Thermodynamics: Gas Expansion with Piston Friction

[Chestermiller]

Chet,

Thanks for the analysis.

Regarding the new example, I had in mind the following:

System 1: massless piston with friction between the piston and cylinder walls, and a pile of small masses sitting on top of the piston, such that the total downward force due to the masses plus the downward force due to atmospheric pressure equals the upward force due to the gas pressure. Or, the pressure upward exerted by the gas equals the pressure downward exerted by the weights.

System 2: Same as system 1 except there is no friction.

An equal amount of weight is abruptly removed from each system so that the external downward pressure exerted by the weights plus the atmosphere is ½ the original external downward pressure.

To solve: What will be the final volume of the two systems?

If we can agree on the problem statement, I will give it a try on my own, with the benefit of the analysis you did on our previous example, and then post it.

Bob

Bob,

Although I don't think that the two cases are really going to be comparable, I think each of them is interesting on its own. So please proceed and let's see what you come up with. (Please note that, in both these cases, the gas experiences a non-quasistatic deformation, and entropy is generated within the gas in each case).

Chet

 

[Chestermiller]

I've been able to analytically solve the differential equation (Eqn. 6) of post #29 for the dimensionless temperature of the gas as a function of the dimensionless volume ratio. The result, which agrees with the numerical solution I presented in post #20 is:
$$\bar{T}=\frac{(1-\xi)}{\bar{V}^{(\gamma-1)}}+\xi \bar{V}$$where $$\xi=\frac{(\gamma-1)}{\gamma}\frac{F}{P_0A}$$

 

[Robert Davidson]

Hi Chet,

Haven’t forgot this. So far only tackled the easy case (system 2). Here’s what I have so far.

Going to assume dealing with ideal gas (air) with the following initial conditions and properties:

$P_{i}=10atm$, $T_{i}=300K$, $n=1$, $C_{v}=0.2053\frac{l.atm}{mol.K}$, $R_{u}=0.08205\frac{l.atm}{mol.K}$, $k=1.4$

Applying the ideal gas equation to get the initial volume:

$$V_{i}=\frac{nRT_{i}}{P_{i}}=\frac{(1)(0.08205)(300)}{10}=2.46 l$$

Weight is abruptly removed to reduce the external pressure by one half, or to 5 atm, and the gas allowed to expand non-quasi-statically at constant pressure until equilibrium is reached.

Since the final pressure is known, we can (1) determine the relationship between the final temperature and volume by applying the general gas equation to the initial and final states and (2) substitute the relationship into the first law equation for a constant pressure adiabatic process.

Applying the general gas equation:

$$\frac{P_{i}V_{i}}{T_{i}}= \frac{P_{f}V_{f}}{T_{f}}$$

$$T_{f}=\frac{P_{f}V_{f}T_{i}}{P_{i}V_{i}}$$

$$T_{f}=\frac{(5)(V_{f})(300)}{(10)(2.46)}=61V_{f}$$

Applying the first law for an adiabatic process:

$$\Delta U=-W$$

$$nC_{v}(T_{f}-T_{i})=-P_{f}(V_{f}-V_{i})$$

Substituting $T_{f}=61V_{f}$ and the other known values

$$(1)(0.2053)[61V_{f}-300]=-5(V_{f}-2.46)$$

Which gives us $V_{f}=4.22l$ and $T_{f}=257K$, and the work done is

$$W=5(4.22-2.46)l.atm=892j$$

Comparing to the final volume and work done for a reversible adiabatic expansion with the same initial conditions and same final pressures:

$$P_{i}V_{i}^{k}= P_{f}V_{f}^{k}$$

$$V_{f}=6l$$

$$W_{rev}=\frac{(P_{f}V_{f}-P_{i}V_{i})}{1-k}$$

$$W_{rev}= 13.5 l.atm=1368j$$

Which, as expected, is greater than the irreversible work.

The next step is to tackle system 1, which adds piston friction. I would like to bring system 2 to the same final pressure and then compare the work on the surroundings to system 2. I recall that in your analysis of the quasi-static expansion with friction it was necessary to apply an external tensile force to overcome static friction in order to reach the same volume as the non-friction system. It appears that will again be the case here, except it would be in order to reach the same final pressure.

I was thinking about a possible alternative approach, such as assuming the surface of the cylinder at some point changes to a smooth service prior to reaching the maximum static friction force so that the expansion could be completed, albeit without friction work done at the end. But that would probably create more complications than it's worth.

 

[Chestermiller]

I approached this problem differently but, of course, arrived at the same results. My focus was on the final temperature, rather than the final volume.
$$nC_v(T_f-T_i)=-P_f(V_f-V_i)=P_f\left(\frac{nRT_f}{P_f}-\frac{nRT_i}{P_i}\right)$$
Solving for $T_f$ then yields:
$$T_f=\left[\frac{1}{\gamma}+\left(1-\frac{1}{\gamma}\right)\frac{P_f}{P_i}\right]T_i$$For $\gamma=1.4$ and $P_f/P_i=0.5$, this yields $T_f=0.8571T_i$. And, for $T_i=300\ K$, $T_f=257.1K$.

The work is given by: $$W=-\Delta U=-nC_v(T_f-T_i)=-(1)(2.5)(8.314)(257.1-300)=892\ J$$
It'll be interesting to compare the approach and results you get for system 2 with those that I obtained.

 

[Chestermiller]

Here's an interesting paradox for you. As $P_f/P_i\rightarrow 0$, $$\frac{T_f}{T_i}\rightarrow\frac{1}{\gamma}$$and then $$W\rightarrow nC_vT_i\left(1-\frac{1}{\gamma}\right)>0$$But how can this be if $P_f/P_i\rightarrow 0$ implies free expansion.

 

[Robert Davidson]

Chet
Interesting. Going away on a two week cruise so won't get back to this till I get back. But I'll be thinking about it on the beaches of Barbados! (Ha Ha).

 

[Chestermiller]

Chet
Interesting. Going away on a two week cruise so won't get back to this till I get back. But I'll be thinking about it on the beaches of Barbados! (Ha Ha).

Don't laugh. You and I both know that you actually will be thinking about it, even on the beaches of Barbados.

 

[Chestermiller]

To complete this thread, here is the beginning of my solution for system 2 in which there is a friction force F.

Equation for the initial force balance on the piston: $$P_{gi}A-F-P_{atm,i}A=0$$where A is the piston cross sectional area, F is the friction force, $P_{gi}$ is the initial equilibrium gas pressure, and $P_{atm,i}$ is the initial atmospheric pressure.

Equation for the force balance on the piston during the expansion: $$F_g-F-\frac{P_{atm,i}}{2}A=0$$
where $F_g$ is the force that the gas exerts on the inside piston face during this irreversible expansion.

 

[Robert Davidson]

Hi Chester,

Finally getting back to this.

Thanks for giving the beginning of the solution for system 2. Before I go too far down the road, I need to make sure I’m on the same page with you.

I have no problem with the force balance during expansion (your second equation). I am having difficulties with the force balance for the initial state (the first equation).

I would like to see the initial gas pressure exactly equal to the externally applied pressure (atmosphere plus weights= 10 atm) as we did for system 1. Since there would be no net externally applied force, there would be no fiction force, $F$, to oppose it. In other words, $F=0$ initially.

Once the externally applied force is halved, the difference between the force of the gas and the externally applied force will be the fiction force $F$, as presented in the second equation (with external pressure = 5 atm).

Do you see any issues with this approach?

 

[Chestermiller]

Hi Chester,

Finally getting back to this.

Thanks for giving the beginning of the solution for system 2. Before I go too far down the road, I need to make sure I’m on the same page with you.

I have no problem with the force balance during expansion (your second equation). I am having difficulties with the force balance for the initial state (the first equation).

I would like to see the initial gas pressure exactly equal to the externally applied pressure (atmosphere plus weights= 10 atm) as we did for system 1. Since there would be no net externally applied force, there would be no fiction force, $F$, to oppose it. In other words, $F=0$ initially.

Once the externally applied force is halved, the difference between the force of the gas and the externally applied force will be the fiction force $F$, as presented in the second equation (with external pressure = 5 atm).

Do you see any issues with this approach?

I find this kind of confusing. We are saying that, initially, there is no friction force, but, as soon as we drop the external pressure, friction kicks in at full force during the entire time that the piston is sliding. And it's still present at the very end. I guess this would be OK, under our assumption that the coefficients of static and kinetic friction are equal. But, during the expansion, the friction force would have to be less than 5 atm times the piston area, or the gas would not be able to expand. OK. Let's see where that takes us.

Chet

 

[Philip Koeck]

Hi all.
I'm switching over from the thread https://www.physicsforums.com/threa...deal-gas-be-irreversible.971054/#post-6172432 to join this discussion.
I'm wondering how certain processes can become irreversible. In the other thread we discussed isothermal processes, but adiabatic processes are just as mysterious to me.
I haven't gone through all the previous posts so I'm entering the discussion somewhat unprepared, but maybe with a new angle.

If I take system 1 depicted in post #6 of this thread and assume a quasistatic, adiabatic expansion of an ideal gas.
Without friction, ΔU of the system will be equal to W done by the system. Q is zero and the process is reversible.

Now I'll assume that the cylinder and piston have zero heat capacity and that the thermal insulation is such that no heat can pass into the environment.
If there's friction between piston and cylinder, the gas has to do extra work to overcome this.
This extra work will be converted directly into heat which goes back into the gas.
I get the impression that nothing has changed compared to the frictionless case.

Does that mean that in this case friction doesn't make the process irreversible?

 

[Philip Koeck]

... if there is a differential change in volume dV, we have
$$PdV=\frac{mg}{A}dV+\frac{F}{A}dV+\frac{f}{A}dV\tag{1}$$

FIRST LAW OF THERMODYNAMICS APPLIED TO SYSTEM 1, THE GAS:
$$dU=dQ-PdV\tag{2}$$
where dQ is the differential heat transferred from the piston to the gas.

FIRST LAW OF THERMODYNAMICS APPLIED TO SYSTEM 2, THE GAS AND PISTON:
In this case, we need to use the more general form of the first law which includes the change in potential energy of whatever is present with the boundaries of the system (in this case, the piston).
$$dE=dU+d(PE)=dU+\frac{mg}{A}dV=-\frac{f}{A}dV\tag{3}$$where $\frac{F}{A}dV$ is the work done by system 2 on its surroundings. Note that, in this case, there is no heat transfer because the combined system is adiabatic. If we substitute Eqns. 1 into this relationship, we obtain:
$$dU=\frac{F}{A}dV-PdV\tag{4}$$From Eqns. 2 and 4, it follows that the heat transferred from the piston to the gas dQ is given by:
$$dQ=\frac{F}{A}dV\tag{5}$$

I'm wondering about two things in your derivation:

Where does the last step of equation 3 come from?

Equation 2 should be dU = dQ - dW in general.

If you write PdV instead of dW, are you not assuming that the expansion is reversible, if P stands for the system's pressure?

 

[Chestermiller]

Hi all.
I'm switching over from the thread https://www.physicsforums.com/threa...deal-gas-be-irreversible.971054/#post-6172432 to join this discussion.
I'm wondering how certain processes can become irreversible. In the other thread we discussed isothermal processes, but adiabatic processes are just as mysterious to me.
I haven't gone through all the previous posts so I'm entering the discussion somewhat unprepared, but maybe with a new angle.

If I take system 1 depicted in post #6 of this thread and assume a quasistatic, adiabatic expansion of an ideal gas.
Without friction, ΔU of the system will be equal to W done by the system. Q is zero and the process is reversible.

Now I'll assume that the cylinder and piston have zero heat capacity and that the thermal insulation is such that no heat can pass into the environment.
If there's friction between piston and cylinder, the gas has to do extra work to overcome this.
This extra work will be converted directly into heat which goes back into the gas.
I get the impression that nothing has changed compared to the frictionless case.

Does that mean that in this case friction doesn't make the process irreversible?

No. It is reversible for the gas, but not for the combination of gas plus piston. The system and surroundings cannot be restored to the original state without causing a change in something else. Please read the previous posts to come up to speed on this.

 

[Chestermiller]

I'm wondering about two things in your derivation:

Where does the last step of equation 3 come from?

$\frac{f}{A}dV$ is the work done by the combination of piston plus gas (the system) on their surroundings (at the outside face of the piston).

Equation 2 should be dU = dQ - dW in general.

No. The general form of the first law of thermodynamics also includes change in the potential energy and kinetic energy of the system: $$\Delta U+\Delta (PE)+\Delta (KE)=Q-W$$The potential energy of the piston changes (case 2).

If you write PdV instead of dW, are you not assuming that the expansion is reversible, if P stands for the system's pressure?

No. The force per unit area at the interface between the system and surroundings (where work is being done by the system, the gas plus piston) is $\frac{f}{A}$.

 

[Philip Koeck]

No. It is reversible for the gas, but not for the combination of gas plus piston. The system and surroundings cannot be restored to the original state without causing a change in something else. Please read the previous posts to come up to speed on this.

I specified that the cylinder and piston have zero heat capacity. To keep things really simple I would also give the piston zero mass or orient the cylinder horizontally so that the piston doesn't change its potential energy.
I distinguish between two situations: In the first all friction heat goes back into gas, in the second all friction heat goes into the surroundings.

If the cylinder and piston have zero heat capacity they can't receive heat and they can't have any inner energy so they shouldn't be part of the thermodynamic state of either the system (gas) or the surroundings.

In the first case the friction heat goes back into the gas. To me that looks like the first case is indistinguishable from the frictionless case and should be reversible. That could be an example of reversibility despite friction.

In the second case all the friction heat goes into the surroundings and the entropy of the surroundings increases. The system behaves just like in the friction-less case. In sum that should mean that the second case is irreversible.

Could it be that whether or not a particular process is irreversible depends only on where the friction heat goes?

 

[BvU]

If the cylinder and piston have zero heat capacity

Meaning they would evaporate due to extremely high temperature ?
If there is friction there is material. If there is material there is non zero heat capacity.

 

[Chestermiller]

I specified that the cylinder and piston have zero heat capacity. To keep things really simple I would also give the piston zero mass or orient the cylinder horizontally so that the piston doesn't change its potential energy.
I distinguish between two situations: In the first all friction heat goes back into gas, in the second all friction heat goes into the surroundings.

If the cylinder and piston have zero heat capacity they can't receive heat and they can't have any inner energy so they shouldn't be part of the thermodynamic state of either the system (gas) or the surroundings.

In the first case the friction heat goes back into the gas. To me that looks like the first case is indistinguishable from the frictionless case and should be reversible. That could be an example of reversibility despite friction.

In the second case all the friction heat goes into the surroundings and the entropy of the surroundings increases. The system behaves just like in the friction-less case. In sum that should mean that the second case is irreversible.

Could it be that whether or not a particular process is irreversible depends only on where the friction heat goes?

In the first case where the frictional heat goes into the gas, the entropy of the gas does increase, and thus, the entropy of the system and surroundings increases. Such a process is, of course, irreversible. If there were no friction, then the entropy of the gas would not increase (but this is a different process, and it has a different final state).

 

[Philip Koeck]

Meaning they would evaporate due to extremely high temperature ?
If there is friction there is material. If there is material there is non zero heat capacity.

It's a thought experiment! Perfect thermal contact with gas, so the cylinder and piston have the same temperature as the gas. Quasistatic expansion, so the friction heat produced per second is zero. Very exotic material! Just don't try to do it in practice!
I completely understand your concerns. I'm just simplifying the description, hopefully not too much.

 

[Chestermiller]

Quasistatic expansion, so the friction heat produced per second is zero.

Quasistatic does not mean that cumulative friction heat produced is zero.

 

[Philip Koeck]

I'll make another attempt, this time with equations.
I'll treat only the ideal gas as system and assume a quasistatic, adiabatic expansion.
I'll assume that the cylinder and piston have negligible mass and heat capacity and that the thermal insulation is such that no friction heat can pass into the environment.
I'll discuss two cases, one without friction and one with friction between cylinder and piston.
I'll use the same letters for the forces as in post #8.

The first law for the gas (the system) alone is
dU = dQ - dW

Frictionless case:
dQ = 0
dW = f/A dV
Therefore: dU = - f/A dV

Case with friction:
Here all the friction work is converted to heat that goes into the gas!
dQ = F/A dV
dW = F/A dV + f/A dV
Therefore: dU = - f/A dV

If I assume that the initial state of the system was the same in both cases, then the final state must also be the same, since U changed by the same amount. Therefore dS for the system is the same in both cases.

In the frictionless case dS = 0 for both system and environment.

In the case with friction dS = 0 for the system because of the reasons given above and for the surroundings it's also 0 because no heat flows into them.

I conclude that the case with friction is also reversible.

 

[Chestermiller]

I'll make another attempt, this time with equations.
I'll treat only the ideal gas as system and assume a quasistatic, adiabatic expansion.
I'll assume that the cylinder and piston have negligible mass and heat capacity and that the thermal insulation is such that no friction heat can pass into the environment.
I'll discuss two cases, one without friction and one with friction between cylinder and piston.
I'll use the same letters for the forces as in post #8.

The first law for the gas (the system) alone is
dU = dQ - dW

Frictionless case:
dQ = 0
dW = f/A dV
Therefore: dU = - f/A dV

Case with friction:
Here all the friction work is converted to heat that goes into the gas!
dQ = F/A dV
dW = F/A dV + f/A dV
Therefore: dU = - f/A dV

If I assume that the initial state of the system was the same in both cases, then the final state must also be the same, since U changed by the same amount. Therefore dS for the system is the same in both cases.

In the frictionless case dS = 0 for both system and environment.

In the case with friction dS = 0 for the system because of the reasons given above and for the surroundings it's also 0 because no heat flows into them.

I conclude that the case with friction is also reversible.

f is different in the frictionless case from the case with friction. In the frictionless case, $$f=PA$$ and $$nC_vdT=-PdV$$In the case with friction, $$f=PA-F$$ and $$nC_vdT=-PdV+\frac{F}{A}dV$$So, for the same change in volume, the temperature change is different.

 

[Philip Koeck]

f is different in the frictionless case from the case with friction. In the frictionless case, $$f=PA$$ and $$nC_vdT=-PdV$$In the case with friction, $$f=PA-F$$ and $$nC_vdT=-PdV+\frac{F}{A}dV$$So, for the same change in volume, the temperature change is different.

That's right. I missed that completely. Obviously f has to be different so that both cases can have the same P and still be quasistatic.

Maybe one should say that piston and cylinder are part of the system in my example, since the thermal insulation is outside both of them.

What I'm wondering now is, whether the case with friction is still an adiabatic process.
There's no heat transfer between environment and system, but inside the system work is converted to heat.

 

[Chestermiller]

That's right. I missed that completely. Obviously f has to be different so that both cases can have the same P and still be quasistatic.

Maybe one should say that piston and cylinder are part of the system in my example, since the thermal insulation is outside both of them.

What I'm wondering now is, whether the case with friction is still an adiabatic process.
There's no heat transfer between environment and system, but inside the system work is converted to heat.

That depends on what you define as your system (a decision that is totally at the discretion of the analyst). If you define your system as the combination of gas plus piston, then this system experiences an adiabatic process. If you define your system as the gas only, then this system experiences a non-adiabatic process, since it receives heat from the piston. The piston alone also experiences a non-adiabatic process, since net work is done on it and an equal amount of heat is discharged to the gas.

 

[Philip Koeck]

That depends on what you define as your system (a decision that is totally at the discretion of the analyst). If you define your system as the combination of gas plus piston, then this system experiences an adiabatic process. If you define your system as the gas only, then this system experiences a non-adiabatic process, since it receives heat from the piston. The piston alone also experiences a non-adiabatic process, since net work is done on it and an equal amount of heat is discharged to the gas.

Back again after a period lacking time (strange as it sounds).

If we include the piston and cylinder into the system and we have friction between piston and column and there is perfect insulation from the environment (anything that's not piston, column or gas), that would mean we have an adiabatic, quasistatic expansion.
Would that not have to be reversible?
A reversible process with friction?

 

[Chestermiller]

Back again after a period lacking time (strange as it sounds).

If we include the piston and cylinder into the system and we have friction between piston and column and there is perfect insulation from the environment (anything that's not piston, column or gas), that would mean we have an adiabatic, quasistatic expansion.
Would that not have to be reversible?
A reversible process with friction?

No way. The presence of friction makes it irreversible. Just calculate the change in entropy of the system for this adiabatic process and see what you get.

 

[Philip Koeck]

No way. The presence of friction makes it irreversible. Just calculate the change in entropy of the system for this adiabatic process and see what you get.

I think I have it. I can't use dS = dQ_rev / T since I can't come up with
a definitely reversible process to replace the expansion with friction so I use T dS = dU + P dV.

Without friction the force balance gives P dV = f₀/A dV as in post 51.
f₀ is the external force needed when there's no friction.
With dQ = 0 I get dU = - P dV and thus dS = 0

With friction the force balance gives
P dV = f₁/A dV + F/A dV.
f₁ is the external force needed when there is friction and is equal to f₀ - F.

dQ = F/A dV
and dU = F/A dV - P dV
and therefore T dS = F/A dV.

Does that make sense?

I think what bothered me was that you wrote that the process is adiabatic if the piston and cylinder are regarded as part of the system. For me an adiabatic process has constant entropy.

Also I wouldn't call it adiabatic if work is converted to heat within the system, even if no work is transferred between surroundings and system. If dQ is not 0 I wouldn't call the process adiabatic.

To me it seems that friction makes the process irreversible and non-adiabatic.

 

[Chestermiller]

I think I have it. I can't use dS = dQ_rev / T since I can't come up with
a definitely reversible process to replace the expansion with friction so I use T dS = dU + P dV.

Without friction the force balance gives P dV = f₀/A dV as in post 51.
f₀ is the external force needed when there's no friction.
With dQ = 0 I get dU = - P dV and thus dS = 0

With friction the force balance gives
P dV = f₁/A dV + F/A dV.
f₁ is the external force needed when there is friction and is equal to
f₀ - F.

dQ = F/A dV

and dU =
F/A dV - P dV

and therefore T dS =
F/A dV.

Does that make sense?

I think what bothered me was that you wrote that the process is adiabatic if the piston and cylinder are regarded as part of the system. For me an adiabatic process has constant entropy.

Also I wouldn't call it adiabatic if work is converted to heat within the system, even if no work is transferred between surroundings and system. If dQ ≠ 0 I wouldn't call the process adiabatic.

To me it seems that friction makes the process irreversible and non-adiabatic.

I hate to say it, but most of this is incorrect.

An adiabatic process as one in which there is no heat transfer between the system and its surroundings, irrespective of whether the process is reversible or irreversible. An adiabatic reversible process has constant entropy.

If the system is regarded as the piston plus gas (plus cylinder), then the change in entropy of the system is equal to the sum of the entropy changes for the gas, the piston, and the cylinder. The cylinder is regarded as insulated (from the gas, the piston, and the surroundings), so its entropy doesn't change. The piston is regard as having negligible heat capacity, so its entropy doesn't change either. So the entropy change for the combined system is just equal to the entropy change of the gas.

Since entropy is a function of state, you do not have to carry out a reversible process on the gas in this exact apparatus involving piston friction to get its entropy change. And the reversible process does not even have to be adiabatic. You can remove the gas, put it into a different cylinder, and allow it to expand reversibly against a frictionless piston, with heat transfer also permitted, to evaluate the entropy change. Since we are dealing with an ideal gas, if you follow this procedure, you will find that the change in entropy of the gas will be given by:
$$\Delta S=nC_p\ln{(T_f/T_i)}-nR\ln{(P_f/P_i)}$$
This result will be the same no matter what reversible path you subject the gas to.

To get a better understanding of how to determine the entropy change for a closed system that has experienced an irreversible process, please see my cookbook tutorial:
https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

 

[Philip Koeck]

Thanks. I'll read.

 

[Philip Koeck]

I hate to say it, but most of this is incorrect.

Since entropy is a function of state, you do not have to carry out a reversible process on the gas in this exact apparatus involving piston friction to get its entropy change. And the reversible process does not even have to be adiabatic. You can remove the gas, put it into a different cylinder, and allow it to expand reversibly against a frictionless piston, with heat transfer also permitted, to evaluate the entropy change. Since we are dealing with an ideal gas, if you follow this procedure, you will find that the change in entropy of the gas will be given by:
$$\Delta S=nC_p\ln{(T_f/T_i)}-nR\ln{(P_f/P_i)}$$

I have to gain some clarity about this.
The equation I use (T dS = dU + P dV) is just the fundamental equation and it's valid for all processes, reversible and irreversible, since it only contains state functions.
If I have the correct expression for dU I should therefore get the correct dS.
Now I took the dU for the cases with and without friction from post 51, so we should agree on that.
P dV is the same for both cases since I'm assuming that the external force is adapted so that f₀ = P A in the frictionless case and
f₁ + F = P A in the case with friction (just as you did, I thought).
What could be wrong with this calculation?
Maybe my result for the case with friction (T dS = F/A dV) is the same as what you write in the quoted text.

 

[Chestermiller]

I have to gain some clarity about this.
The equation I use (T dS = dU + P dV) is just the fundamental equation and it's valid for all processes, reversible and irreversible, since it only contains state functions.
If I have the correct expression for dU I should therefore get the correct dS.
Now I took the dU for the cases with and without friction from post 51, so we should agree on that.
P dV is the same for both cases since I'm assuming that the external force is adapted so that f₀ = P A in the frictionless case and
f₁ + F = P A in the case with friction (just as you did, I thought).
What could be wrong with this calculation?
Maybe my result for the case with friction (T dS = F/A dV) is the same as what you write in the quoted text.

The pressure is not the same in both cases because the gas temperature is higher in the case with friction. Have you tried solving this problem using the Recipe in my insights article?