# Thermodynamics: Gas Expansion with Piston Friction

Mentor
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.

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.

Mentor
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.

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?

Mentor
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.

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 = 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 is not 0 I wouldn't call the process adiabatic.

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

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Mentor
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 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/

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 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.

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Mentor
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?

I haven't tried your methods yet, but I'll give it a try soon.

Here's another go at my method:

The fundamental relation: T dS = dU + P dV.
I use subscript 0 for the frictionless case and 1 for the case with friction.

Without friction the force balance gives P0 dV = f
0/A dV as in post 51.
f0 is the external force needed when there's no friction.

With dQ = 0, I get dU = - P0 dV and thus dS = 0

With friction the force balance gives
P1 dV = f1/A dV + F/A dV.

f1 is the external force needed when there is friction and P1 is the gas pressure.
Both T and P differ from the friction-less case.

dQ = F/A dV
and dU =
F/A dV - P1 dV
and therefore T1 dS = F/A dV.

The next step would be to work out T as a function of V for the case with friction and integrate.

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Mentor
Well, from the equation for an ideal gas, since the gas is experiencing an "internally reversible" expansion in this process, $$dS=nC_v\frac{dT_1}{T_1}+nR\frac{dV}{V}$$So, $$T_1dS=nC_vdT_1+\frac{nRT_1}{V}dV=nC_vdT_1+P_1dV$$Substituting into your equation then gives:$$nC_vdT_1=\frac{F}{A}dV-PdV=-\frac{f_1}{A}dV$$which is certainly correct.

So I stand corrected. Your equation does apply in this situation (and will yield a result identical to the standard ideal gas result I gave), although using it will require a more complicated integration.

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.

Mentor
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.
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.

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:
$$dS=\frac{F}{A}\frac{dV}{T}$$
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?

Or is it because f = (P-F-mg)? 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?

Thanks,
An

Mentor
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?
If the system is taken as the combination of gas and piston, the external pressure of the surroundings to this system is f/A.
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?
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.