Work during free expansion of a gas

[FoxBox]

Hello,

I have a small problem concerning the free expansion of a gas.
https://www.physicsforums.com/attachment.php?attachmentid=20388&stc=1&d=1252066499

The two chambers are separated by a movable piston (without friction). Obviously the gas will expand because it puts pressure p on the piston. It will move the piston against the right chamberwall.

I think the gas has done work: W=p*delta(V), but apparently it hasn't: Chemical principles by Steven S. Zumdahl page 418.

What's my mistake??

-Greetings from FoxBox

 

[Doc Al]

How much pressure must the gas apply to move the piston and compress the vacuum? How much work is required to move the (presumably massless) piston?

 

[FoxBox]

(the piston is indeed massless) A is the area of the piston. So, the gas puts a force F of p*A to the piston. It moves the piston all the way(x) to the right wall. So it does work: W=p*A*s=p*delta(V)

But Zumdahl says no work is done. (Page 418: http://books.google.be/books?id=2Ox...e=gbs_v2_summary_r&cad=0#v=onepage&q=&f=false)

What's my mistake?

 

[Doc Al]

Your mistake is thinking that the gas exerts a force of p*A against the piston. You cannot exert a force on something that cannot push back. Usually "free expansion" is shown by unplugging a hole in the barrier and letting the gas leak out and expand into the vacuum (or by suddenly removing the barrier). Such a visualization might make it easier to understand that no force is exerted (or work done) in a free expansion.

 

[FoxBox]

But what is making the piston move then? Only a force can do that? It's the particular situation with the piston that i don't understand...

 

[Dadface]

But what is making the piston move then? Only a force can do that? It's the particular situation with the piston that i don't understand...

I think the author of the book has used a poor example.Since the piston is frictionless and massless we can say that in effect it does not exist at all.The situation then is similar to free expansion in a vacuum.

 

[FoxBox]

Yes, in fact, forces have to exert in pairs. Here, the piston CANNOT exert a force if it's frictionless... Even if it would have a mass, it wouldn't be able to exert a force, would it?

Thanks in advance...

 

[Doc Al]

But what is making the piston move then? Only a force can do that? It's the particular situation with the piston that i don't understand...

How much force is required to move a massless, frictionless piston?

As I said in my last post, letting the gas expand through a hole in the barrier (or suddenly removing the barrier) in vacuum is a much better way to illustrate free expansion. (Although it is technically equivalent, I agree with Dadface that the book chose a poor example.)

Even if it would have a mass, it wouldn't be able to exert a force, would it?

Sure you can exert a force against a massive piston, but then you get into all sorts of complications about transferring kinetic energy to the piston, all of which only serve to distract you from the basic point of free expansion.

 

[FoxBox]

Hello,

Thank you for your information. Much appreciated ;)

-FoxBox

 

[Count Iblis]

You can actually have a nonzero mass piston and/or friction between the piston and the walls in here. If you take the system to be the two chambers taken together, then the work performed by the system is not the work performed by the gas, because the external parameters if the system do not change (the position of the piston counts as an internal degree of freedom)

The total internal energy of the system says the same, because the system as a wole does not perform any "external" work and no heat is supplied to the system. This conclusion thus does not depend at all on whether or not the gas performed any "internal" pressure-volume work inside the system.

 

[RohansK]

I agree with the Free Exapnsion theory where the wall ( piston) is punctured or completely removed and the gas expands freely in the vacuum. This is what any Thermodynamics book illustrates for free Expansion.

But if we consider the gas part ( left chamber) as the system alone then can we say that work is done, as there is a force ( p.a) and a displacement ( s) as now the work can be called as ususal p(dv) work against the surounding ie the right chamber( vacuum).

 

[Doc Al]

You can actually have a nonzero mass piston and/or friction between the piston and the walls in here.

But I wouldn't call that free expansion.

If you take the system to be the two chambers taken together, then the work performed by the system is not the work performed by the gas, because the external parameters if the system do not change (the position of the piston counts as an internal degree of freedom)

The total internal energy of the system says the same, because the system as a wole does not perform any "external" work and no heat is supplied to the system. This conclusion thus does not depend at all on whether or not the gas performed any "internal" pressure-volume work inside the system.

The total internal energy of any isolated system stays the same. But if the piston is neither massless nor frictionless, then the expanding gas does work and thus the internal energy of the gas decreases.

 

[Count Iblis]

But I wouldn't call that free expansion.

Intuitvely it looks very differently, but it is exactly the same process. It is a good example to give to students when they ask questions like in the OP.

The total internal energy of any isolated system stays the same.

I agree, but this is the explanation why in the free expansion the internal energy stays the same. The "p=0" explanation that is sometimes given is not the correct argument. To be 100% sure, I looked it up in the book by F. Reif and he indeed does not give the "p=0 argument" and instead gives my argument: The whole system comprising of the gas and the enclosure does not perform any work and the whole system does not absorb heat, so the internal energy does not change.

So, what is wrong about the "p=0 argument"? Of course this has to do with the fact that the pressure is not defined the moment you let the gas escape. Reif does not mention this directly, but in a remark after the discussion he mentions that the free expansion is a complicated irreversible process involving turbulence and gross nonuniformities of pressure and temperature (to the extent that these quantities can be defined at all). And then he goes on to say that because the intitial state and the final state are states of thermal equilibrium, we can still use thermodynamics to compute the outcome of the process, we don't need to do detailed computations of what happens during the process.

But if the piston is neither massless nor frictionless, then the expanding gas does work and thus the internal energy of the gas decreases.

Yes, but then you have a non-equilibrium situation in the enclosure. E.g. due to friction the system has been heated in a nonuniform way. Then, if you wait until that heat is dissipated in the system, the internal energy is the same as it was. Now, you may object by saying that the heat capacity of the walls will now be a factor, however, that's always the case in the free expansion experiment and that's why Joule and Thomson proposed the steady state throtling epxeriment, in which this isn't a complicating factor.

 

[FoxBox]

Hi guys,

I took a introductory thermodynamics course (to prepare for university) when I wondered what happened when one performs a free expansion... (like I mentioned in the first post of this thread: with a piston and two chambers, one vacuum, one filled with gas)

Would it be reasonable if I stated this?

At the start: the system is in equilibrium. The piston is fixed. When we release the piston it will begin to move and the gas expands (no pressure on the other side of the piston). In that way, the piston (non-massless: i assume a massles piston does not exist, we also assume the piston moves horizontal) will gain kinetic energy and |KE| = |W| with W the work done by the gas = n.R.T.ln(|V₂|/|V₁|) and KE the kinetic energy of the piston just before it hits the right wall. When it hits the wall it entirely loses its kinetic energy and it is transformed to heat.

Internally, in the system: the two chambers and the piston, work is done. The piston gets kinetic energy. This kinetic energy becomes heat: the internal energy stays the same...
Delta(E) = q + W = 0

One thing I'm not sure about is what energy is required to do this work, but I think it must be heat. To make my entire reasoning correct (including the work formula W = n.R.T.ln(|V₂|/|V₁|) ), you can perform this experiment in a heat bath.

Is this a correct reasoning or do I lack basic or advanced knowledge of this? (my knowledge is very basic, as you may notice...)

-Thanks in advance, FoxBox

 

[Mapes]

EDIT: This post is addressed to FoxBox.

A couple things stand out as improper to me:

First, your definition of the system varies. At one point, you consider the entire box as a system, but then you talk about work done by the gas. Work is defined as a type of energy transfer to or from a system (specifically, without a corresponding entropy transfer), so at this point you're treating the gas as your system. This is sloppy. Count Iblis made this point above when discussing how the larger system does no work.

For example, this part is cringe-inducing:

Internally, in the system: the two chambers and the piston, work is done.

This point could be cleared up if you clearly defined all the systems you want to refer to.

Second:

One thing I'm not sure about is what energy is required to do this work, but I think it must be heat.

The internal energy of the gas moves the piston. There is no "work" in the gas, just as there is no "heat" (correctly we would say that the gas contains thermal energy). Work and heat are energy in transit.

 

[conceptmaker]

hello everyone. All of you discussing this with a wrong concept. The equation including the term pdv is valid only for reversible process . Free expansion is highly irreversible process. In free expansion as no heat is supplied and work done is zero because In thermodynamics work is said to be done only when its crosses the boundary of the system

 

[hello!!]

@FOXBOX

i guess its too late...but still,

suppose the gas at pressure P applies force on the piston .then

PA-friction=ma.where A is area and m is mass of piston,a is its acceleration.
as m=0,f=0 implies PA=0,P=0 and no force is required to move piston.W=0.

AND piston doesn't gain any K.E coz' K.E=1/2mv(squared) and as again m=0 K.E=0(there will be a constant velocity of piston but as m=0,KE=0).

SO no P,no Force applied by gas on expansion.no K.E,no work done by gas,no heat was given so no change in internal energy,so delta T=0 and thus free expansion is isothermal process.

 

[kwm]

fair enough - I was not going to reply to this but I guess I will try again. Now I must state that the following has little to do with the concept of work as currently taught. Hence should be judged accordingtly.
To me work has to be done onto something. Therefore, we must define the surroundings with clarity. If the piston apparatus is surrounded by nothing i.e. a massless vacuum then no work can be done during expansion because work has to be done onto something. Analogy lifting a massless rock into the air requires no work.
The above differs from work done on Earth wherein a piston apparatus is surrounded by our atmosphere. i.e. On the Earth's surface an expanding piston apparatus must displace the weight of the Earth's atmosphere against the Earth's gravity, hence it must do work in order to expand. Hence the expanding gas must experience a temperature decrease, or have an influx of thermal energy in order to be isothermal.
Now this is where my thoughts differ from the standard model: The same cannot be said of the piston apparatus expanding into a volume of nothingness (vacuum) because there is nothing to displace against the forces of gravity, hence no work is done during expansion, hence no temperature change is experienced by the expanding gas. Understand: Such an argument goes against certain principles beheld by many (i.e. entropy and the second law then come into question). so please judge my response accordingly = I am probably going to get crucified again by this forum for speaking my mind but so be it.
Kent