Free Expansion of Ideal Gas: Isothermal or Adiabatic?

[kulkajinkya]

Let's consider free expansion of an Ideal gas into vacuum. We have 2 (insulated) partitions, one with the gas, another vacuum, separated by a stop cock. I lift the cock, letting the gas expand into vacuum. I am aware that in free expansion of a gas into vacuum, the change in internal energy i.e ΔU is zero. Hence temp remains constant. Now if I take the system as the gas initially, then since the process is isothermal, we have temp constant. And now since the partitions are insulated, we have ΔQ=0. Hence the process is adiabatic. Now it confuses me, whether the process is isothermal, or adiabatic? it can't surely be both!

 

[Andrew Mason]

Let's consider free expansion of an Ideal gas into vacuum. We have 2 (insulated) partitions, one with the gas, another vacuum, separated by a stop cock. I lift the cock, letting the gas expand into vacuum. I am aware that in free expansion of a gas into vacuum, the change in internal energy i.e ΔU is zero. Hence temp remains constant. Now if I take the system as the gas initially, then since the process is isothermal, we have temp constant. And now since the partitions are insulated, we have ΔQ=0. Hence the process is adiabatic. Now it confuses me, whether the process is isothermal, or adiabatic? it can't surely be both!

Why not? The expansion is against 0 pressure. This means there is no work done by the gas so W = -ΔU = 0, so it is isothermal. Since Q = 0, it is, by definition, adiabatic.

Technically, only the first bit of gas does no work. After that, the gas expands into a space that is occupied by some gas molecules. But the work it does is on the gas itself and not the surroundings. The end result is that whatever work it does comes from the internal energy of the gas and simply adds back to internal energy, so the net change in internal energy is 0.

AM

 

[kulkajinkya]

I'm sorry but I still don't get it. My point is you cannot have adiabatic and isothermal process because for isothermal we have PV=C, and for adiabatic, we have PV^n=const. These 2 curves will never be the same (i.e fit) between any 2 state points (here initial and final state)..So it has to be either iso or adiabatic

 

[Jano L.]

The contradiction is due to poor use of the word isothermal. As a whole, the gas does not receive any additional energy, so in the end, its temperature indeed will be the same as in the beginning.

But this does not mean that the temperature of the gas does not change at all. During the process, the gas in the first partition does work on the gas in the second partition, roughly adiabatically. This means that locally for any pocket of gas, the equation pV^n - const. is valid. The temperature of the gas in the filled partition will decrease, while the temperature of the gas in the partition being filled will increase. Only after a long time will the conduction equilibrate temperature all over the container to its original value (if the gas is ideal).

 

[Andrew Mason]

I'm sorry but I still don't get it. My point is you cannot have adiabatic and isothermal process because for isothermal we have PV=C, and for adiabatic, we have PV^n=const. These 2 curves will never be the same (i.e fit) between any 2 state points (here initial and final state)..So it has to be either iso or adiabatic

The adiabatic condition PV^γ = constant applies only to quasi-static adiabatic processes. It does not apply to irreversible processes. The assumption in deriving the condition is that the incremental work done by the gas on the surroundings is PdV where P is the internal pressure of the gas (ie. the internal and external pressures are the same). This is not the case in a free expansion.

AM

 

[Jano L.]

The adiabatic condition PVγ = constant applies only to quasi-static adiabatic processes. It does not apply to irreversible processes. The assumption in deriving the condition is that the incremental work done by the gas on the surroundings is PdV where P is the internal pressure of the gas (ie. the internal and external pressures are the same). This is not the case in a free expansion.

That would be true for expansion into vacuum. But we are releasing the gas to another partition of similar volume. If the gas is released to the other partition at a moderate speed (we use throttled valve to regulate the flow), the changes in the pressures will be slow enough and the equation pV^n will be quite accurate for any small pocket of gas (locally), even if overall the process is irreversible.

 

[Andrew Mason]

That would be true for expansion into vacuum. But we are releasing the gas to another partition of similar volume. If the gas is released to the other partition at a moderate speed (we use throttled valve to regulate the flow), the changes in the pressures will be slow enough and the equation pV^n will be quite accurate for any small pocket of gas (locally), even if overall the process is irreversible.

?? Quasi-static does not just mean slow. It means that the system is in equilibrium during the process. A free expansion is definitely not an equilibrium process no matter how slow or how fast it goes. PV^γ=constant does not apply between the beginning and end states.

 

[Jano L.]

True, the process is not quasi-static, but that does not mean thermodynamics is useless. Such processes can be described treated locally (thermodynamics of continuum). The basic description of adiabatic process is roughly valid if the time of the process is much greater than the relaxation time of the process of pressure equalization in the partition (which is a short instant).

 

[Andrew Mason]

True, the process is not quasi-static, but that does not mean thermodynamics is useless. Such processes can be described treated locally (thermodynamics of continuum). The basic description of adiabatic process is roughly valid if the time of the process is much greater than the relaxation time of the process of pressure equalization in the partition (which is a short instant).

But the question is whether PV^γ=constant applies to the equilibrium states before and after the free expansion. It does not. When are you suggesting that PV^γ=constant applies?

I agree that a small amount of gas dn occupying a volume dV in the original container (dn = PdV/RT) expanding into a large volume of empty space will quickly reach an equilibrium state. But that does not mean that PV^γ=constant will apply when it settles down.

AM

 

[kulkajinkya]

Thanks people, things are much clear now :)

 

[Jano L.]

When are you suggesting that PVγ=constant applies?

It applies to any pocket of the expanding fluid small enough that pressure is uniform over it.

 

[Andrew Mason]

It applies to any pocket of the expanding fluid small enough that pressure is uniform over it.

That is true if the expanding fluid is doing work on the surroundings and if the amount of work it does is ∫PdV where P is the pressure of the gas. That would be the case in, say, in a sound wave propagating through air. But that is not the case in a free expansion. PV^γ=constant does not apply in a free expansion.

AM

 

[Jano L.]

Yes, but the OP mentioned that the expansion is to another partition. That is not free expansion, because after the very first moment the pressure in the partition being filled is non-zero. There will be continuous drop of pressure from the filled partition towards that being filled, and the pocket of gas coming from the first to the second will continuously expand and perform work on the surrounding gas. When this pocket is taken small enough, the above formula can be applied to estimate roughly the temperature drop. The filled partition does work on the partition being filled; but of course, the gas as a whole does not do any work on the surrounding walls of the container.

 

[Andrew Mason]

Yes, but the OP mentioned that the expansion is to another partition. That is not free expansion, because after the very first moment the pressure in the partition being filled is non-zero. There will be continuous drop of pressure from the filled partition towards that being filled, and the pocket of gas coming from the first to the second will continuously expand and perform work on the surrounding gas. When this pocket is taken small enough, the above formula can be applied to estimate roughly the temperature drop. The filled partition does work on the partition being filled; but of course, the gas as a whole does not do any work on the surrounding walls of the container.

I think you will find that this is considered a free expansion. Since this is an ideal gas and there is no external work done, PV is the same before and after expansion, not PV^γ.

I am not sure how you are defining a "pocket" of gas - in particular, the volume of such a "pocket"? So I am not sure how you can say that PV^γ=constant applies locally during the process.

AM

 

[Jano L.]

I am not sure how you are defining a "pocket" of gas - in particular, the volume of such a "pocket"? So I am not sure how you can say that PVγ=constant applies locally during the process.

Think of air as of compressible fluid, as of continuum. Air under common circumstances behaves as such a fluid, for the mean free path of molecules is around micrometers.

 

[Chestermiller]

I think that the confusion between Jano L. and Andrew Mason stems from the fact that Andrew is focusing primarily to the initial and final equilibrium states of the system, while Jano L. is focusing primarily on what happens to the temperature and pressure of the gas in the two chambers during the transition between the initial and final states. In my judgement, they are both correct in what they are saying. In the end, the temperature of the gas in the final state will be the same as in the initial state.

Chet

 

[Andrew Mason]

I think that the confusion between Jano L. and Andrew Mason stems from the fact that Andrew is focusing primarily to the initial and final equilibrium states of the system, while Jano L. is focusing primarily on what happens to the temperature and pressure of the gas in the two chambers during the transition between the initial and final states. In my judgement, they are both correct in what they are saying. In the end, the temperature of the gas in the final state will be the same as in the initial state.

Chet

I agree that during the process there will be work done by a quantity of gas dn on the gas already in the low pressure chamber. See my first post in this thread.

But in order for the adiabatic condition PV^γ=constant to apply when that dn element of gas flows through the open valve and expands into the low pressure chamber, the amount of work it does has to be: δW = P_intdV where P_int is the internal pressure of the gas element dn.

This is not the case.

The work done as dn expands is P_extdV where P_ext is the pressure of the gas already in the low pressure side, which is definitely less than the internal pressure of the gas element dn. Since dn does some work on its surroundings (i.e. the gas already in the low pressure side), the temperature of dn decreases.

Consequently, during the process, neither PV = constant nor PV^γ=constant applies. Rather PVⁿ=constant where 1 < n < γ (polytropic). At the end of the process after equilibrium is reached, PiVi = PfVf = nRTi = nRTf

So, Jano L is quite right that calling it an isothermal process is a bit inaccurate because T is not strictly constant while the process is occurring.

AM

 

[Jano L.]

The work done as dn expands is PextdV where Pext is the pressure of the gas already in the low pressure side, which is definitely less than the internal pressure of the gas element dn.

The distinction between the internal pressure and external pressure is important only for macroscopic portion of gas, like all the gas inside the valve. What I meant is that the formula pV^γ applies to a portion of gas of volume V small enough that the gradient of the pressure is negligible over it. Then there is one pressure P and locally pV^γ = const. In mathematical treatments this is more often written as dS = 0 for any such volume, i.e. the flow is assumed isentropic. Of course, this is just a simplification, and the flow and other transport processes may very well result in a different behaviour, but the adiabatic process is the closest simple thing.

 

[Chestermiller]

The distinction between the internal pressure and external pressure is important only for macroscopic portion of gas, like all the gas inside the valve. What I meant is that the formula pV^γ applies to a portion of gas of volume V small enough that the gradient of the pressure is negligible over it. Then there is one pressure P and locally pV^γ = const. In mathematical treatments this is more often written as dS = 0 for any such volume, i.e. the flow is assumed isentropic. Of course, this is just a simplification, and the flow and other transport processes may very well result in a different behaviour, but the adiabatic process is the closest simple thing.

I agree with this assessment. I've done some analysis of gas dynamics involving the removal of a partition between a high pressure chamber and a low pressure chamber, and this is consistent with what my analysis led to. It is worth mentioning that the pressure will not be uniform in each chamber, and that pressure waves can travel down each chamber (at roughly the speed of sound) forming a boundary between a low pressure region and an higher pressure region. All the adiabatic reversible compression will take place at the wave front. This, at least, is the idealized situation that prevails in the absence of viscous dissipation and heat conduction.

Chet

 

[Andrew Mason]

In mathematical treatments this is more often written as dS = 0 for any such volume, i.e. the flow is assumed isentropic. Of course, this is just a simplification, and the flow and other transport processes may very well result in a different behaviour, but the adiabatic process is the closest simple thing.

And my point is that the gas flow is NOT isentropic. It is adiabatic, sure. But not isentropic. As the pressure in the two sides get close to being equal, the flow may approach isentropic flow.

AM

 

[Jano L.]

Well, it depends on how fast is the expansion. If it is very fast, then the state of the gas will be very far from equilibrium and it may be even impossible to assign it entropy.

If the expansion is slower, but a lot of friction and dissipation occurs, then the formula pV^γ does not apply; perhaps some other exponent could do better.

But if the expansion is slow and friction is negligible, even for high pressure difference, the flow may be modeled approximately as quasistatic process. Then the expansion is isentropic. Of course, in the end, the entropy has to increase, since the volume increased, but this is more properly assigned to subsequent conduction of heat that will equilibrate the temperature throughout the whole system. This process is much slower than the expansion, so within good approximation, it can be separated from the the latter.

Of course, in reality the expansion will not be exactly isentropic, but if we want to estimate e.g. temperatures right after the expansion, it seems that isentropic process is the simplest and quite reasonable approximation.

 

[Chestermiller]

This is a very interesting discussion, and I'd like to add to it. In my posting #19, I alluded to a gas dynamics analysis I did to try to get a better understanding of what is going on in situations like these. I'd like to elaborate a little. In the analysis, I deliberately set the viscosity and thermal conductivity of the gas equal to zero to see what happens in the limit without heat conduction and viscous dissipation. I assumed a perfectly insulated cylinder, with a frictionless massless, insulating piston initially separating two chambers of equal volume. There were equal number of moles in the two chambers, but the temperature (and pressure) in one of the chambers was initially higher than in the other chamber. Prior to time zero, the piston was held in place, but, at time zero, the piston was released. Here are some of the results of the analysis:

1. Entropy was conserved for every parcel of mass within the system.
2. Mechanical energy was conserved even though the deformations were very rapid.
3. The system could never attain thermodynamic equilibrium because there were no dissipative processes operating.
4. The deformation and flow continued forever. The system behavior was analogous to a spring-mass system. The massless piston oscillated back and forth without ever coming to an equilibrium position.
5. Compression and expansion waves traveled up and down each of the cylinders at roughly the speed of sound (at the average temperature).
6. The kinetic energy of the gas was significant and oscillatory.

If anyone is interested in seeing a write up on the analysis, please contact me via a private message.

Chet

 

[vivesdn]

Reversibility is not about speed but about small differences. Of course, small differences account for slow processes. But an expansion against the vacuum, even when it is slow (small nozzle), it is irreversible as pressure difference is not differentially small between the parts. So the change is not isentropic (entropy will increase), and being adiabatic, the temperature will decrease.