# Thermodynamics -- Free expansion of gas

#### Vibhor

We have 2 (insulated) partitions, one with the gas, another vacuum, separated by a stop cock. We remove the cock, letting the gas expand into vacuum .

I understand that since there is no opposing force on the expanding gas ,the work done by the gas is zero .

But work done by gas is given by PgasΔV . Now Pgas≠0 and ΔV≠0 , so PgasΔV≠0 , so mathematically how is work done by gas zero ?

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#### 256bits

Gold Member
Well, you could say that there has been some work done, because the centre of mass of the gas has moved, but that can easily conceptually be taken care of by arrangement of the gas container and the vacuum container.

The simple explanation is that Pgas is the pressure on the pressurized side, right. If a little bit of that gas goes through the stop cock does it encounter a resistance from the vacuum? What would the pressure of that little bit of gas be then? Certainly not Pgas.

Normally, when one calculates the work of a gas there is an opposing resistant force on the other side of the piston, so that the expansion is controlled. What you are interested in a lot of times is the value of that force, either with an expansion or compression, and one way to do that is through the P deltaV formula.

#### Chestermiller

Mentor
We have 2 (insulated) partitions, one with the gas, another vacuum, separated by a stop cock. We remove the cock, letting the gas expand into vacuum .

I understand that since there is no opposing force on the expanding gas ,the work done by the gas is zero .

But work done by gas is given by PgasΔV . Now Pgas≠0 and ΔV≠0 , so PgasΔV≠0 , so mathematically how is work done by gas zero ?

Are the two chambers insulated from one another, or are they in thermal contact with one another? Does the gas escape very rapidly, or does it seep through the stopcock?

Chet

#### Vibhor

The simple explanation is that Pgas is the pressure on the pressurized side, right.
Right.
If a little bit of that gas goes through the stop cock does it encounter a resistance from the vacuum? What would the pressure of that little bit of gas be then? Certainly not Pgas.
If not Pgas ,then what is the pressure ?

Normally, when one calculates the work of a gas there is an opposing resistant force on the other side of the piston, so that the expansion is controlled. What you are interested in a lot of times is the value of that force, either with an expansion or compression, and one way to do that is through the P deltaV formula.
Is work of a gas different from the standard definition of work in physics i.e FΔx ? I am trying to understand the work done by gas using this standard definition .

#### Vibhor

Are the two chambers insulated from one another, or are they in thermal contact with one another? Does the gas escape very rapidly, or does it seep through the stopcock?

Chet
Yes,the chambers are thermally insulated . I guess ,the gas escapes very rapidly . Does gas seeping through the stopcock make any difference ? I am going with the standard setup taken to explain free expansion.

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).

#### Chestermiller

Mentor

Yes,the chambers are thermally insulated . I guess ,the gas escapes very rapidly . Does gas seeping through the stopcock make any difference ?
Ultimately no unless the two chambers are thermally isolated from one another.
I am going with the standard setup taken to explain free expansion.

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).
What you're looking at is the same thing as if you have a cylinder, with the two chambers separated by a massless frictionless piston. And, at time zero, you suddenly release the piston, and allow the system to re-equilibrate.

Now, my questions to you are:

1. in this situation, is the pressure and temperature of the gas uniform within the cylinder, or does it vary with spatial position during the rapid deformation?

2. If the piston is massless and frictionless, from Newton's 2nd law, if the pressure on one side of the piston is zero, what does this tell you about the force per unit area exerted by the gas on the other side of the piston.

3. If the force per unit area exerted by the gas on the piston is zero, what is the work done by the gas?

4. Is the viscosity of an ideal gas zero?

5. In a rapidly deforming fluid, are the stresses simply isotropic pressure, are are there viscous contributions to the stresses (i.e., the forces per unit area)?

Chet

#### Vibhor

Ultimately no unless the two chambers are thermally isolated from one another.
Are you saying a "yes" or a "no" to whether the two situations are different ? By the way the chambers are thermally insulated .

What you're looking at is the same thing as if you have a cylinder, with the two chambers separated by a massless frictionless piston. And, at time zero, you suddenly release the piston, and allow the system to re-equilibrate.
Ok

Now, my questions to you are:

1. in this situation, is the pressure and temperature of the gas uniform within the cylinder, or does it vary with spatial position during the rapid deformation?
I think the pressure and temperature of the gas are uniform within the left partition (assuming the gas is initially in left partition) but vary when it moves to the right partition .

2. If the piston is massless and frictionless, from Newton's 2nd law, if the pressure on one side of the piston is zero, what does this tell you about the force per unit area exerted by the gas on the other side of the piston.
It has to be zero . I think here lies the problem in my thinking . Isn't the pressure at the left side of piston Pgas (at the left interface)?

3. If the force per unit area exerted by the gas on the piston is zero, what is the work done by the gas?
Zero . But Isn't force F =Pgas A (Area of piston) ? The PA term is non zero .

What I do not understand is that how is F=0 even when Pgas ≠0 ?

4. Is the viscosity of an ideal gas zero?

5. In a rapidly deforming fluid, are the stresses simply isotropic pressure, are are there viscous contributions to the stresses (i.e., the forces per unit area)?

Chet
Sorry . No idea . I am looking for a high school level explanation.

#### Chestermiller

Mentor
Are you saying a "yes" or a "no" to whether the two situations are different ? By the way the chambers are thermally insulated .
If the two chambers are thermally isolated and the gas seeps through the stopcock, there is a significant difference between this and free expansion.
I think the pressure and temperature of the gas are uniform within the left partition (assuming the gas is initially in left partition) but vary when it moves to the right partition .
Remember, we are now talking about the case where a frictionless massless piston separates the two chambers, and there is never any gas on the vacuum side. In any event, during the rapid gas expansion that occurs, the temperature and pressure are not uniform within the gas, until the final equilibrium state is attained. This is partly because the gas has distributed mass, and different parts of the gas are experiencing different accelerations and expansion rates. And the different expansion rates also cause the temperature to vary spatially.

It has to be zero . I think here lies the problem in my thinking . Isn't the pressure at the left side of piston Pgas (at the left interface)?
No. Basically, what is happening is that the rapid movement of the piston away from the bulk of the gas affects the rate of momentum transfer from the gas to the piston. Only a negligible fraction of the gas molecules in the vicinity of the interface actually strike the piston (just enough to keep the "massless" piston moving).

Zero . But Isn't force F =Pgas A (Area of piston) ? The PA term is non zero .
You just said that the force of the gas on the piston has to be essentially zero.
What I do not understand is that how is F=0 even when Pgas ≠0 ?
The rapid movement of the piston away from the gas modifies the momentum transfer at the piston face such that only a negligible fraction of the molecules actually strike the "massless" piston.

• Vibhor

#### DrDu

T
Is work of a gas different from the standard definition of work in physics i.e FΔx ? I am trying to understand the work done by gas using this standard definition .
No, it isn't. Even if the partitions would be filled by a gas under the same pressure, the work would be zero. The point is that the relevant area is the area perpendicular to the displacement Delta x. If your cock is e.g. a thin metal plate, then this area is negligible. The work done on the piston is in deed p Delta V, but Delta V is not the volume of the second partition, but the volume of your metal plate, which is negligible in idealized settings.

#### Vibhor

No, it isn't. Even if the partitions would be filled by a gas under the same pressure, the work would be zero. The point is that the relevant area is the area perpendicular to the displacement Delta x. If your cock is e.g. a thin metal plate, then this area is negligible. The work done on the piston is in deed p Delta V, but Delta V is not the volume of the second partition, but the volume of your metal plate, which is negligible in idealized settings.
What about the case of massless piston ? There area of the piston face is not negligible .

Why is F=0 despite non zero value of PgasA ?

#### DrDu

There is no really massless piston. A very light piston will get accelerated more than a heavier piston, but its final kinetic energy will be the same.

#### Vibhor

If the two chambers are thermally isolated and the gas seeps through the stopcock, there is a significant difference between this and free expansion.

Remember, we are now talking about the case where a frictionless massless piston separates the two chambers, and there is never any gas on the vacuum side. In any event, during the rapid gas expansion that occurs, the temperature and pressure are not uniform within the gas, until the final equilibrium state is attained. This is partly because the gas has distributed mass, and different parts of the gas are experiencing different accelerations and expansion rates. And the different expansion rates also cause the temperature to vary spatially.

No. Basically, what is happening is that the rapid movement of the piston away from the bulk of the gas affects the rate of momentum transfer from the gas to the piston. Only a negligible fraction of the gas molecules in the vicinity of the interface actually strike the piston (just enough to keep the "massless" piston moving).

You just said that the force of the gas on the piston has to be essentially zero.

The rapid movement of the piston away from the gas modifies the momentum transfer at the piston face such that only a negligible fraction of the molecules actually strike the "massless" piston.
Are you suggesting that Pgas is uniform until the piston is held fixed and once it is released Pgas is no longer uniform .The pressure at the left interface is essentially zero i.e Pgas,left interface = 0 which makes F = ( Pgas,left interface )( A ) = 0 ??

Is it correct ?

#### Chestermiller

Mentor
Are you suggesting that Pgas is uniform until the piston is held fixed and once it is released Pgas is no longer uniform .The pressure at the left interface is essentially zero i.e Pgas,left interface = 0 which makes F = ( Pgas,left interface )( A ) = 0 ??

Is it correct ?
Yes.

• Vibhor

#### DrDu

The nice thing with thermodynamic is that results don't depend on details of the process. You can also consider a massive piston. It will speed up and gain kinetic energy equal to $p\Delta V$. But finally it wil choque against the wall of the container and the kinetic energy will be transformed back into internal energy of the gas so that it's change in the end is zero.
What I had in mind in my earlier post was a plate which is pulled out through a slid.

#### Vibhor

Thank you so much .I was really struggling with the concept of free expansion.

Why do we relate temperature to only average translational kinetic energy of the molecules , and not to the internal energy of the molecules ? In case of monoatomic molecules ,it makes sense as the internal energy comprises only translational KE .

But what if we consider diatomic molecules. There we have rotational energy as well .The RKE also depends on the temperature .

So why do we say that temperature is related to only translational KE ?

Again does that mean if two substances having same translational KE of molecules but different internal energy , no heat would flow between them ?

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