Don't laugh. You and I both know that you actually will be thinking about it, even on the beaches of Barbados.Robert Davidson said: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).
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.Robert Davidson said: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 said:... 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:
$$dQ=\frac{F}{A}dV\tag{5}$$
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.Philip Koeck said: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?
##\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).Philip Koeck said:I'm wondering about two things in your derivation:
Where does the last step of equation 3 come from?
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).Equation 2 should be dU = dQ - dW in general.
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}##.If you write PdV instead of dW, are you not assuming that the expansion is reversible, if P stands for the system's pressure?
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.Chestermiller said: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.
Meaning they would evaporate due to extremely high temperature ?Philip Koeck said:If the cylinder and piston have zero heat capacity
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 said: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?
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!BvU said: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.
Quasistatic does not mean that cumulative friction heat produced is zero.Philip Koeck said:Quasistatic expansion, so the friction heat produced per second is zero.
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 said: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.
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.Chestermiller said: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 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 said: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.
Back again after a period lacking time (strange as it sounds).Chestermiller said: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.
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 said: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?
I think I have it. I can't use dS = dQrev / T since I can't come up withChestermiller said: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 hate to say it, but most of this is incorrect.Philip Koeck said:I think I have it. I can't use dS = dQrev / 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 have to gain some clarity about this.Chestermiller said: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:
$$\Delta S=nC_p\ln{(T_f/T_i)}-nR\ln{(P_f/P_i)}$$
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?Philip Koeck said: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.
Entropy is a function of state, which means that, like internal energy, specific volume, heat capacity, etc., it is a property of the material that is being processed, and not a function of any process that the material is subjected to. Eqns. 8 in the Insights article (and the equation given in post #7) is the general equation for the entropy change of an ideal gas between any two thermodynamic states. So it applies to any and all changes of an ideal gas.Philip Koeck said:I've read your text from post 57 now, and I'm especially interested in the second example and equation (8) which you derive for adiabatic process.
It looks to me like this equation could be applicable to a whole class of processes:
For an isochoric process (of an ideal gas or anything else) the second term is zero and you get only the entropy change due to the temperature change.
For a reversible isothermal process or a free expansion the first term is zero.
For an adiabatic process (with and without friction, reversible or irreversible) both terms differ from zero, but for a reversible adiabatic expansion they must add up to zero.
Is equation (8) always correct, would you say, or are there processes that can be described by changes in V, P and T, but don't obey this equation?
The other thing I'd be interested in is to get from my final equation in post 61 to your equation (8) or vice versa. I haven't had time to try that yet.
Chestermiller said:Hi Bob.
First of all, WELCOME TO PHYSICS FORUMS!
Secondly, if you want to edit a previous post, go to the bottom of the page and hit the edit button. This will reopen the editing window. Afterwards, you can hit the SAVE EDITS button. If you want to reply to a post in sections, you hit the reply button at the bottom of the post.
I had some trouble following the details of your analysis, so I am going to present my own short development. I think this will answer most of your questions.
As in your analysis, everything starts with the FORCE BALANCE ON THE PISTON:
$$PA=mg+F+f$$or equivalently
$$P=\frac{mg}{A}+\frac{F}{A}+\frac{f}{A}$$
And, 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:
$$dQ=\frac{F}{A}dV\tag{5}$$
FIRST LAW OF THERMODYNAMICS APPLIED TO THE PISTON ALONE:
We can obtain the equation for the first law applied to the piston alone by subtracting Eqn. 2 from Eqn. 3 to yield:
$$\frac{mg}{A}dV=-dQ-\left(\frac{f}{A}-P\right)dV\tag{6}$$
The left hand side is the change in potential energy of the piston. Again, dQ is the heat transferred from the piston to the gas. The work done by the piston on its surroundings is ##\left(\frac{f}{A}-P\right)dV##. Note that the piston does not do any work on the cylinder, since the displacement of the cylinder is zero.
FURTHER DISCUSSION
For an ideal gas, Eqn. 4 becomes:
$$nC_vdT=\frac{F}{A}dV-\frac{nRT}{V}dV\tag{7}$$This equation does not have a simple analytic solution that I am aware of, and it would probably have to be integrated numerically. In any event, from this equation, it follows from this equation that the change in entropy of the gas is obtained from:
$$dS=\frac{F}{A}\frac{dV}{T}$$
If the system is taken as the combination of gas and piston, the external pressure of the surroundings to this system is f/A.anphan96 said:Hi Chet,
In the equation (3), why did you use -(f/A)dV for the work done by the system on the surrounding? Isn't it supposed to be the external pressure to the surrounding (that is the pressure of the system P) that matter?
What else would you include in the force balance if the piston is taken as massless? And what has irreversibility got to do with the force balance on a system.If this is the case, how do you know that the force balance is maintained throughout the process, given this is irreversible due to the friction?