Adiabatic Expansion: Solve for Final Volume of Gas

In summary: What you have tried to do is to show us that you know the basics and that you're able to use them. But I'm sorry to say that you have just shown the opposite.In summary, a frustrated student asks for help with an ideal monatomic gas expansion problem. They tried using the ideal gas law but it didn't work, and now they are unsure if they are approaching the problem correctly. Another user suggests using the adiabatic relation PV^gamma = constant, which the student is familiar with but is not sure how to apply in this situation. A discussion ensues about the correct approach and the importance of understanding the underlying principles. The student ultimately receives a solution using the adiabatic relationship for temperature and volume.
  • #1
espnaddict014
7
0
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?
 
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  • #2
espnaddict014 said:
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 [itex] PV^\gamma [/itex] is a constant. Try writing [itex] V^\gamma [/itex] as [itex] V*V^{(\gamma-1)} [/itex] and replacing the product PV with what you know from the ideal gas law. I think there is no escaping the [itex] \gamma [/itex] in this problem, so you will have to get a reasonable value from your tables
 
  • #3
Did you use the proper units...convert celcius to kelvin?
 
  • #4
espnaddict014 said:
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

[tex]PV^{\gamma} = constant[/tex]

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

[tex]TV^{\gamma -1} = constant[/tex]

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.
 
  • #5
I would imagine that P is constant. Internal energy is directly converted to [tex]P \Delta V [/tex] work, work done by expanding against a constant pressure atmosphere/constant pressure situation.
 
  • #6
GCT said:
I would imagine that P is constant. Internal energy is directly converted to [tex]P \Delta V [/tex] 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.
 
  • #7
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.
 
  • #8
nevermind...I was unfamiliar with the concept, confused it with a vague recollection of another situation.

espnaddict, what course is this for?
 
  • #9
Dr.Brain said:
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: [itex]PV^\gamma[/itex] per Dan and Zapperz rather than develop it from first principles every time you use it?

AM
 
  • #10
^^^^
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.
 
  • #11
Dr.Brain said:
^^^^
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.
 
  • #12
Dr.Brain said:
^^^^
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:

[tex]nc_vdT=-PdV[/tex] 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.
 
  • #13
Dr.Brain said:
^^^^
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 [itex]dU = -dW[/itex] and [itex]nC_v = dU/dT[/itex]:

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

[tex]\frac{Pdv + Vdp}{n(C_p-C_v)} = -\frac{Pdv}{nC_v}[/tex]

[tex]Pdv + Vdp = - Pdv\frac{(C_p-C_v)}{C_v} = - Pdv({\frac{C_p}{C_v} - 1})[/tex]

[tex]Vdp = - Pdv\frac{C_p}{C_v}}[/tex]

[tex]\frac{dp}{P} + \frac{dv}{V}\frac{C_p}{C_v}} = 0[/tex]

Integrating:

[tex]\int\frac{dp}{P} + \int \frac{dv}{V}\frac{C_p}{C_v}} = K[/tex]

Integrating and letting [itex]\gamma = C_p/C_v[/itex]:
[tex]lnP + \gamma lnV= K[/tex]

[tex]PV^\gamma= e^K = constant[/tex]

AM
 
  • #14
Clausius2 said:
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
 
  • #15
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.
 
  • #16
siddharth said:
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 [itex]PV^\gamma[/itex]. 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:

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

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:

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

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
 
  • #17
Andrew Mason said:
(BTW the work done is never [itex]PV^\gamma[/itex]. 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 [itex]PV^\gamma[/itex] 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

Andrew Mason said:
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:
[tex] P_2(V_2-V_1) < \int_{V_1}^{V_2}Pdv \implies Q \ne 0[/tex]

Because the value of the inital pressure P1 is rapidly changed to P2 before the expansion actually occurs the value of
[tex] \int_{V_1}^{V_2}Pdv = P_2(V_2-V_1) [/tex]
so q=0 is still valid
 
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  • #18
.

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:

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

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

[tex] 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}) [/tex]

where:

[tex] 3: \ \ \ \ \gamma \, \ = \, \ \frac{C_{p}}{C_{v}} \, \ = \, \ 1 \, + \, \frac{nR}{C_{v}} \, \ > \, \ 1 [/tex]

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.


~~
 
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  • #19
Andrew Mason said:
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 [tex] P_1[/tex] with a free piston over it. The system is submerged into an atmosphere of pressure [tex]P_o<P_1[/tex]. 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 [tex]\delta Q=0[/tex], or which is the same [tex] dU=\delta W[/tex]. So that, the decreasing of internal energy is the same than the work done by the system over the atmosphere. Of course [tex] W=-P_o\Delta V[/tex]. And [tex]PV^{\gamma}=const[/tex] 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.
 
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  • #20
siddharth said:
Because the value of the inital pressure P1 is rapidly changed to P2 before the expansion actually occurs the value of
[tex] \int_{V_1}^{V_2}Pdv = P_2(V_2-V_1) [/tex]
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.

[tex] P_2(V_2-V_1) < \int_{V_1}^{V_2}Pdv + \int_{P_1}^{P_2}Vdp = \Delta U \implies Q \ne 0[/tex]

AM
 
  • #21
Perhaps the heat issue is resolved by taking into account chemical reactions...or not.
 
  • #22
xanthym said:
.

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:

[tex]P_1V_1 = P_2V_2 + \frac{1}{2}\rho v^2[/tex]

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 [itex]PV^\gamma = constant[/itex] 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
 
  • #23
Andrew Mason said:
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.
 
  • #24
ZapperZ said:
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
 
  • #25
Andrew Mason said:
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.
 
  • #26
ZapperZ said:
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
 
  • #27
Andrew Mason said:
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.
 
  • #28
ZapperZ said:
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: [itex]\Delta U = P\Delta V[/itex] so [itex]T_f \ne T_i[/itex].

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
 
  • #29
Andrew Mason said:
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: [itex]\Delta U = P\Delta V[/itex] so [itex]T_f \ne T_i[/itex].

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.
 
  • #30
Clausius2 said:
Imagine there is a closed cylinder filled with an ideal gas at pressure [tex]P_1[/tex] with a free piston over it. The system is submerged into an atmosphere of pressure [tex]P_o<P_1[/tex]. 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 [itex]V2-V1 = Ah[/itex] and the distance the piston moves is h. But the work done is not:

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

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

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

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
 
  • #31
Adrew Mason said:
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
:smile:
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. :wink:

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.
 
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  • #32
Clausius2 said:
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
 
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  • #33
Andrew Mason said:
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 [tex] dS=\delta Q_{reversible}/T[/tex]. 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 [tex] \delta Q_{reversible}=0[/tex] and [tex] dS=0[/tex]. 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 [tex]\delta Q_{reversible}>0[/tex], 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 [tex]d\sigma[/tex]. Then entropy can be reshaped as: [tex]dS=\delta Q_{irreversible}/T+d\sigma[/tex]. 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.


Andrew Mason said:
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.


Andrew Mason said:
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 :smile: .
 
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  • #34
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

[tex]\int Fdx = 0[/tex]

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.
 
  • #35
Clausius2 said:
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 [tex] dS=\delta Q_{reversible}/T[/tex].
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 [tex] \delta Q_{reversible}=0[/tex] and [tex] dS=0[/tex]. 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)? [itex]\int P_{gas}dv [/itex] is the work done by the gas which is [itex]\int P_{ext}dv[/itex] + 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 [tex]\delta Q_{reversible}>0[/tex], 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 :smile: .
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
 

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