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Weird ideal gas equation question

  1. Sep 2, 2011 #1
    if a gas expands isothermally but irreversibly from V=1litres to V=2litres under constant external 1atm pressure,

    if i want to know the workdone by the gas,

    should i be using

    1) WD = integral pdV = nRT ln V

    or

    2) WD = pexternal(Vfinal-Vinitial)

    i realise they give different answers, since T was not specified, i assumed it to be 298K

    thanks!
     
  2. jcsd
  3. Sep 3, 2011 #2
    sorry but there is an error in your problem, a quasi static process cannot be both isothermal and isobaric. if it happens then volume will be constant as pV=nRT ,condition if number of moles not changes(closed system).
     
  4. Sep 3, 2011 #3
    :(

    weird......

    ok thanks!
     
  5. Sep 4, 2011 #4

    Andrew Mason

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    The internal gas pressure is not constant, just the external pressure. So you can have an isothermal expansion against a constant external pressure. You just have to start with an internal gas pressure that is much higher than the external pressure and add heat as it expands to keep the temperature constant. The work done does not depend on the temperature or pressure of the gas (so long as the gas pressure is higher than the external pressure). [Note: It is not possible to have an isothermal isobaric compression, however, for the reason given by Multiverse].

    In this case, the work done is the integral of PdV over the change in volume where P is the external pressure. Since P is constant what does this integral work out to?

    AM
     
  6. Sep 6, 2011 #5
    i see thanks everyone

    just pext(v2-v1) i suppose,

    btw, i was reading my notes and it said that for expansion,

    the WD(irreversible) is always < WD(reversible)

    but why for an adiabetic expansion, the T(irreversible) > T(reversible)?

    shouldn't it follow WD(irreversible) < WD(reversible) ? since WD is proportional to T?
     
  7. Sep 7, 2011 #6

    Andrew Mason

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    The work done BY the gas in an expansion is [itex]\int P_{ext}dV[/itex]. The work done in a non-quasi static (irreversible) expansion is less than [itex]\int P_{int}dV[/itex] since [itex]P_{ext}<P_{int}[/itex]. However, [itex]\int P_{ext}dV = \int P_{int}dV[/itex] is the work done in a reversible expansion where the internal and external pressures are equal (ie differ by an infinitessimal amount). If the external pressure is any greater then there is no expansion, so the reversible expansion is the maximum amount of work you can get.

    This applies to all expansions. In an adiabatic expansion, the internal pressure constantly decreases. So, to have a reversible adiabatic expansion the external pressure has to be constantly decreasing so as to be constantly just a bit lower than the internal pressure of the gas. Since the irreversible adiabatic expansion does less work than the reversible adiabatic expansion (when expanding the same amount) the decrease in internal energy is less for the irreversible expansion: first law. Since, for an ideal gas the internal energy is proportional to temperature, the final temperature of an ideal gas will be higher in the irreversible process than in the reversible one.

    AM
     
    Last edited: Sep 7, 2011
  8. Sep 7, 2011 #7

    Ken G

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    I don't think so. It's a little hard to interpret exactly what the conditions are, but in my view, what is happening is that the gas is being held at fixed temperature T, and so allowed to have a pressure proportional to 1/V as V increases. So the work done by the gas is the integral of PdV, which is integral of dV/V = ln 2 times the initial value of PV. So I get work done by the gas is PV ln 2. It shouldn't make any difference if it is done reversibly or not (as long as the irreversibility is due only to heat flow and the gas stays in equilibrium with itself), that just relates to the work done on the environment, which is not what was asked so something seems strange about the question.

    Another interpretation is that if the irreversibility means the gas is not even maintained in thermodynamic equilibrium with itself, then I can easily give a case where the work done by the gas is zero-- instantaneous expansion of the container by an external force. In that case, the isothermality can be maintained without any input of heat, it would also be perfectly adiabatic. But such a gas wouldn't have a temperature at all in the interim period, so my first interpretation seems the most natural to me.
     
  9. Sep 7, 2011 #8

    Andrew Mason

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    The pressure of the gas cannot remain constant if n and T are held constant during an expansion: P=nRT/V. I don't think the question contemplates adding more gas. You can't maintain constant internal pressure and gas temperature if V changes.

    So this is an isothermal expansion where the gas pressure starts off much higher than the 1 atm. external pressure. Heat flows into the gas to maintain constant temperature during expansion but the pressure decreases as it expands until it reaches 1 atm (external pressure) when it stops expanding.

    So the answer [itex]W = P_{ext}(V_f-V_i)[/itex] is correct.

    The scenario you have suggested can only apply if there is effectively 0 external pressure (free expansion). That is not the case here. One would have to expand the container faster than the fastest molecule in order to achieve a free expansion!

    AM
     
    Last edited: Sep 7, 2011
  10. Sep 7, 2011 #9

    Ken G

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    That's why I said the pressure scales like 1/V, and used that in my calculation. The outside air pressure stays the same, but that is quite irrelevant to the work done by the gas-- another strange thing about the question, it gives unnecessary information.

    I assume there is no added gas, yes.

    I don't see anything in the question that requires that the internal gas pressure remain constant. Indeed, the irreversibility tends to make me expect the internal gas pressure should be allowed to be different from the external gas pressure, but that's one of the unclear things about the question-- the reason that the expansion is irreversible. That reason will matter to the work done, for example free expansion does no work.
    Again, the external pressure is irrelevant, as is the heat flow. All you need is that the expansion is isothermal, and you know that the work done is PV ln 2. Since we don't know P unless it equals the external pressure, my interpretation is that it starts out at 1 atm, but something is happening to the container to make the gas expand irreversibly. Or maybe you're right, and we're supposed to assume it ends up at 1 atm instead of starting there-- in which case it still does PV ln 2 work, but the P is the initial pressure, which would have to then be 2 atm instead of 1 atm. So if P=2 and V=1 initially, the answer would then be 2 ln 2 instead of just ln 2, in those units.
    No, that is the work done on the outside, it is not the work done by the gas. If you are right that the pressure starts off at 2 atm (which seems reasonable), then for the container to cease expanding when the volume is 2 liters, the kinetic energy associated with the expanding container will need to be removed somehow. That's why the work done by the gas will be more than the work done on the surroundings, in that scenario.
    No, the scenario I suggest doesn't care at all what the external pressure is-- we are asked the work done by the gas, so all we need is the initial P and V of the gas, and then the work done by the isothermal gas is PV ln 2.
     
  11. Sep 7, 2011 #10

    Andrew Mason

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    ?? What are you suggesting that the gas does work ON if not against the outside air pressure? The work done by the gas in an expansion is not determined by the internal gas pressure. It is determined by the external pressure. The precise internal pressure is not relevant. It just has to be greater than the external pressure. That is why internal pressure is not given but external pressure is.
    The two are the same in an expansion! The work done BY the gas in an expansion is [itex]\int P_{ext}dV[/itex]. Seehttp://www.transtutors.com/chemistry-homework-help/chemical-thermodynamics/expansion-of-an-ideal-gas.aspx" [Broken]. Be careful, there are many incorrect ones. This issue tends to confuse a lot of people.

    Knowing the internal pressure is essential if you are asked to find the work done ON the gas in a compression. So you may be thinking of positive work done ON the gas in a compression, rather than positive work done BY the gas in an expansion. The work here is done BY the gas is done on its surroundings, so knowing the pressure exerted by the surroundings is essential.

    If the gas pressure is greater than the external pressure by a finite amount, the expansion is not quasi-static and is, therefore, not reversible (the direction cannot be reversed by an infinitessimal change of conditions).

    AM
     
    Last edited by a moderator: May 5, 2017
  12. Sep 8, 2011 #11
    thanks i understand,

    but with regards to Ken G and AM debate,

    i have a new question , namely, what is the definition of WD?

    for irreversible expansion, Pint > Pext

    so assume Fint(500N) > Fext(200N)

    so lets say i am the gas. if i need to use 500N(Fint) to push the lid 1m,

    but since newton's third law requires action reaction forces to be equal

    will the external force now be 500N? (this is the idea for reversible reaction right?)

    or will the external force now be 200N ? (this is irreversible right?)




    so if i spend 500N(internal force) to push lid 1m , but the lid only has 200N(external force from constant ext pressure) to counter,

    then what is the work done by the gas? 500J or 200J?

    newton's third law is wrong here? since it says the Pint = Pext, and hence Fint = Fext should be 500N opposite direction

    so if we use 200N to calculate the work done by the gas, aren't we implying that newton 3L is wrong?

    or is time now a deciding factor?

    so that if i use 500N to fight a -200N , my net F = 300N

    so shouldn't the work done be the difference between internal and external pressure? net pressure?

    but my internal pressure is not constant, since it scales with 1/V like what Ken G says?

    so i need to integrate?
     
    Last edited: Sep 8, 2011
  13. Sep 8, 2011 #12

    Andrew Mason

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    A very good question. This is why this area can be so confusing! The problem stated that the external pressure was a constant 1 atm. This would be the case in blowing up a limp balloon with a tank of pressurized helium, for example (at least while the balloon remained limp). In the case of piston, the external pressure is not constant due to the acceleration of the piston, as you note.

    In this case the piston/cylinder has mass. (We will assume friction is negligible to make the analysis simpler). So the mass accelerates as the gas expands since there is a net force on it: F = ma = (Pgas x A - Pext x A). Since it accelerates, there is a reaction force on the gas that is equal and opposite, so the effective external pressure is Pext + ma/A = Pext + (Pgas - Pext) = Pgas. So you would think that the work done is [itex]\int P_{ext}dV = \int P_{int}dV[/itex], the same as in a reversible expansion. Not quite.

    The reason is that the gas is in a dynamic ie non-equilibrium state during this process. The internal pressure is not given by P=nRT/V. The pressure is less due to the fact that there is a gas flow (ie. Bernoulli principle).

    When expansion stops (eg. the piston hits the end of the cylinder) the kinetic energy of the centre of mass of the gas will become thermal kinetic energy, the gas will return to equilibrium and the pressure will increase so that P will equal nRT/V.

    AM
     
  14. Sep 8, 2011 #13

    Ken G

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    That depends a lot on the ambiguities of the question. The big issue is, what is keeping the gas in thermodynamic equilibrium? This is a necessary requirement to state that the gas is "isothermal". Free expansion of a 2 atm gas against a 1 atm environment is not going to be an isothermal gas-- in fact, very quickly much of the kinetic energy in that gas is going to be bulk flow energy at near the sound speed, and the same for its environment. So we need some kind of container for this gas. The next question is, can the container support forces (and therefore have work done on it)? If the answer is "yes", then the internal gas may indeed be maintained in an isothermal state, and will obey the ideal gas law. If the answer is "no", then again the internal gas cannot be maintained in thermodynamic equilibrium, and cannot be said to be isothermal.

    My solution is the only one consistent with an "isothermal" ideal gas-- it must have pressure P that scales with 1/V (yes?) and it must do work PdV (yes?). There is no other possibility consistent with that information. Note your solution does not depend at all on the temperature of the gas, so why are we told that the gas is isothermal?
    If that were true, no information about the internal gas would be needed, especially not that it was isothermal.

    It certainly seems like there is something weird with the question, I'd like to have a better idea about the overall scenario that is imagined. I can easily come up with situations where you have an isothermal equilibrium in the internal gas, and it is irreversibly expanded, and it does the work I mentioned above, not the work done on the external gas. For example, we can have gas expanding against a piston in thick molasses. The piston in molasses is how you can sustain a pressure difference between inside and outside, and the piston will gradually expand until the pressure equalizes. This will be an irreversible expansion, and the molasses will dissipate the extra work between what was done by the gas (the issue of the question) and what was done on the surrounding air.
     
  15. Sep 9, 2011 #14

    Andrew Mason

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    You raise some valid points. However, while it may be difficult to fashion an apparatus that fulfills the requirements of the problem, it is not impossible. We just have to accept the question and work out the solution based on the information given.

    Could the apparatus not simply be a cylinder of gas at a pressure greater than 1 atm raising a frictionless piston whose mass x g /area + external pressure = 1 atm. The gas would expand until the piston reaches the end of the cylinder when it suddenly stops. The external pressure would be constant. There would have to be heat flow into the gas as it expands to keep it at constant temperature. Would that not fit the information provided?

    I don't think we are supposed to ask why the question is posed that way. The information may be important or it may not be. The fact that the process is isothermal is not needed in order to solve the problem. Perhaps it was put there to see if the student could determine that it is not relevant.
    Since the force required to move the piston would be greater with the friction than without, if the external air pressure is 1 atm. adding friction would increase the external pressure while it expanded, would it not?

    AM
     
  16. Sep 9, 2011 #15
    if thats the case, why for irreversible rxn, we take the external pressure to calculate work done? shouldn't the work done by gas be equal to the internal pressure of gas x dV ? although it is not given by P=nrT/ V as you said.


    this is the point about the gas equalizing with the surroundings so that its final pressure is the external pressure right?

    so as per my question above, why do we take the work done by gas to be the constant external pressure? shouldn't it be the changing internal pressure?
     
  17. Sep 9, 2011 #16

    Ken G

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    But you are not using the information given. Where do you use that the gas is isothermal?
    What suddenly stops it? That will involve dissipation. If you do anything that maintains the ideal gas law inside the piston, then the work done by the internal gas is what I gave, and some dissipation will account for the difference between that and the work done on the external gas.
    But it is relevant. If you know the gas is isothermal, then it has a T, then it is in thermodynamic equilibrium, then it does the work I gave. The work on the external gas is something different.
    The external air pressure is still 1 atm. Does that count as expansion "under 1 atm"? Perhaps not, I really don't know what the question has in mind, it's not clear where the extra work is ending up. Maybe they mean that in the local frame of the fluid, the gas maintains a Maxwellian with that T, and you are right that essentially free expansion is envisaged, but then the expansion speed will be of order the sound speed, perhaps faster. When the gas reaches 2 liters, it will have all kinds of kinetic energy, which is where the extra work will be at that time. The new situation will not be a static state-- when the forces balance, there will then have to be overshoot. There might even be a shock! In the spirit of the thermodynamics chapter they seem to be on, I would still expect the work done by the gas to be the PdV of the gas doing the work, and that's the only assumption I made in my solution.
     
    Last edited: Sep 9, 2011
  18. Sep 9, 2011 #17

    Ken G

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    I should think the pressure of the internal gas would indeed be P=nrT/V, that should correspond to the meaning of an "isothermal" gas. That's why I feel P should scale like 1/V, and the work done like ln V.

    That's my read of the situation, but I feel there could be some dependence on just how this system is being pictured. "Irreversible expansion" could mean a lot of things-- for example, we could have a box that the gas is in, and we suddenly grow the box from 1 liter to 2 liters, and then the work done by the internal gas would be zero! So there needs to be more of a sense of just what the scenario is here.
     
  19. Sep 9, 2011 #18

    Andrew Mason

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    Perhaps there are other parts to the question - eg. what is the heat flow into the gas?
    The only way to maintain the ideal gas law is to have a quasi-static process which is a reversible process. This is not a quasi-static process. It is irreversible.
    You are correct in saying that temperature is only defined for a gas in thermal equilibrium. It is also true that any dynamic change in the gas causes the distribution of speeds of the gas molecules to deviate from a Maxwell-Boltzmann distribution and, therefore, no real process can be isothermal. In that respect the only truly isothermal process is a reversible process.

    In presenting a scenario in which the gas is expanding isothermally and irreversibly, the question is essentially stating that the internal energy of the gas does not change during the process.

    It doesn't say that the work is done on a gas.
    Would it be any different if it said the irreversible expansion was "adiabatic" instead of "isothermal"? There would be no T defined while it was expanding. There would just be no heat flow into the gas. The work would still be Pext(V2-V1).

    AM
     
  20. Sep 9, 2011 #19

    Andrew Mason

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    Because in this case it is said that the gas expands against a constant external pressure. In your example it does not.

    For example a gas in a cylinder of cross-sectional area A lifting a weight, mg, on a piston through a distance h. There is also substantial friction force, F, between cylinder and piston as it expands until the internal gas pressure is equal to 1 atm. The external pressure is constant: Pext = (mg+F)/A = 1 atmosphere. The volume change is Ah. The work done is [itex]W= (mg+F)h = PAh = P_{ext}\Delta V[/itex].

    AM
     
  21. Sep 9, 2011 #20

    Ken G

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    The heat input into the gas depends on whether or not we have a free expansion (so there's a lot of kinetic energy in the gas when the size is 2 liters) or quasi-steady expansion through a dissipative system (like the piston in molasses). So we can't answer that until we know this.
    Not necessarily, itt depends on the source of irreversibility. There are actually three ways for a process to be irreversible that I can think of-- it can move heat across a temperature difference (which can happen in a quasi-static way, like in a Carnot cycle), it can convert work into heat (again this can be quasi-static, and can be a result of a Carnot cycle if desired), or it can be something other than quasi-static. So which is the source of irreversibility in this problem? We aren't told, but we are told the gas is "isothermal." So that sounds like the second option to me-- like the piston in molasses. If the gas is isothermal, then the work done by the gas has to be balanced by the heat input into the gas, there just isn't any other option, unless we are allowing the gas to end up with a lot of high-speed flow energy that is not counted in the temperature because it is not thermalized (but then the final state of 2 liters is not a static state, and the work done by the gas will still be 2 ln 2 because it will partly show up in this kinetic flow energy).
    The expansion against a piston in molasses can be truly isothermal, you just need good thermal contact with a reservoir and very thick molasses to insure that any flow energy is immediately dissipated as heat.
    If you take that interpretation, then there can be no bulk flow energy in the gas, so you are ruling out your own free expansion scenario. I am frankly not sure what the question intends, but it seems like PdV of the internal gas is the safest course.
    It asks for the work done by the internal gas.

    If the expansion were adiabatic, then we would not have P ~ 1/V in the internal gas, we'd have constant entropy, and that would give us P(V), but the approach would still be to integrate PdV of the internal gas, not the external gas.
     
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