Thermodynamics: Gas Expansion with Piston Friction

[Philip Koeck]

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 = f0/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.

 

[Chestermiller]

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.

 

[Philip Koeck]

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]

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.

 

[anphan96]

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

 

[Chestermiller]

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.