restores to a new equilibrium position.Thus to obtain work using integration we need a series of equilibrium states to specify the path upon which work is to be calculated hence work is practically always an inexact differential.

So it's always appropriate to write δw=Fdx or δw=PdV as it demands the application of mathematical integration to obtain total work.

Hope this helps
Best regards

 

[Chestermiller]

Hello,
If I am not wrong the issue is why displacement work is an inexact differential whereas mechanical work is not always denoted as an exact differential?

Short answer: W=∫Fdx is the generalized equation for mechanical work ,where,

It's most appropriate to write W=∫δw and NOT ∫dw as work is a path dependent process.
So, writing W=∫dw or dw=Fdx is a serious malpractice.

Explanation: You need to specify the properties pressure and volume for a series of quasi static or equilibrium states to specify the path upon which work can be calculated. As work is defined as the area under the curve between the properties P and V in a displacement process one needs to add up all the elemental work for each equilibrium states to get total work done.As there can be more than one quasi static paths possible between two end states of a thermodynamic process different work transfers are possible.Thus work is essentially a path dependent process.

Even when considering mechanical extension or compression of a spring you need intermediate equilibrium positions to be achieved after each oscillation restores to a new equilibrium position.Thus to obtain work using integration we need a series of equilibrium states to specify the path upon which work is to be calculated hence work is practically always an inexact differential.

So it's always appropriate to write δw=Fdx or δw=PdV as it demands the application of mathematical integration to obtain total work.

Hope this helps
Best regards

For an irreversible process, you can't assume that the work can be obtained by integrating over a quasistatic sequence of equilibrium states. This is only correct to do for a reversible process.

Chet

 

[ffia]

Hi!

I've read part of the discussion, and I'd have a related question.
I believe you can always (?) choose the environment so that its changes occur reversibly. Why is this i.e. why can you choose it that way?

 

[Chestermiller]

Hi!

I've read part of the discussion, and I'd have a related question.
I believe you can always (?) choose the environment so that its changes occur reversibly. Why is this i.e. why can you choose it that way?

I don't quite understand this question. Are you saying that you can always find a reversible path between the initial and final equilibrium states of a (closed) system? Or, are you saying that, even for an irreversible path between the initial and final equilibrium states of a closed system, you can always somehow view the irreversible path from some perspective as being a reversible path?

Have you read my Physics Forums blog?

Chet

 

[ffia]

I don't quite understand this question. Are you saying that you can always find a reversible path between the initial and final equilibrium states of a (closed) system? Or, are you saying that, even for an irreversible path between the initial and final equilibrium states of a closed system, you can always somehow view the irreversible path from some perspective as being a reversible path?

Have you read my Physics Forums blog?

Chet

Thanks for the fast reply!
I'm actually looking at your blog right now.
And sorry. But I feel like I can't be more specific with the question before I understand the answer.
But I'll try, this is something I read from a book by Reiss. Reiss uses the wording "the environment (surroundings, outside the system in which there might occur an irreversible change) has been chosen to always behave reversibly". He refers to one of his chapters for a proof of this, but in that chapter I only find an example of a single process, hardly a proof.

Now that I think of it, I am not sure whether the statement requires for the system to be closed or not (the system is of course not isolated). But in the situations in which this is supposed to apply, this is supposed to help the examination of the system, if we know anything about the surroundings. For if the changes in the surroundings are reversible, then it is supposed to have well-defined quantities, and I can always define the entropy change, work/heat change, related to the possibly irreversible changes of the system.

I would appreciate any clarification:
-is that statement true (surroundings behave reversibly), and when?
-if so, how can you motivate/prove that it's true

EDIT: The idea of a reversible behaviour of the surroundings sounds like a very general concept, as he uses this, for example, to show that the work done reversibly is always greater than work done irreversibly (for the same change in a system)
[Here, he assumes an irreversible change in a system. For this change, there is a corresponding change in heat, Dq (he uses D for not necessarily reversible changes). Here he states that as the surroundings behave reversibly, Dq = dq_{surr} = TdS_{surr}. From here he goes onto writing the entropy change for system + surroundings, and so on..]

 

[Chestermiller]

Thanks for the fast reply!
I'm actually looking at your blog right now.
And sorry. But I feel like I can't be more specific with the question before I understand the answer.
But I'll try, this is something I read from a book by Reiss. Reiss uses the wording "the environment (surroundings, outside the system in which there might occur an irreversible change) has been chosen to always behave reversibly". He refers to one of his chapters for a proof of this, but in that chapter I only find an example of a single process, hardly a proof.

Now that I think of it, I am not sure whether the statement requires for the system to be closed or not (the system is of course not isolated). But in the situations in which this is supposed to apply, this is supposed to help the examination of the system, if we know anything about the surroundings. For if the changes in the surroundings are reversible, then it is supposed to have well-defined quantities, and I can always define the entropy change, work/heat change, related to the possibly irreversible changes of the system.

I would appreciate any clarification:
-is that statement true (surroundings behave reversibly), and when?
-if so, how can you motivate/prove that it's true

EDIT: The idea of a reversible behaviour of the surroundings sounds like a very general concept, as he uses this, for example, to show that the work done reversibly is always greater than work done irreversibly (for the same change in a system)
[Here, he assumes an irreversible change in a system. For this change, there is a corresponding change in heat, Dq (he uses D for not necessarily reversible changes). Here he states that as the surroundings behave reversibly, Dq = dq_{surr} = TdS_{surr}. From here he goes onto writing the entropy change for system + surroundings, and so on..]

I'm not familiar with Reiss' book, so I can't comment intelligently. I'm going to try to articulate what you are saying to see if I understand correctly.

He is saying that, even if a (closed) system experiences an irreversible process, it is possible to identify one or more reversible paths for the surroundings that imposes the exact same irreversible path on the system. Correct?

Chet

 

[ffia]

I'm not familiar with Reiss' book, so I can't comment intelligently. I'm going to try to articulate what you are saying to see if I understand correctly.

He is saying that, even if a (closed) system experiences an irreversible process, it is possible to identify one or more reversible paths for the surroundings that imposes the exact same irreversible path on the system. Correct?

Chet

Yes! Or, rather (the bolded part) irreversible change (I think). Actually quite like you suggested earlier, I think Reiss says that between any two points on the state space one can find the reversible path between them.
The most counterintuitive part is that we would be able to always "find" that path from the surroundings.
[Now that I think of it it's actually kind of funny that we always "end up" on a point in the state space even with an irreversible change. I guess that's because of the "static" nature of thermodynamics - in the end of a process we are at some kind of equilibrium].

 

[Soumalya]

For an irreversible process, you can't assume that the work can be obtained by integrating over a quasistatic sequence of equilibrium states. This is only correct to do for a reversible process.

Chet

Hi friend,
You are correct!But I already mentioned about the reversible path when I say quasi static process nevertheless I was speaking solely about reversible work.All quasi static processes are reversible.

For an irreversible process we cannot use integration.

 

[Chestermiller]

Hi friend,
You are correct!But I already mentioned about the reversible path when I say quasi static process nevertheless I was speaking solely about reversible work.All quasi static processes are reversible.

For an irreversible process we cannot use integration.

All quasi static processes are not reversible, because you can still have irreversible heat transfer (involving temperature gradients) occurring within the system. Just consider the transient heating of a solid by conduction. The solid is not deforming, so the process is quasi static, but it definitely is not reversible.

Chet

 

[ffia]

Chestermiller: So, do you know whether the statement holds?
And moreover, why does/doesn't it?
["He is saying that, even if a (closed) system experiences an irreversible process, it is possible to identify one or more reversible paths for the surroundings that imposes the exact same irreversible path on the system."]