Adiabatic Expansion: Solve for Final Volume of Gas

[espnaddict014]

Sorry about the repost, but I am frustrated with this problem.

An ideal monatomic gas, consisting of 2.47 mol of volume 0.0890 m3, expands adiabatically. The initial and final temperatures are 22.3oC and -64.3oC. What is the final volume of the gas?

I've tried solving for P in terms of T using PV = nRT and it didn't seem to work out. Am I even starting in the right place?

 

[OlderDan]

Sorry about the repost, but I am frustrated with this problem.

An ideal monatomic gas, consisting of 2.47 mol of volume 0.0890 m3, expands adiabatically. The initial and final temperatures are 22.3oC and -64.3oC. What is the final volume of the gas?

I've tried solving for P in terms of T using PV = nRT and it didn't seem to work out. Am I even starting in the right place?

In your earlier discussion I think you got as far as understanding that PV^\gamma is a constant. Try writing V^\gamma as V*V^{(\gamma-1)} and replacing the product PV with what you know from the ideal gas law. I think there is no escaping the \gamma in this problem, so you will have to get a reasonable value from your tables

 

[GCT]

Did you use the proper units...convert celcius to kelvin?

 

[ZapperZ]

Sorry about the repost, but I am frustrated with this problem.

An ideal monatomic gas, consisting of 2.47 mol of volume 0.0890 m3, expands adiabatically. The initial and final temperatures are 22.3oC and -64.3oC. What is the final volume of the gas?

I've tried solving for P in terms of T using PV = nRT and it didn't seem to work out. Am I even starting in the right place?

The problem with using PV=nRT to solve this is that there's nothing in here that's a constant. So even while it is valid, it is not a very helpful form. However, from what I've gathered based on what Olderdan has said, you should already arrive at a very useful form for any adiabatic process, which is

PV^{\gamma} = constant

You should KNOW this cold whenever an adiabatic process is involved. Now, using this, and the PV=nRT relations, you can substitute P as nRT/V, which gives you another useful adiabatic relation, which is

TV^{\gamma -1} = constant

This is the form that is relevant to your question, because you are given the two values of the temperature, and the initial volume of the gas.

Moral of the story: you need to not only know what to do, but also be aware of WHY an approach doesn't work. PV=nRT is not a suitable starting point here because there's no quantity or relationship here that's constant. So you have to go hunt for a more appropriate form.

Zz.

 

[GCT]

I would imagine that P is constant. Internal energy is directly converted to P \Delta V work, work done by expanding against a constant pressure atmosphere/constant pressure situation.

 

[ZapperZ]

I would imagine that P is constant. Internal energy is directly converted to P \Delta V work, work done by expanding against a constant pressure atmosphere/constant pressure situation.

Come again?

If you look at an adiabatic curve on a PV diagram, P isn't a constant (it is not a flat, horizontal straight line). Look at the Carnot cycle, for example.

Zz.

 

[Dr.Brain]

tHE EXACT SOLUTION:

As per First law of Thermodynamics:

dQ=dU+dW

here dQ=0 (adiabatic)

Therefore dU=-dW

dU=nC(dT)

W=p(dV)

where C= heat capactity at constant volume
for monoatomic gas

dT=change in temp.

dV=change in volume

Put the forms and do the calculations . a Correct answer is assured.

 

[GCT]

nevermind...I was unfamiliar with the concept, confused it with a vague recollection of another situation.

espnaddict, what course is this for?

 

[Andrew Mason]

tHE EXACT SOLUTION:

As per First law of Thermodynamics:

dQ=dU+dW

here dQ=0 (adiabatic)

Therefore dU=-dW

dU=nC(dT)

W=p(dV)

where C= heat capactity at constant volume
for monoatomic gas

dT=change in temp.

dV=change in volume

Put the forms and do the calculations . a Correct answer is assured.

This is a non-trivial calculation. Why not just use the resulting adiabatic relationship: PV^\gamma per Dan and Zapperz rather than develop it from first principles every time you use it?

AM

 

[Dr.Brain]

^^^^
ANDREW MASON,

PV^ lamda will not give the correct answer to the above question for sure.Its by the basics that physics has developed. My method is correct.

PV^lamda will give an answer which is independent of the number of moles being used and change in temperature taking place.Therefore it is wrong method as it is independent of the constraints which can make a huge difference.

 

[ZapperZ]

^^^^
ANDREW MASON,

PV^ lamda will not give the correct answer to the above question for sure.Its by the basics that physics has developed. My method is correct.

PV^lamda will give an answer which is independent of the number of moles being used and change in temperature taking place.Therefore it is wrong method as it is independent of the constraints which can make a huge difference.

Eh?

Why would the "number of moles" matter in THIS case? The amount of gas doesn't change - it is a constant at both temperatures. And simply via substitution using the ideal gas law, one arrives at the OTHER form for an adiabat with a T-V relationship, which is ALL that is needed in solving the original problem.

.. unless of course you're claiming that this form is also "incorrect".

Zz.

 

[Clausius2]

^^^^
ANDREW MASON,

PV^ lamda will not give the correct answer to the above question for sure.Its by the basics that physics has developed. My method is correct.

PV^lamda will give an answer which is independent of the number of moles being used and change in temperature taking place.Therefore it is wrong method as it is independent of the constraints which can make a huge difference.

Well, I think the initial question is not clear enough. The main question is:

-Is it that adiabatic expansion reversible?

If it is reversible, then Andrew, Zz and Dan are right, and the result is independant of mass.

If it is irreversible, then you're right, but you don't actually know how to solve the problem. Let's see:

nc_vdT=-PdV is the equation for an irreverisible adiabatic curve as you have stated. Now try to integrate it. You don't know how pressure evolve with T nor V. So that, you cannot integrate it without additional information.

I recall have done similar problems with non-reversible adiabatic process, but all of them consisted in a cylinder with a piston in mechanical equilibrium with atmosphere, so that it can be considered that pressure remains constant across that adiabatic curve. Remind all that an irreversible adiabatic process can have the pressure constant.

 

[Andrew Mason]

^^^^
ANDREW MASON,

PV^ lamda will not give the correct answer to the above question for sure.Its by the basics that physics has developed. My method is correct.

PV^lamda will give an answer which is independent of the number of moles being used and change in temperature taking place.Therefore it is wrong method as it is independent of the constraints which can make a huge difference.

I don't think that is correct. Work it out from dU = -dW and nC_v = dU/dT:

dU = d(PV) = Pdv + Vdp= nRdT = n(C_p-C_v)dT = -dW = -Pdv = nC_vdT

\frac{Pdv + Vdp}{n(C_p-C_v)} = -\frac{Pdv}{nC_v}

Pdv + Vdp = - Pdv\frac{(C_p-C_v)}{C_v} = - Pdv({\frac{C_p}{C_v} - 1})

Vdp = - Pdv\frac{C_p}{C_v}}

\frac{dp}{P} + \frac{dv}{V}\frac{C_p}{C_v}} = 0

Integrating:

\int\frac{dp}{P} + \int \frac{dv}{V}\frac{C_p}{C_v}} = K

Integrating and letting \gamma = C_p/C_v:
lnP + \gamma lnV= K

PV^\gamma= e^K = constant

AM

 

[Andrew Mason]

Well, I think the initial question is not clear enough. The main question is:

-Is it that adiabatic expansion reversible?

How can it not be? The same amount of work is required to compress it back to its original state as the gas does in expanding. So if you store the work done (eg by lifting a weight) you can reverse the process without going outside of the system.

AM

 

[siddharth]

Adiabatic expansions need not be reversible. Consider this example. I have some moles of a gas in an insulated container .Now the gas adiabatically expands against a constant external pressure (p). In this case work done while expanding, that is, in the intergral pdv, is simply -p(v2-v1) as the pressure is constant and equal to external pressure. In this case the work done will not be Pv^gamma though the process is adiabatic but will be -p(v2-v1). The reason is that in a reversible adiabatic expansion, the system and surrondings are almost in equilibirium in every stage. But in an irrerversible expansion we first rapidly increase the pressure to the external pressure and then expand against that pressure.

 

[Andrew Mason]

Adiabatic expansions need not be reversible. Consider this example. I have some moles of a gas in an insulated container .Now the gas adiabatically expands against a constant external pressure (p). In this case work done while expanding, that is, in the intergral pdv, is simply -p(v2-v1) as the pressure is constant and equal to external pressure. In this case the work done will not be Pv^gamma though the process is adiabatic but will be -p(v2-v1). The reason is that in a reversible adiabatic expansion, the system and surrondings are almost in equilibirium in every stage. But in an irrerversible expansion we first rapidly increase the pressure to the external pressure and then expand against that pressure.

I disagree, especially with your last two sentences. (BTW the work done is never PV^\gamma. That is just a constant).

If a gas expands adiabatically, this means that the work done by the gas is equal to the change in internal energy. If a gas expands from P1V1 to P2V2 and does work of only P2(V2-V1) then heat is lost to the system because:

P_2(V_2-V_1) < \int_{V_1}^{V_2}Pdv \implies Q \ne 0

So, by definition, this process cannot be adiabatic.

What happens in a case where a gas at P1V1 does work against an external pressure of P2 where P2<P1, is that the work done by the expanding gas produces kinetic energy as well as doing P2(V2-V1) of work:

\Delta U = P_2(V_2-V_1) + KE = \int_{V_1}^{V_2}Pdv

The work done by the gas is greater than P2(V2-V1) by the amount of this kinetic energy (that area in the PV diagram below the adiabatic gas curve and above the line P=P2).

If this kinetic energy is stored along with the P2(V2-V1) work (for example, by raising a weight and giving it KE and its kinetic energy allowing the weight to move higher after expansion ceases), the gas can be compressed with that stored work back to its original state.

So, I would argue, every truly adiabatic expansion/compression is reversible.

AM

 

[siddharth]

(BTW the work done is never PV^\gamma. That is just a constant).

Oops. Well your absoulutley right there of course. My mistake.

What i should have said was that to calculate the final temperature using PV^\gamma and then substituting P=nRT/V will not work in case of irreversible adiabatic process . This is because the gas law holds only when the gas is in equilibirium. In my example, the final temperature can be found by the dU=q+w. As the process is adiabatic, q=0. Therefore nCv(T2-T1)= -p(v2-v1) and T2 can be found

If a gas expands adiabatically, this means that the work done by the gas is equal to the change in internal energy. If a gas expands from P1V1 to P2V2 and does work of only P2(V2-V1) then heat is lost to the system because:
P_2(V_2-V_1) &lt; \int_{V_1}^{V_2}Pdv \implies Q \ne 0

Because the value of the inital pressure P1 is rapidly changed to P2 before the expansion actually occurs the value of
\int_{V_1}^{V_2}Pdv = P_2(V_2-V_1)
so q=0 is still valid

 

[xanthym]

.

The classic example of an IRReversible adiabatic process is that of "adiabatic free expansion". Let an Ideal Gas be confined in the Left compartment of a 2 compartment completely insulated container. The Right compartment is separated from the Left by a thin massless plate (currently held by a stop) and contains a vacuum.

The separator plate is now released from its stop. The gas in the Left compartment chaotically expands, filling the entire Right compartment (pushing the thin massless plate to the end of the Right compartment). The Ideal Gas now occupies the total volume of the Left and Right compartments.

By the FLT:
ΔU = ΔQ - ΔW

The process is adiabatic so that {ΔQ = 0}. Furthermore, since the gas expanded into a vacuum, no work was done and {ΔW = 0}. Thus:
ΔU = ΔQ - ΔW = (0) - (0) = (0)

Because we have an Ideal Gas:

1: \ \ \ \ \ \Delta U \ = \ 0 \ \ \, \ \Longrightarrow \ \ \, \ \color{red}T_{final} \ = \ T_{initial}

The above Eq #1 contradicts the following commonly used formula for "adiabatic processes":

2: \ \ \ \ T_{final} \, \ = \, \ T_{initial} \left ( \frac{V_{initial}}{V_{final}} \right )^{(\gamma \, - \, 1)} \ \ \color{red}\neq \ \ 1 \ \ \ \ \ \ \ \ \color{black}(\mbox{for} \, \ \ V_{initial} \ \neq \ V_{final})

where:

3: \ \ \ \ \gamma \, \ = \, \ \frac{C_{p}}{C_{v}} \, \ = \, \ 1 \, + \, \frac{nR}{C_{v}} \, \ &gt; \, \ 1

The reason is that Eq #2 applies only to REVERSIBLE adiabatic processes and always predicts non-zero temperature changes dependent only on initial and final volumes (and γ). The case of adiabatic free expansion, where {ΔT = 0}, clearly indicates Eq #2 is NOT applicable to IRReversible adiabatic processes.


~~

 

[Clausius2]

How can it not be? The same amount of work is required to compress it back to its original state as the gas does in expanding. So if you store the work done (eg by lifting a weight) you can reverse the process without going outside of the system.

AM

I think Siddharth has answered you yet. But anyway I'll try to clear it up.

Imagine there is a closed cylinder filled with an ideal gas at pressure P_1 with a free piston over it. The system is submerged into an atmosphere of pressure P_o&lt;P_1. Besides there is a mass put onto the piston. The whole system is adiabatic. At some instant, the mass is instantaneously removed and the piston goes up. Assuming the mass of the piston is negligible, then the mechanical equilibrium must yield internal pressure equals atmospheric pressure. Due to the rapidity of the process, it must be considered irreversible. Such irreversible trajectory is represented by the equation \delta Q=0, or which is the same dU=\delta W. So that, the decreasing of internal energy is the same than the work done by the system over the atmosphere. Of course W=-P_o\Delta V. And PV^{\gamma}=const has no sense for this problem because it represents an adiabatic reversible trajectory (isentropic).

In order to make this process be reversible, the mass would be removed very very slowly, such that internal pressure evolves yielding mechanical equilibrium in each infinitesimal variation.

Also it can be demonstrated that the temperature of internal gas is greater after irreversible adiabatic expansion than after a reversible adiabatic (isentropic) expansion.

Turning back to the original problem, I don't know which is the real case, by the way we have not obtained any feedback of the thread author as usually.

 

[Andrew Mason]

Because the value of the inital pressure P1 is rapidly changed to P2 before the expansion actually occurs the value of
\int_{V_1}^{V_2}Pdv = P_2(V_2-V_1)
so q=0 is still valid

But if the initial pressure is rapidly changed before the expansion, without doing work, heat must leave the system. There is no other way that can happen. By definition that would not be adiabatic.

P_2(V_2-V_1) &lt; \int_{V_1}^{V_2}Pdv + \int_{P_1}^{P_2}Vdp = \Delta U \implies Q \ne 0

AM

 

[GCT]

Perhaps the heat issue is resolved by taking into account chemical reactions...or not.

 

[Andrew Mason]

.

The classic example of an IRReversible adiabatic process is that of "adiabatic free expansion". Let an Ideal Gas be confined in the Left compartment of a 2 compartment completely insulated container. The Right compartment is separated from the Left by a thin massless plate (currently held by a stop) and contains a vacuum.

The separator plate is now released from its stop. The gas in the Left compartment chaotically expands, filling the entire Right compartment (pushing the thin massless plate to the end of the Right compartment). The Ideal Gas now occupies the total volume of the Left and Right compartments.

By the FLT:
ΔU = ΔQ - ΔW

I don't think you are describing adiabatic expansion. In this scenario, is it correct to say that the gas does no work in accelerating the itself to the right?

If the gas in the left side is allowed to rush to the right it has to push off to the left which results in the centre of mass moving to the right with speed (average) v. The work required to create this kinetic energy comes from the internal energy of the gas as shown by Bernouilli's equation:

P_1V_1 = P_2V_2 + \frac{1}{2}\rho v^2

So there has to be a drop in temperature while the gas is expanding (while the centre of mass is moving).

Then there is a compression phase on the right side, while the centre of mass of the gas comes to a stop (relative to its box). Again, the gas is doing work (on itself) converting kinetic energy back to internal energy (pressure).

So what you really have here are two adiabatic processes: adiabatic expansion producing work and lowering temperature followed by adiabatic compression of the gas (using the work generated by the expansion) and an increase in temperature. The process will result in a simple harmonic oscillation of compression/expansion unless the system somehow gets rid of this energy. But if the system gets rid of the energy, there is a loss of heat to the gas and the process is not adiabatic.

I would suggest that each of these processes (expansion/compression) are reversible adiabatic processes. For example, one could connect the box to a ratcheted spring on the left. The sudden push by the gas in accelerating to the right would cause the box to move to the left and compress the spring (which would stay compressed because of the ratchet mechanism). This would leave the gas fully expanded but cooler and the adiabatic condtion PV^\gamma = constant would apply.

One could devise a system that would use that spring energy to compress the gas back to its original volume and pressure, again preserving the adiabatic condition.

AM

 

[ZapperZ]

I don't think you are describing adiabatic expansion. In this scenario, is it correct to say that the gas does no work in accelerating the itself to the right?

This is correct and a common source of confusion of many people in Thermo classes. An adiabatic free expansion does NO work. While there is an increase in volume (implying a translation), there is no "force" - the gas is expanding into vacuum. So it does no work during the expansion.

Zz.

 

[Andrew Mason]

This is correct and a common source of confusion of many people in Thermo classes. An adiabatic free expansion does NO work. While there is an increase in volume (implying a translation), there is no "force" - the gas is expanding into vacuum. So it does no work during the expansion.

So where does the translational kinetic energy of the gas (ie motion of its centre of mass) come from if not from the internal (PV) energy of the gas? Does it not take work to create that translational motion?

AM

 

[ZapperZ]

So where does the translational kinetic energy of the gas (ie motion of its centre of mass) come from if not from the internal (PV) energy of the gas? Does it not take work to create that translational motion?

AM

Not if no force is applied to cause the translation. Don't forget that work, by definition, is the sum of the applied force with the resulting translation. A simple translation alone is not a sufficient criteria to automatically indicate that work is done.

In any case, I thought the fact that adiabatic free expansion does no work is a "textbook" stuff.

Zz.

 

[Andrew Mason]

Not if no force is applied to cause the translation. Don't forget that work, by definition, is the sum of the applied force with the resulting translation. A simple translation alone is not a sufficient criteria to automatically indicate that work is done.

In any case, I thought the fact that adiabatic free expansion does no work is a "textbook" stuff.

Sometimes textbooks gloss over the subtleties. If the free expansion is infinitesimally slow, I would agree. If it is sudden, I would not agree, for the reasons stated above.

AM

 

[ZapperZ]

Sometimes textbooks gloss over the subtleties. If the free expansion is infinitesimally slow, I would agree. If it is sudden, I would not agree, for the reasons stated above.

AM

You are not agreeing to which one? The explanation for no work done in an adiabatic free expansion? Or that free expansion does no work?

If you disagree with the explanation, then you are the one with some explaining to do on why adiabatic free expansion does no work. To me, an "infinitesimally slow" expansion automatically implies a net force preventing the "liberation" of the gas particles. This does NOT constitute a free expansion, but rather a deliberate confinment of the rate of expansion. The gas particles are pushing against something during the expansion. So how would this be any different than the reversible adiabatic expansion, which actually HAS a work done.

Unless I misrepresent something, I believe I have not stated anything new at all in here. You are welcome to check this in any Thermo textbooks to make sure I'm not stating something based on my opinion or preference. And since this is the homework help section and not the TD or the main physics section, I thought we are supposed to confine responses to questions being asked within accepted standard explanation.

Zz.

 

[Andrew Mason]

You are not agreeing to which one? The explanation for no work done in an adiabatic free expansion? Or that free expansion does no work?

If you disagree with the explanation, then you are the one with some explaining to do on why adiabatic free expansion does no work. To me, an "infinitesimally slow" expansion automatically implies a net force preventing the "liberation" of the gas particles. This does NOT constitute a free expansion, but rather a deliberate confinment of the rate of expansion.

I was referring to the incremental free expansion of a gas. This would be achieved, for example, by putting a small hole in the plate in xanthym's example rather than lifting the plate suddenly. In that case you end up with almost no translational kinetic energy of the gas (v, speed of the centre of mass, is very small).

When the expansion is sudden, the centre of mass moves rapidly and this sudden expansion causes a dynamic oscillation (all of which is adiabatic) within the chamber, as I described in my post above. I can't see how this dynamic oscillation can stop without the loss of some of this energy to the surroundings (ie the gas doing work on the surroundings). So you end up with a gas at lower temperature than when you started: \Delta U = P\Delta V so T_f \ne T_i.

The gas particles are pushing against something during the expansion. So how would this be any different than the reversible adiabatic expansion, which actually HAS a work done.

The point that I make is that if they all push off in one direction, the gas performs work on itself: translational kinetic energy of the gas amounts to work. Do you disagree with that? Perhaps I am wrong on that, but it seems obvious to me.

AM

 

[ZapperZ]

I was referring to the incremental free expansion of a gas. This would be achieved, for example, by putting a small hole in the plate in xanthym's example rather than lifting the plate suddenly. In that case you end up with almost no translational kinetic energy of the gas (v, speed of the centre of mass, is very small).

When the expansion is sudden, the centre of mass moves rapidly and this sudden expansion causes a dynamic oscillation (all of which is adiabatic) within the chamber, as I described in my post above. I can't see how this dynamic oscillation can stop without the loss of some of this energy to the surroundings (ie the gas doing work on the surroundings). So you end up with a gas at lower temperature than when you started: \Delta U = P\Delta V so T_f \ne T_i.

The point that I make is that if they all push off in one direction, the gas performs work on itself: translational kinetic energy of the gas amounts to work. Do you disagree with that? Perhaps I am wrong on that, but it seems obvious to me.

AM

No, I'm not referring to ANY example in particular. I'm referring to the "classic" adiabatic free expansion. What xanthym described isn't the classic free expansion. And there is no "incremental" free expansion. You open a throttle, and you have it. It is the whole definition of "sudden expansion". Anything more complicated than that and the whole "adiabatic free expansion" scenario breaks down and you can no longer call the process by that name.

Zz.

 

[Andrew Mason]

Imagine there is a closed cylinder filled with an ideal gas at pressure P_1 with a free piston over it. The system is submerged into an atmosphere of pressure P_o&lt;P_1. Besides there is a mass put onto the piston. The whole system is adiabatic. At some instant, the mass is instantaneously removed and the piston goes up. Assuming the mass of the piston is negligible, then the mechanical equilibrium must yield internal pressure equals atmospheric pressure. Due to the rapidity of the process, it must be considered irreversible.

I guess this is the part that I am not getting. Why is it irreversible simply because of the rapidity of the process? We seem to be ignoring the fact that the gas has mass, and therefore kinetic energy.

Consider your example but use a smaller mass and keep it on the piston but not mechanically connected to the piston shaft. The downward force is mg/A. The volume of the gas expands by V2-V1 = Ah and the distance the piston moves is h. But the work done is not:

\int_{V1}^{V2} Pdv = \int_{V1}^{V2}mg/A(dv) = \int_{V1}^{V2}mg/A(Adh) = mgh

This is because the mass has kinetic energy when the gas is finished expanding and the mass moves a further distance: \Delta h_2 = KE/mg higher. The work done is:

\int_{V1}^{V2} Pdv + KE_{system} = mgh + \frac{1}{2}mv^2 = mg(h + h2)

When that mass comes back down and hits the piston shaft, the piston will compress back to its original pressure and volume - all without any addition of heat or work to the system. That describes a reversible adiabatic process. It only seems irreversible if we ignore the KE part of the work. Of course if you let KE escape, you cannot get back to the original state without adding energy.

This is a very interesting issue and one that I see has caused some consternation. See for example: https://www.physicsforums.com/showthread.php?t=54129

I think the original poster of that question hits the issue directly and the answers he was provided do not fully answer this question.

AM

 

[Clausius2]

I think the original poster of that question hits the issue directly and the answers he was provided do not fully answer this question.
AM

Forget about original poster, maybe we are learning more about this issue than he, this discussion provides us the oportunity to exchange the views about this stuff and clarify and correct our thinkings about it.

Having read your last post:

I) we agree the expansion I referred to and you have quoted, is irreversible. I am not saying I am absolutely correct. I am only saying there are several books which calls this kind of rapid expansions to be adiabatically irreversible. The main problem here is how I justify they're irreversible and you understand it.

II) the main problem with your formulation is you are doing integrals where there is no trajectory to integrate over it. You cannot writte an integral of a non defined or discontinuous function. Please don't get surprised and keep on reading. The PV diagram represents Equilibrium States. A line in a PV diagram represents a secquence of Equilibrium States. You can integrate the pressure respect to volume if the function P(V) can be plotted in the PV diagram. In such a rapid process when we remove instantaneously the mass which was firstly placed on the piston, there it could be a sucession of thermodynamical points (P,V) (probably could be measured by some sensor inside cylinder), but these states have not reached thermal equilibrium.

III) In some way Zz has mentioned how thermodynamic books treat this kind of problems. I think this problem we are talking about cannot be justified only by usual thermo equations. Moreover, I think its irreversibility is ultimately justified by the proper Flow Phenomena. If we want to state deeper commentaries, I think we should beging to talk about how is the real flow field inside the cylinder during such fast adiabatic expansion. When the mass is instantaneously removed and the piston begins to go upwards, there must be some kind of depression in zones near the piston surfaces, and a detachment of the boundary layer is assured. Such detachment would cause turbulence and instabilities near the piston surface, provoking strong unsteady effects and viscous dissipation inside the gas. You ask: why is the process irreversible? Well, the last usual cause of non isentropic behavior is viscous dissipation which degenerates mechanical energy into heat. This heat cannot be rejected to surroundings so that it must be absorbed in thermal energy form.

Finally, if the mass would be removed infinitesimaly slowly, then the proper flow field evolution would be "smoothed" enough to avoid boundary layer separation and viscous dissipation.

I don't know if you agree. Let me know.

EDIT: the link you show me to a similar problem goes to a solved question. We answered Kistos widely and there is no place to doubt.

 

[Andrew Mason]

I) we agree the expansion I referred to and you have quoted, is irreversible. I am not saying I am absolutely correct. I am only saying there are several books which calls this kind of rapid expansions to be adiabatically irreversible. The main problem here is how I justify they're irreversible and you understand it.

My understanding is that reversibility and irreversibility depends on entropy. Entropy is [Edit as per Clausius:] dQ/T. Since there is no flow of heat into or out of the system in either a rapid or a slow adiabatic expansion/compression, there is no change in entropy: dQ = 0. So, by definition, every adiabatic process has to be reversible.

I think that a 'reversible expansion' signifies that the gas can be compressed back to its original state with an amount of work equal to the work done in expanding. If the system does work and the work is lost to the system, it is still an adiabatic, reversible process. It just can't be actually reversed until the lost energy is replaced.

II) the main problem with your formulation is you are doing integrals where there is no trajectory to integrate over it. You cannot writte an integral of a non defined or discontinuous function.

Where is the discontinuity? Pressure of the gas is a continuous function with volume during even a rapid expansion.

Just because the ambient pressure (ie. the pressure that the gas is working against) is less than the gas' internal pressure does not mean that the gas pressure drops instantaneously to the ambient pressure. The gas pressure x area creates an unbalanced force causing system mass to accelerate. This creates kinetic energy. The kinetic energy represents work done by the gas. So the work done by the gas is always the integral of its own internal pressure x dv, not the external pressure x dv.

III) In some way Zz has mentioned how thermodynamic books treat this kind of problems. I think this problem we are talking about cannot be justified only by usual thermo equations. Moreover, I think its irreversibility is ultimately justified by the proper Flow Phenomena. If we want to state deeper commentaries, I think we should beging to talk about how is the real flow field inside the cylinder during such fast adiabatic expansion. When the mass is instantaneously removed and the piston begins to go upwards, there must be some kind of depression in zones near the piston surfaces, and a detachment of the boundary layer is assured. Such detachment would cause turbulence and instabilities near the piston surface, provoking strong unsteady effects and viscous dissipation inside the gas. You ask: why is the process irreversible? Well, the last usual cause of non isentropic behavior is viscous dissipation which degenerates mechanical energy into heat. This heat cannot be rejected to surroundings so that it must be absorbed in thermal energy form.

This is the key, I think, to understanding adiabatic transitions. When the work done by the gas in an adiabatic expansion, flows back into the gas in the form of heat, you no longer have a reversible process. But that simply means that there is an adiabatic process followed by a non-adiabatic process: adiabatic expansion followed by a conversion of work into heat and a flow of heat into the gas. The latter part is, obviously, not adiabatic.

AM

 

[Clausius2]

My understanding is that reversibility and irreversibility depends on entropy. Entropy is dQ/dT. Since there is no flow of heat into or out of the system in either a rapid or a slow adiabatic expansion/compression, there is no change in entropy: dQ = 0. So, by definition, every adiabatic process has to be reversible.

In order to enhance a better comprehension, Andrew, I beg you to read the entire post, no matter it is a bit long. I think it is worthy. My english is not very good, but don't leave any line without being understood. Let me know if you don't understand something.

I keep on disagreeing with you. Entropy is not dQ/dT. On the other hand, entropy differential variation of some system is dS=\delta Q_{reversible}/T. I will try to summarize logically why you're wrong:

I) The entropy change between two states (1 and 2) is defined as the integral of the REVERSIBLE differential exchange of heat divided by the temperature. We agree that this integral doesn't depend on the trajectory chosen between 1 and 2, so that entropy is a function of state.

II) What happens if the process is adiabatically reversible? Then \delta Q_{reversible}=0 and dS=0. As entropy surroundings change is zero, then the whole process is reversible.

III) What happens if the process is adiabatically irreversible?. An example of this is the problem we have been discussing (we will reafirm later it is reversible again). Imagine you can produce this experiment in a laboratory. If you have a pressure-volume sensor maybe you'll obtain a serie of data (P,V) during the short process. Well, what pressure are you measuring? answer: who knows?. Is the pressure uniform inside the chamber? Answer: no. Is there heat flux inside gas due to temperature gradients during the process? Answer: Yes, there is. Therefore, is there thermodynamic equilibrium inside the chamber? Answer: No, there isn't. Could I plot (P,v) data in a P-V diagram? Answer: Yes, sure you could. You could plot also a Picasso draw, but you are not respecting the thermodynamic equilibrium, you aren't plotting equilibrium points and so you are out of the original sense of a thermodynamic sucession of equilibrium state points. There aren't differential variations between those points, so you cannot use differential forms and integrate them to obtain any area between any curve. In order to obtain a physical meaning of a PV trajectory it is needed a "relax time" such that molecules reach "statiscal" equilibrium, the fundamental hypothesis of the kinetic theory which is just behind state equation and thermodynamic coefficients. Anyway, you have obtained a change of state after the process, i.e state 2. Experimentally, plotting ONLY 1 and 2 states in a PV diagram (both states DO are states of equilibrium), you should check that they are NOT connected by an isentropic curve. Although the whole system is adiabatic, the process is not isentropic. It can be argued that \delta Q_{reversible}&gt;0, although there have not been any heat exchange. The fact is that if both points are not over the same isentropic curve, then there must be some kind of heat we don't know what it is apparently. Such definition of entropy is only a superficial one. Inside a system, it can be an increasing of entropy due to heat exchange and by means of Entropy Internal Generation, which I'll call d\sigma. Then entropy can be reshaped as: dS=\delta Q_{irreversible}/T+d\sigma. Here the net exchange of irreversible heat with surroundings is zero. But there is internal entropy generation by means of internal viscous dissipation as I described in my last post. I mean, there must be some "virtual" process that conect states 1 and 2 with a reversible trajectory which exerts a "virtual" reversible heat exchange which models that entropy generation. That's the real sense of the entropy definition. I have just demonstrated you what happens experimentally, and so what happens theoretically because thermodynamic theory is constructed just over experimental issues.

I think that a 'reversible expansion' signifies that the gas can be compressed back to its original state with an amount of work equal to the work done in expanding. If the system does work and the work is lost to the system, it is still an adiabatic, reversible process. It just can't be actually reversed until the lost energy is replaced.

Sure it is so, you're right. But in fast processes as the one we are concerned about in our last posts, there is a work done when we compress the gas putting the mass over it but this work is not entirely recovered when the gas has been expanded again removing rapidly the mass. In fact, it is called a "destruction of Availability". Availability Function measures the amount of mechanical energy storaged in any system. Availability is destructed partially by means of irreversibilities.

This is the key, I think, to understanding adiabatic transitions. When the work done by the gas in an adiabatic expansion, flows back into the gas in the form of heat, you no longer have a reversible process. But that simply means that there is an adiabatic process followed by a non-adiabatic process: adiabatic expansion followed by a conversion of work into heat and a flow of heat into the gas. The latter part is, obviously, not adiabatic.

AM

You have quoted my description of the internal fluid flow when you wrote this. Why? In my description I mentioned internal viscous dissipation. In order to have irreversibilities it is not needed a non-adiabatic system. I mean, there could be an adiabatic system with internal viscous dissipation. The heat produced is not exchanged with surroundings, but absorbed as Internal Energy.


I hope you have understood the purpose of this thread. Obviously not every adiabatic expansion must be reversible by definition, but have I achieved your understanding of why you're mistaken?.

Regards.

Javier.

BTW: Be sure, Andrew, that I have learned a lot by means of this discussion. I had many doubts about entropy, but answering your puzzle I have cleared them up. Thank you by the way .

 

[OlderDan]

I've seen the reply number growing and growing on this thread, and finally had to read through it. I thought I would just throw out an observation about free expansion without taking sides in the debate.

So for what its worth, I turned on my microscope and looked at a contained ideal gas as an ensemble of interacting particles undergoing elastic collisions among themselves and with the walls of the conatiner. The walls do no work on individual particles because

\int Fdx = 0

in every elastic collision. The particles do work on each other resulting in energy transfer from one to another, but of course no net work is done; one particle gains what the other particle loses. In the half empty container separated by a rigid massless barier, the momentum distribution is isotropic, with the Maxwell speed distribution.

http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/kintem.html

What happens when the barrier is released, or suddenly vaporizes is that the particles already moving toward that barrier no longer encounter reflection and continue moving into the vacuum until they reach the new wall. There is no sudden movement of every gas particle toward the vacuum. There is a gradual (on some time scale) migration of particles into the vacuum analogous to the evaporation of a liquid into the air, except there is no energy threshold to overcome. This is a microscopic view of xanthym's "chatotic expansion" idea. I don't think that can result in an oscillating center of mass, but does lead to the conclusion of no change in internal energy and no termperature change. What is happening is a gradual loss of isotropy of the energy/momentum distribution that is gradually restored as the particle density within the box redistributes toward uniformity. I see this as more likely being a center of mass approaching the middle of the box asymptotically than setting up endless oscillations. From that point of view, there is no sudden change of pressure across the entire volume of the gas either. There is a sudden change in the pressure gradient, but it certainly will not be initially uniform.

 

[Andrew Mason]

In order to enhance a better comprehension, Andrew, I beg you to read the entire post, no matter it is a bit long. I think it is worthy. My english is not very good, but don't leave any line without being understood. Let me know if you don't understand something.

I keep on disagreeing with you. Entropy is not dQ/dT.

Quite right. Entropy is dQ/T. dQ/dT is heat capacity at constant volume. The point I was making was that dQ = 0 in an adiabatic process, so there can be no change in entropy.

On the other hand, entropy differential variation of some system is dS=\delta Q_{reversible}/T.

If you are only considering the system by itself and ignoring surroundings, I agree. I can compress a gas from P1V1 to P2V2 reversibly or irreversibly. In the first case, there is no change in entropy of the system and surroundings; in the second there is. It does not just depend on the beginning and end points of the system.

I) The entropy change between two states (1 and 2) is defined as the integral of the REVERSIBLE differential exchange of heat divided by the temperature. We agree that this integral doesn't depend on the trajectory chosen between 1 and 2, so that entropy is a function of state.

That is the change in entropy of the system only, not system + surroundings. As long as that is clear, I agree.

II) What happens if the process is adiabatically reversible? Then \delta Q_{reversible}=0 and dS=0. As entropy surroundings change is zero, then the whole process is reversible.

I agree.

III) What happens if the process is adiabatically irreversible?. ... There aren't differential variations between those points, so you cannot use differential forms and integrate them to obtain any area between any curve.

Here is where I lose you. There is pressure in the system. It can be measured. It is effectively constant. The system reaches effective statistical equilibrium extremely quickly (pressure changes propagate at the speed of sound and the density remains effectively constant). Why does this mean you cannot integrate Pdv to find the work done (P = (average) internal gas pressure)? \int P_{gas}dv is the work done by the gas which is \int P_{ext}dv + KE of the system.

Experimentally, plotting ONLY 1 and 2 states in a PV diagram (both states DO are states of equilibrium), you should check that they are NOT connected by an isentropic curve.

I don't see why not. Can you show me some actual data on this? It seems to me that since the decrease in internal energy of the gas is converted entirely to external work, so long as that work is not being sent back to the gas in the form of heat, the rapidity of the expansion does not affect entropy. The only way the gas can fall off the isentropic curve is if some of the external work done by the gas is converted to heat and let back into the system.

Although the whole system is adiabatic, the process is not isentropic. It can be argued that \delta Q_{reversible}&gt;0, although there have not been any heat exchange. The fact is that if both points are not over the same isentropic curve, then there must be some kind of heat we don't know what it is apparently.

Precisely my point. But if that happens, can we still say the process is adiabatic? Perhaps we should define adiabatic more carefully: adiabatic process is a process in which work but no heat is exchanged with the surroundings; a conversion into system heat of work done by the gas shall be considered a heat exchange with the surroundings.

If work is done by the gas but is converted to heat and allowed to enter the system, I would say that the process is not adiabatic. I think you would say that it is. An example would be where a gas in an insulated container (so that it is completely isolated from its surroundings) expands adiabatically and very slowly. As the gas expands and does work, the gas turns an electrical generator. The generator powers a heating coil inside the gas. As the gas expands and does work turning the generator, the electricity in the coil warms up the gas. I think you would say that this is an adiabatic process, since there is no exchange of heat with the surroundings. I would say this is not adiabatic.

Sure it is so, you're right. But in fast processes as the one we are concerned about in our last posts, there is a work done when we compress the gas putting the mass over it but this work is not entirely recovered when the gas has been expanded again removing rapidly the mass. In fact, it is called a "destruction of Availability". Availability Function measures the amount of mechanical energy storaged in any system. Availability is destructed partially by means of irreversibilities.

Not necessarily though. It could be removed but stored as potential energy.

You have quoted my description of the internal fluid flow when you wrote this. Why?

Because, in my view, the internal viscous flow (produced by the work of the fluid) being converted to heat makes the process non-adiabatic.

In my description I mentioned internal viscous dissipation. In order to have irreversibilities it is not needed a non-adiabatic system. I mean, there could be an adiabatic system with internal viscous dissipation. The heat produced is not exchanged with surroundings, but absorbed as Internal Energy.

Let me ask you this: if the gas expanded because of an increase in internal energy from chemical energy (eg. combustion) would you still call it adiabatic expansion? My understanding is that this is considered and exchange of heat with the surroundings and is not adiabatic.

BTW: Be sure, Andrew, that I have learned a lot by means of this discussion. I had many doubts about entropy, but answering your puzzle I have cleared them up. Thank you by the way .

Javier, I am quite enjoying this exchange too. It is a very useful and pleasant way to clarify some of these very interesting ideas, especially that very difficult concept of entropy.

AM

 

[Andrew Mason]

What happens when the barrier is released, or suddenly vaporizes is that the particles already moving toward that barrier no longer encounter reflection and continue moving into the vacuum until they reach the new wall. There is no sudden movement of every gas particle toward the vacuum. There is a gradual (on some time scale) migration of particles into the vacuum analogous to the evaporation of a liquid into the air, except there is no energy threshold to overcome.

It certainly wouldn't be gradual. It would be extremely fast.

Consider xanthym's box but with two plates with the gas confined between them. The plates are lifted at the same time and the gas expands into a vacuum on both sides. The ends of the container, some distance from the centre are connected to ratcheted springs with large spring constants. As the gas hits the ends, the springs are compressed by the translational kinetic energy of the gas which store all that translational energy by compressing only a very small distance. What is the temperature of the gas? I would say it has to be:

T_f = T_i(\frac{V_i}{V_f})^\gamma

AM

 

[OlderDan]

It certainly wouldn't be gradual. It would be extremely fast.

Consider xanthym's box but with two plates with the gas confined between them. The plates are lifted at the same time and the gas expands into a vacuum on both sides. The ends of the container, some distance from the centre are connected to ratcheted springs with large spring constants. As the gas hits the ends, the springs are compressed by the translational kinetic energy of the gas which store all that translational energy by compressing only a very small distance. What is the temperature of the gas? I would say it has to be:

T_f = T_i(\frac{V_i}{V_f})^\gamma

AM

I agree with you if the ends of the box are free to move. The gas particles would then transfer energy to the walls. Work would be done and the temperature would drop. I just don't think the idea of an oscillating center of mass works. There would certainly be a wave front as the faster particles moved ahead into empty space, and they are certainly going to hammer the ends when they get there. But a wave front does not mean there is going to be a net movement of the center of mass past the middle of the box.

I'm not arguing against you conclusion. In fact I have some some serious heartburn with the concept of a massless barrier confining the gas in the first place. Such an idea is preposterous on its face. The notion is a violation of conservation of momentum. What it is saying is that the particles in the middle of the box are magically reversing direction without interacting with anything to absorb their lost momentum. Can't happen; never will happen; and such a concept probably has no place in defending any argument about how even an ideal gas will actually behave. The only thing that can confine a gas is a real barrier, and expansion against that barrier is going to move it and do work.

 

[Clausius2]

I think I know where are you getting at. But to say the truth we are lost in semantic issues. I have discovered you're very difficult to get convinced.

That is the change in entropy of the system only, not system + surroundings. As long as that is clear, I agree.

As dS=\delta Q_R/T we understand the entropy of a system. Of course there is a change of surroundings entropy, so that the change of universe (surroundings+system) entropy is dS_{universe}=dS_{system}+dS_{surroundings}. We will assume dS_{surroundings}=0 hereinafter, therefore the process is so-called adiabatic. This is the definition of an adiabatic process. You have mentioned processes of internal chemical reactions (combustion), or electrical heat dissipation in your example. Both of them had an insulated (adiabatic) container. After thinking of it for a while:

I) we agree that the example of electrical-heat dissipation is not an adiabatic process. It is because dS_{surroundings}&lt;0 from the electrical generation system.

II) the example of chemical reaction and internal viscous dissipation is not as clear as the last one. Although dS_{surroundings}=0 the release of chemical reaction heat such as in a combustion process, or the heat released by internal fluid friction, both of these effects have the same behavior of an inflow of external heat. Semantically, we would say they are systems with adiabatic walls, but thermodynamically their behaviours are not similar to an adiabatic process, so that we can conclude that they are not adiabatic processes. But they have an essential difference with an usual non-adiabatic process: there is no heat inflow and so none impact on the surroundings. The irreversibility doesn't come from external heat inflow issues dS_{surroundings}, but it comes from internal entropy generation. The combustion process is an irreversible one which is enclosed to an increasing of entropy due to chemical reaction, while internal viscous dissipation is a source of entropy due to the heat released.

As a first conclusion, I think you're right when you said internal viscous dissipation is not an adiabatic process. Maybe it is the key of this whole discussion. Let's try to summarize it:

i) the irreversible adiabatic process can be described as an adiabatic one because the walls are adiabatic, there is no entropy exchange with the surroundings.

ii) such process can be viewed as a non-adiabatic one, because it is a heat internal release and such heat release behavior is mathematically similar to a heat external inflow. The main difference between both behaviors is that the first one doesn't make any effect on the surroundings, all the entropy generated is done inside the system; while the second one has an effect on the surroundings, the whole irreversibility is generated by means of the sum of the surroundings effect and system effect.

Maybe the two points of view are right. The fact is that I have always believed that the quantity \delta Q in entropy was referred to a heat exchanged through the boundaries, and that it is the true and original sense of the entropy definition. My opinion is that such effects of internal viscous dissipation are so difficult to treat them with "thermodynamic usual equations" that physicists and engineers model them as some kind of additional entropy generation ad hoc, as I made in my last post. After thinking of this, I realize your point of view is quite right, but I think my interpretation of the event is right too. I mean, it could be said the irreversibility comes from a non effective adiabatic behavior as you said, but it could be said to that the irreversibility comes from an internal entropy generation due to a heat release although the whole system can be considered adiabatic from the usual point of view of dS_{surroundings}=0.

I don't see why not. Can you show me some actual data on this? It seems to me that since the decrease in internal energy of the gas is converted entirely to external work, so long as that work is not being sent back to the gas in the form of heat, the rapidity of the expansion does not affect entropy. The only way the gas can fall off the isentropic curve is if some of the external work done by the gas is converted to heat and let back into the system.

Sorry, I don't have any piston filled with gas at hand to check it .
I have said to you that the irreversibility (and so the not coincidence with an isentropic trajectory) comes from the proper fluid field. If the process is too fast, sure there will be internal dissipation and viscous effects. Such entropy increasing (which we agree it exists) has the effect of taking away the final equilibrium point from the same isentropic curve. That's because dS_{surroundings}=0 and then dS_{system}&gt;0. There must be a loose of Availability somewhere, and in order to reinstaurate the original state there must be employed more work than the exerted in the expansion.

As a second conclusion, any rapid (non-cuasiestatic) process is irreversible due to the proper flow field produced. Such flow field might enhance internal viscous dissipation, which could be modeled as an external heat inflow but taking into account that dS_{surroundings}=0 (which right now seems to me puzzling and without any sense), or it can be modeled too as a volumetric heat release which provokes an additional source of internal entropy generation (which I called d\sigma).


What do you think? Have we come to some agreement?

 

[Andrew Mason]

I think I know where are you getting at. But to say the truth we are lost in semantic issues.

Perhaps. We seem to disagree over the meaning of adiabatic. But it is an interesting point.

I have discovered you're very difficult to get convinced.

Yeah. part of my legal training.

As dS=\delta Q_R/T we understand the entropy of a system. Of course there is a change of surroundings entropy, so that the change of universe (surroundings+system) entropy is dS_{universe}=dS_{system}+dS_{surroundings}. We will assume dS_{surroundings}=0 hereinafter, therefore the process is so-called adiabatic. This is the definition of an adiabatic process. You have mentioned processes of internal chemical reactions (combustion), or electrical heat dissipation in your example. Both of them had an insulated (adiabatic) container. After thinking of it for a while:

I) we agree that the example of electrical-heat dissipation is not an adiabatic process. It is because dS_{surroundings}&lt;0 from the electrical generation system.

In my example, the electrical generation system and heating coil is inside the insulated container. Why would the entropy of the surroundings increase?

As a first conclusion, I think you're right when you said internal viscous dissipation is not an adiabatic process. Maybe it is the key of this whole discussion. Let's try to summarize it:

i) the irreversible adiabatic process can be described as an adiabatic one because the walls are adiabatic, there is no entropy exchange with the surroundings.

If that is your definition of adiabatic, then ok. But this is not how adiabatic is normally defined. As I would understand and use the term, "Adiabatic" refers to an absence of heat inflow/outflow into/out of the system. ie. a process that occurs without heat entering or leaving the system.

Heat can be added to the system in at least two ways:

1. heat being transferred from surroundings to system;
2. Energy in a form other than heat being converted to heat inside the system (whether or not the energy is transferred from the surroundings).

In the first case (1.) there is an exchange of heat with the surroundings causing a decrease in entropy of the surroundings and an increase of the entropy of the system. (1.) is NOT adiabatic. We both agree.

In the second case (2.), there is NO EXCHANGE OF HEAT with the surroundings. There should be no decrease in the entropy of the surroundings while there is an increase in the entropy of the system. (2.) is adiabatic according to your definition (dS_{surroundings} = 0) but is NOT adiabatic under my definition as there is a flow of heat into the system.

Maybe the two points of view are right. The fact is that I have always believed that the quantity \delta Q in entropy was referred to a heat exchanged through the boundaries, and that it is the true and original sense of the entropy definition.

I think this may be the essential difference between us.

I don't think the dQ (in dS=dQ/T) refers to heat transferred through a boundary. dQ is the heat added to the system. It need not be transferred as heat from the surroundings. The heat can be added to the system from an internal supply of energy (eg. chemical, gravitational, nuclear) converted to heat and entropy will increase with no decrease in entropy of the surroundings. And if that occurs,the process cannot be adiabatic.

My opinion is that such effects of internal viscous dissipation are so difficult to treat them with "thermodynamic usual equations" that physicists and engineers model them as some kind of additional entropy generation ad hoc, as I made in my last post. After thinking of this, I realize your point of view is quite right, but I think my interpretation of the event is right too. I mean, it could be said the irreversibility comes from a non effective adiabatic behavior as you said, but it could be said to that the irreversibility comes from an internal entropy generation due to a heat release although the whole system can be considered adiabatic from the usual point of view of dS_{surroundings}=0.

I think we change the meaning of adiabatic if we say that an adiabatic process can result in a change in entropy. If there is a change in entropy, there must be a flow of heat into/out of the system. And that means, by definition, that ds \ne 0

What do you think? Have we come to some agreement?

Almost. The difference between us appears to be whether a process that results in a change of the entropy of a system without any change in the entropy of the surroundings can be adiabatic.

AM

 

[Q_Goest]

Adiabatic is a process without heat transfer, which is not the same as isentropic (no change in entropy).

For a process which is both adiabatic and reversible, such as the compression or expansion of a gas in a cylinder, the process is also isentropic.

For a process which is adiabatic and irreversible, such as an expansion of gas in a pipeline, the process is not isentropic. Note that as the gas flows through a pipeline and expands, there is no conservation of mechanical energy. In this case, the process would be isenthalpic.

 

[Clausius2]

We are nearer each other right now.

In my example, the electrical generation system and heating coil is inside the insulated container. Why would the entropy of the surroundings increase?
[\quote]

Ah! is the generator inside the container also?. Well, I forgot it. I thought it was outside.

If that is your definition of adiabatic, then ok. But this is not how adiabatic is normally defined. As I would understand and use the term, "Adiabatic" refers to an absence of heat inflow/outflow into/out of the system. ie. a process that occurs without heat entering or leaving the system.

That's the same definition as mine. If there is no heat inflow/outflow then dS_{surroundings}=0. I think it has been the traditional definition of an adiabatic process.

Heat can be added to the system in at least two ways:

1. heat being transferred from surroundings to system;
2. Energy in a form other than heat being converted to heat inside the system (whether or not the energy is transferred from the surroundings).

In the first case (1.) there is an exchange of heat with the surroundings causing a decrease in entropy of the surroundings and an increase of the entropy of the system. (1.) is NOT adiabatic. We both agree.

In the second case (2.), there is NO EXCHANGE OF HEAT with the surroundings. There should be no decrease in the entropy of the surroundings while there is an increase in the entropy of the system. (2.) is adiabatic according to your definition (dS_{surroundings} = 0) but is NOT adiabatic under my definition as there is a flow of heat into the system.

I think this may be the essential difference between us.

I agree.

I don't think the dQ (in dS=dQ/T) refers to heat transferred through a boundary. dQ is the heat added to the system. It need not be transferred as heat from the surroundings. The heat can be added to the system from an internal supply of energy (eg. chemical, gravitational, nuclear) converted to heat and entropy will increase with no decrease in entropy of the surroundings. And if that occurs,the process cannot be adiabatic.

I think we change the meaning of adiabatic if we say that an adiabatic process can result in a change in entropy. If there is a change in entropy, there must be a flow of heat into/out of the system. And that means, by definition, that ds \ne 0

As I was hoping, you're stating the definition of an isentropic process right now. But for you isentropic=adiabatic.

Keep in touch with this thread, I will post later an example of such a problem, and we will play with the equations to extract the final conclusions, ok?

 

[Clausius2]

Adiabatic is a process without heat transfer, which is not the same as isentropic (no change in entropy).

For a process which is both adiabatic and reversible, such as the compression or expansion of a gas in a cylinder, the process is also isentropic.

For a process which is adiabatic and irreversible, such as an expansion of gas in a pipeline, the process is not isentropic. Note that as the gas flows through a pipeline and expands, there is no conservation of mechanical energy. In this case, the process would be isenthalpic.

That's our discussion subject. I am agree with you, but Andrew argues that in the sudden expansion of a gas in pipeline the process is not adiabatic. Why? because the irreversibility comes from internal viscous dissipation originated inside the gas. Such dissipation is heat transformed further in internal energy. So that it seems it has been an inflow of "some kind" of heat inside the system. Do you understand his point?.

 

[Andrew Mason]

As I was hoping, you're stating the definition of an isentropic process right now. But for you isentropic=adiabatic.

Exactly:
http://scienceworld.wolfram.com/physics/Adiabatic.html

Keep in touch with this thread, I will post later an example of such a problem, and we will play with the equations to extract the final conclusions, ok?

Ok.

AM

 

[Q_Goest]

I guess we're all in agreement pretty much, but thought I'd toss out a few thoughts here to make sure...

the sudden expansion of a gas in pipeline the process is not adiabatic.

A real gas which expands as the pressure drops traveling down a pipeline can be modeled as adiabatic if
1) the pipe is well insulated
2) the temperature of the gas does not increase or decrease due to this expansion but stays constant and is equal to the temperature of the environment.

Note that when a gas expands isenthalpically, it can either increase in temperature (ex: Helium), decrease in temperature (ex: Nitrogen), or stay the same (ex: any gas at it's inversion temperature such as hydrogen at -183 C). It all depends on how the molecules react to the expansion.

Adiabatic only refers to heat transfer from the environment, not any kind of viscous dissipating forces induced by flow which can affect the temperature. Hope that helps.

 

[Andrew Mason]

I guess we're all in agreement pretty much, but thought I'd toss out a few thoughts here to make sure...


A real gas which expands as the pressure drops traveling down a pipeline can be modeled as adiabatic if
1) the pipe is well insulated
2) the temperature of the gas does not increase or decrease due to this expansion but stays constant and is equal to the temperature of the environment.

Note that when a gas expands isenthalpically, it can either increase in temperature (ex: Helium), decrease in temperature (ex: Nitrogen), or stay the same (ex: any gas at it's inversion temperature such as hydrogen at -183 C). It all depends on how the molecules react to the expansion.

Adiabatic only refers to heat transfer from the environment, not any kind of viscous dissipating forces induced by flow which can affect the temperature. Hope that helps.

Here is the conundrum:

1. What would you call an expansion of gas that:

a) does work without any heat exchange with its surroundings,
b) in which the work is in the form of electrical energy,
c) in which that electrical energy is passed through an electrical resistance inside the gas container and used to heat the gas?​

2. If you say that 1. is adiabatic, how would you characterize and expansion of the same gas, doing the same work where: that electrical energy is passed through an electrical resistance outside the gas container and converted to heat and the heat is then allowed to flow back to raise the temperature of the gas?

The second is clearly non-adiabatic. I am having trouble differentiating between the two.

AM

 

[Q_Goest]

AM,
Put a control volume around the gas, and around the gas only. This could also be called a "control mass" since there is no mass entering or exiting the volume.

For case 1, if there is no heat flux across the control surface, the process is adiabatic. If the gas is doing work, and assuming the work it does is completely reversible (which will not be perfectly true in a real world situation, but a close approximation) then the process is also isentropic.

For case 2, I'm not sure what you're getting at regarding the use of that work to create electric energy. The gas expands to do work, and that's a mechanical process, so if it expands through a turbine, or pushes on a piston, it must do that first, and it is that motion which is used to create electrical energy. Regardless of how the electrical energy is created though, the process is still adiabatic and isentropic.

For case 3, there is heat flux passing through the control surface and entering the control volume. For the case you created, the work done is equal to the heat flux. So energy in the form of work leaving the control volume is equal to the energy in the form of heat entering the control volume. If this is a closed volume, you could use the first law to describe this process and what you'd find is that, assuming there is no mass flow into or out of the control volume, the change in internal energy of the control volume will be zero. As you point out, this is not an adiabatic process since there is heat flux across the control surface.