# Homework Help: Adiabatic piston expansion with friction

1. Jul 18, 2011

### MACH2

An ideal gas is contained in a piston-cylinder arrangement. The walls of the cylinder and the piston itself are "adiabatic", i.e. no heat energy can be exchanged between the piston-cylinder and its surroundings. The piston experiences friction as it moves. All the heat generated by the friction however is trapped inside the cylinder (and therfore it adds heat energy to the gas). By the firts law of theromodynamics the energy balance can be written in differential form as:

dU=dQ-dW

dU - change of the gase's internal energy,
dQ - heat energy added to the gas, (for adiabatic case dQ=0, however for the case presented here dQ would come from the friction of the piston with the cylinder wall)
dW - work done by gas.

Assuming that the inital vol, temp. and pressure are V0, T0 and P0 respectively, how could you express the relation petween pressure and volume of this system at any given point of the expansion?

For example, the relation between the pressure and volume in a similar scenario but WITHOUT friction is given by the well known thermodynamic equation of:

PV^k = Constant, where k=Cp/Cv

(Here Cp and Cv are the gas's molar heat capacity under constant pressure and constant volume respectively).

Last edited: Jul 18, 2011
2. Jul 18, 2011

### Philip Wood

The system you describe is adiabatic, so $\overline{d}$Q = 0.

Two remarks...

(1) The frictional work doesn't count as a heat input onto the system; it's not coming from outside.
(2) The symbol I've written as $\overline{d}$ is supposed to be d crossed through, as $\overline{d}$Q is not an exact differential of any function of state.

So dU = - $\overline{d}$W,

$\overline{d}$W is work done on the outside world by the piston, so-called external work.

However $\overline{d}$W is less than PdV, where P is the gas pressure. This because as the piston moves it does work against friction, as well as external work. If the piston moves out a distance dx the work it does against a frictional force F is Fdx = F(dV/A) = (F/A)dV, in which A is the piston area.

So dU = - {P - (F/A)}dV

Provided F is constant, you can now proceed in a very similar way to the frictionless case. I finished up with (P - P0) V$\gamma$ = constant, in which P0 = (R/cp)(F/A), but I may well have made slips.

Last edited: Jul 18, 2011
3. Jul 18, 2011

### MACH2

dU = - {P - (F/A)}dV,
multiply out
dU = -PdV + (F/A)dV, (F/A is constant)

By the ideal gas law PV = nRT or P= nRT/V (n=number of moles of gas)

Replace "P" in equation for "dU"
dU = -(nRT/V)dV + (F/A)dV

But dU = (n Cv) dT (gas' internal energy depends on temperature only)

Replace "dU" in equation
(n Cv) dT = -(nRT/V)dV + (F/A)dV

devide both sides by dV
(n Cv) dT/dV = -(nRT/V) + (F/A)

now divide both sides by (n Cv)
dT/dV = -(nRT/V)/(n Cv) + (F/A)/(n Cv)
dT/dV = -(nR)/(n Cv) (T/V) + (F/A)/(n Cv)

simplify an re-arrange
dT/dV = -R/Cv (T/V) + F/(n Cv A) where F/(n Cv A) is a constant

Now solve this differential equation to get a relation between T and V, i.e. T=T(V) then use the ideal gas law to get a relation between P and V i.e. P=P(V)

Last edited: Jul 18, 2011
4. Jul 19, 2011

### Philip Wood

So all is well, now?

It's worth pointing out that this is a good example of an irreversible change. If we push the piston inwards, compressing the gas, the frictional force changes its direction, because it always opposes relative motion (between itself and the cylinder). This means that

dU = - {P + (F/A)}dV

Note the change of sign of the frictional term.

5. Jul 19, 2011

### MACH2

Yes friction will produce irreversibilities. However solving the diff. eq.

dT/dV = -R/Cv (T/V) + F/(n Cv A)

and then obtaining the relation P=P(V) as I explained before will produce results which,
I think are inconsistent.

6. Jul 19, 2011

### Philip Wood

I've re-read your posts, and can't see any previous reference to an inconsistency. Can you explain more fully what the problem is?

7. Jul 21, 2011

### Philip Wood

Still don't know what your difficulty is, but....

Bearing in mind that you want to find the relationship between P and V, it's less cumbersome to get rid of T at an early stage, by writing

dU = ncvdT = ncvd(pV/nR) = ncv(1/nR)d(pV) = ncv(1/nR)(pdV + Vdp) = (cv/R)(pdV + Vdp)

Now you have a DE in just p and V. Solve it by separating variables, just as in the frictionless case.

Last edited: Jul 21, 2011
8. Jul 25, 2011

### MACH2

Indeed, getting rid of T an an earlier stage makes things more simple.

Last edited: Jul 26, 2011
9. Jul 27, 2011

### Philip Wood

In the post which appeared briefly you said that the DE could be solved, but that the result of integrating pdV to find the work done by the gas was problematic.

Firstly, did you agree with the solution of the DE that I gave in post 2, namely

(P - P0) Vγ = constant, in which P0 = (R/cp)(F/A) ?

This applies when the piston is moving out. The easiest way to solve the DE is probably by separating variables. If you need help here, shout!

Now the question about work... It's important to realise that the work [STRIKE]d[/STRIKE]w which features in the first law is the external work done by the system. For your irreversible system, in which the piston encounters friction, this external work is not pdV, but (P - P0)dV.

If you care to reveal more about your difficulty, I expect it can be resolved...

Last edited: Jul 27, 2011
10. Jul 28, 2011

### MACH2

I believe that the DE corresponding to this problem is:

p + V(dp/dV) = (R/Cv)(F/A) - (R/Cv)p, (where (R/Cv) and (F/A) are constants )

solving this will get you p=p(v), however the solution is a bit more complicated than
(P-P0)V^k = Const.

My mistake was in the limits if integration when calculating the work done the gas when going from initial volume V0 and final volume Vf
(V0<Vf). Since p=p(V), the final pressure pf (correponding to the final volume Vf) has to be greater or equal to F/A. Since at this point the piston will stop moving. Of course given pf=F/A you have to calculate from p=p(V) the corresponding Vf.

Integrating p(v)dV between V0 and Vf will give the maximum work done by the gas. Of course the work transmitted to the surroundings will be less due to friction.

Last edited: Jul 28, 2011
11. Jul 28, 2011

### Philip Wood

What's your solution to the DE, then?

[I agree with your DE, by the way.]

Last edited: Jul 28, 2011
12. Jul 28, 2011

### MACH2

DE is:

p + V(dp/dV) = (R/Cv)(F/A) - (R/Cv)p,

then the solution to the DE is (I hope no mistakes):

p = (R/Cp)(F/A) [1 - (V0/V)^k] - P0(V0/V)^k (where k=Cp/Cv)

Last edited: Jul 28, 2011
13. Jul 29, 2011

### MACH2

I think I managed to get the answer wrong in the last post.

I think the correct answer is (I hope this one is right):

p = (RF)/(kACv) - {(RF)/(kACv) - P0} (V0/V)^k

Last edited: Jul 29, 2011
14. Jul 29, 2011

### Philip Wood

Well done, Mach2 ! I thought you might not rise to the challenge.

Well done, too, because your answer is exactly the same as mine (which I'd checked by back-substitution into the DE). My answer looks a lot simpler, though!

To see the equivalence, start by noting that RF/kAcv = RF/cpA, which is what I called 'p0'. I'll now call it 'pc' instead, leaving p0 to mean initial pressure, that is I'll (graciously!) come into line with your notation. But instead of your k, I'll stick with $\gamma$, as this is absolutely standard notation for cp/cv. So your solution now becomes

p = pc - {pc - p0} (V0/V)$\gamma$

Using a line of algebra this can be re-arranged as

(p - pc)V$\gamma$ = {p0 - pc} (V0)$\gamma$

In other words, (p - pc)V$\gamma$ = constant.

which was my solution, as given in post 2.

Best wishes.

15. Jul 29, 2011

### MACH2

Very good Philip! Thanks for your interest.

Last edited: Jul 29, 2011