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 celsius 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 independent 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