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Thermodynamics: External vs Internal Pressure

  1. Sep 16, 2012 #1
    Suppose in a piston in which there is a gas, the gas exerts pressure P on surroundings, whereas the surrounding exerts a pressure P[ext] on the gas.

    In order for the gas to expand, P must be greater than P[ext]. So far I understand.

    Suppose initially that P>P[ext] and the gas expands until P=P[ext]

    This is what I don't understand is that when you calculate the work done by the gas, you use the external pressure and not the gas pressure. This makes no sense to me. After all, if you calculate the work done on something, you should use the force or pressure applied on that object, or not the other way around, correct? So why do we use external pressure instead of internal pressure when calculating the work done by the gas?

    I have heard some people say that the internal pressure of the gas over the process is undefined since the process is not quasi-static, but that should make the work done impossible to be calculated rather than just use external pressure which has no reason to be used in the problem?

    BiP
     
  2. jcsd
  3. Sep 17, 2012 #2

    morrobay

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    How are you calculating the work ?
    What is wrong with : Work = nRT ln V2 - ln V1
    for isothermal expansion. Since dV is based on internal pressure.
     
  4. Sep 17, 2012 #3
    Hello bipolarity, I see you are asking lots of questions, that's good.

    :biggrin:

    Here is a viewpoint that may help with this and other questions.

    Thermodynamics divides the universe into two parts.
    The system and the surroundings.
    They are divided by a boundary.

    For both system and surroundings state variables refer exclusively to either the system or the surroundings.
    As you are probably aware, state variable changes are only dependent upon the start and end points, not the intermediate path.

    This would be pretty boring and useless if there was no interaction between the system and its surroundings so further variables are defined by the system process.
    These variables account for exchanges across the boundary. None of these are state variables. They do not describe the state of anything.

    It is a prime assumption of thermodynamics that if a quantity defined by one of these exchange variables is passed across the boundary it is the same on both sides of the boundary.

    For example, work done by surroundings = work received by system (and vice versa)

    wsystem = wsurroundings = w

    This gives us two opportunities to calculate the value of exchange variables - either form the system or from the surroundings - because they refer to the same quantity.

    Sometimes one calculation is easier and sometimes one calculation is impossible.

    With regard to you specific question it is assumed that the surroundings are so large that there is no change to the pressure during the work exchange.

    This is not true of the system. So the system pressure changes during the work exchange.

    For the system the work, w = ∫pdv if the system pressure can be defined at all times during the process.

    For the surroundings w = ∫pdv again, but since the pressure is constant w = p∫dv = p(v2-v1)

    Which calculation is easier?

    This brings me to my final point.

    Of the state variables we can usually claim that the volume change in the system = volume change in the surroundings.

    This is, of course not generally true of other state variables.

    Does this help?
     
    Last edited: Sep 17, 2012
  5. Sep 17, 2012 #4
    This is a question which has driven thermo students crazy over the generations, usually without a satisfactory explanation. But, by looking at the problem in greater detail, you can get a better understanding of what is happening. The basic answer is that, during the expansion, the pressure within the cylinder is not uniform spatially, and is lower at the piston than away from the piston. Here is how to dope out what is happening:

    Problem statement: initially the gas pressure within the cylinder is higher than the pressure outside the cylinder, and the two regions are initially separated by a massless piston which is held in place. At time zero, the massless piston is released, and the volume of gas within the cylinder begins to increase. We assume that the external pressure on the piston is held constant throughout the expansion.

    Now, if the piston is massless, that must mean that the pressures on both sides of the piston (after release) must be equal to one another (by Newton's 2nd law). But how can that be, if one instant earlier, the pressure inside the cylinder was much higher? What roughly happens is that, immediately adjacent to the piston, a small pressure dilitation region begins to form within the cylinder in which the pressure is equal to the external pressure. But, at a certain distance from the piston, the pressure undergoes a jump change from the external pressure to the initial internal pressure. Within the dilitation region the gas is expanded to accommodate the upward movement of the piston. As time progresses, the dilitation region expands at the speed of sound until it reaches the bottom of the cylinder. At that point, the pressure throughout the cylinder is the external pressure. Since the pressure on both sides of the piston is roughly equal to the external pressure throughout the expansion, the work done by the gas and the work done on the surroundings are both equal to the external pressure times the change in volume.

    The key to all this is to recognize that, within the cylinder, the pressure is not uniform while the expansion is occurring, and the pressure at the massless piston interface must be equal to the external pressure (as required by Newton's second law) on both sides of the piston.
     
  6. Sep 18, 2012 #5

    Andrew Mason

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    A gas in a non-quasi-static process can still have a pressure and temperature that are pretty uniform. So that is really not the answer to this question. Here is how I suggest you look at it:

    The gas can do work on two things: the the system or the surroundings.

    When it undergoes a quasi-static thermodynamic process all of the work is done on the surroundings.

    If the process is dynamic (irreversible) it does some work on the system so it does not do all of its work on the surroundings. Ultimately, when equilibrium is reached, the work done on the system (including that gas itself) ends up as thermal energy in the system. In other words, thermal energy is converted into mechanical work some of which returns to the system as thermal energy.

    Since we are looking at the system in its initial and final equilibrium states (thermodynamics is necessarily a study of changes occurring between equilibrium states), the work done on the system that ends up back as thermal energy in the system is a wash - ie treated the same as if it had remained as heat. The only work done that affects the state of the system and surroundings is the work done against an external pressure.

    A good example would be the adiabatic free expansion of an ideal gas through a nozzle into an empty chamber. There is no external work done. The process is dynamic. The internal unbalanced gas pressure causes a stream of gas molecules to exit the nozzle at high velocity. Once some gas has entered the empty chamber it provides a pressure that the incoming steam has to do work against. So the gas does work - it just does work on itself. When the gas eventually reaches equilibrium, the work that it has done on itself ends up as thermal energy in the gas and its final temperature is the same as its initial temperature. It is as if no work had been done at all by the gas.

    AM
     
  7. Sep 18, 2012 #6

    DrDu

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    What drives students crazy is that their teachers have never heard of volume viscosity or second viscosity:
    http://en.wikipedia.org/wiki/Volume_viscosity
     
  8. Sep 18, 2012 #7
    As I mentioned in my earlier post #4, for purposes of answering the question posed by the OP, the details of the non-equilibrium gas dynamics occurring within the cylinder are not relevant to calculating the work done by the system on the surroundings in this particular situation. You only need to know that

    1. Within the cylinder, the pressure (or, more precisely, the stress tensor) is non-uniform spatially during the expansion
    2. As a result of the assumed zero mass for the piston, the pressure exerted by the gas within the cylinder on the piston interface (or, more precisely, minus the normal stress exerted by the gas on the piston) is equal to the external pressure throughout the expansion. This is a constraint imposed by Newton's second law applied to the piston as a free body.

    Therefore, the work done by the gas within the cylinder on the piston will necessarily be equal to the external pressure times the change in volume, irrespective of the shear viscosity and the volumetric viscosity of the gas. Of course, the other details of the non-equilibrium gas dynamics occurring within the cylinder do depend on the shear viscosity and volumetric viscosity, subject to the stress constraint imposed at the piston interface.
     
  9. Sep 18, 2012 #8

    DrDu

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    Chestermiller, I think that now we perfectly agree. What I wanted to point out is that even for changing the volume of a gas there exists a kind of friction which has to be taken into account when considering the equilibrium conditions of the free massless piston. As it is friction, it will dissipate energy and thus make a contribution to heat rather than work.
     
  10. Sep 18, 2012 #9
    Thanks everyone for the replies but I don't think we are done quite yet.
    What I a gather from reading the replies is that
    1) any type of gas expansion/compression requires the gas molecules closest to the piston to be isobaric with the external pressure, whether or not the process is reversible
    2) in a reversible thermodynamic process, all the gas molecules are allowed to equilibrate with one another so that at any given time all the molecules have the same temperature and pressure
    3) in an irreversible thermodynamic process, there is always a pressure gradient within the gas although I didn't completely understand why this happens (it's ok though)

    chester, you mentioned that due to Newton's second law, the gas pressure at the surface must be in equilibrium with the constant external pressure. I didn't understand the reason for this and how Newton's law is relevant. Are we assuming that the piston moves at constant velocity? I see no reason why this must be assumed.

    Also, Andrew, you mention that the gas can either do work on the system or on the surroundings. I thought the gas was part of a system, so how can a system do work on itself? What does it mean to do work on one's self?

    Apart from that I am making good progres! Thank you all!

    BiP
     
  11. Sep 18, 2012 #10

    DrDu

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    Bipolarity,
    your conclusion 1 is not correct. The external force on the piston has to be in equilibrium with the sum of the pressure and the internal friction in the gas. During expansion, p_int > p_ext, but the friction counteracts volume increase.
     
  12. Sep 18, 2012 #11
    Continue to ask searching questions and test the validity of statements made to you.

    +1 :wink:

    No I don't agree that the gas inside (or outside) need be isobaric with anything.
    We have not been told the mechanisms either inside or outside. One of these could be a jet or simply a circulating airstream. We assume that the external pressure is even and constant on the piston. But what if it is a vacuum? Expansion into a vacuum produces one of the simplest examples of an irreversible process.

    No I don't agree we need to take friction into account - this is a perfect gas we are discussing - in a thought experiment like this.


    I have already described the work as the measure of mechanical energy transferred acrossed the boundary.
    Usually this happens when the system boundary is moved (by the system or surroundings) in the course of the process.
    As a result of the movement of the system boundary one side gains volume the other looses it. The work on each side is ∫pdv but this may be difficult or even impossible to calculate on one side of the boundary. However the work integral is the same on both sides, as already noted.
    So yes Andrew is correct.

    Can't see where you got that idea but remember that thermodynamic processes includes heat.

    One of the simplest examples of a reversible process is a bucket of melting ice (the system).
    The system remains at constant temperature, the pressure may or may not be definable.
    The surroundings may be at all sorts of temperatures.

    It is a good idea to differentiate between the gas laws and the laws of thermodynamics since the latter are universal.
     
  13. Sep 18, 2012 #12

    Andrew Mason

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    Why? How is that possible? What if the external pressure is 0?

     
  14. Sep 18, 2012 #13

    DrDu

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    One of the charming points in considering a "massless" or at least very light piston is that it does not form a thermodynamical system on its on and also has to be in equilibrium with the external forces if we want to exclude infinite acceleration. Further boundary conditions require that the velocity of the gas has to vanish at the surface (this will be the case for arbitrary small friction already). Then a pressure difference between inside and outside can arise as friction is never negligible in the Prantl layer at the surface although this layer becomes thinner and thinner with decreasing coefficient of viscosity. For a "perfect gas" -whatever that may be- it is simply not possible to maintain a pressure difference.

    The other situation is that of a massive piston. It can store at least kinetic energy and we have to decide whether we count it to the system or to the surrounding. As it can have finite acceleration, there is no need for pressure (or more generally stress) to be equal on both sides.
    If we calculate the energy using the integral over the external pressure, we calculate the work done on the system including the piston. If we use the internal pressure (or stress in general), we calculate the work done by the system excluding the piston.

    The decisive difference is that in case of a massive piston we can maintain at least for some time a pressure difference although both the system and the surrounding are in thermodynamic equilibrium. In the case of a massless piston, a pressure difference is only possible once friction is taken into account.
     
  15. Sep 18, 2012 #14
    OK, so here is now my refined understanding of the situation:
    - Work is done when the system boundary is moved, whether by the system or surroundings.
    - If the boundary is moved by the system towards the surroundings, the system does positive work W and the surroundings does negative work equal to -W.
    - If the boundary is moved by the surroundings towards the system, the surroundings does positive work W on the system, the work done by the system is simply the negative of that.

    So ultimately is this correct:
    Let's say that initially the gas pressure is at P and the external pressure is a constant P(ext). The pressure P > P(ext) so that the gas expands until its pressure P = P(ext).

    The work done by the gas (in infinitely small interval) will be the PdV and this will be a positive quantity, but this pressure is undefined because the process is not quasi-static, so that PdV is instrinsically impossible to calculate. On the other hand, the work done by the surroundings will equal the negative of the work done by the system, but having a defined value -P(ext)dV, this can be easily calculated.

    But does this not ultimately mean that the [itex] |W_{sys}| = |PdV| = |W_{sur}| = |-P_{ext}dV| [/itex] so that [itex] ||P|| = ||P_{ext}|| [/itex] ?
    In other words, the pressures ultimately end up being equal throughout the expansion?

    BiP
     
  16. Sep 18, 2012 #15
    @Dr Du

    Where did the OP require a massless piston?

    'My piston' is actually an infinitely stretchable membrane, further the OP's "pressure exerted by the surroundings" was not defined to be gas pressure.

    We could play all night swopping fantasies.

    The purpose of this thread is to help someone at the beginning of a thermodynamics course.

    So thought experiments and idealised models are acceptable, because they are simple and lay bare the bones of the subject.

    It is important to realise that the laws of thermodynamics apply even to these models. That is the whole beauty of the subject. It can be applied to systems that only exist in the imagination and cannot be realised in reality.
     
  17. Sep 18, 2012 #16
    Yes I will go with that , but be careful about sign conventions some (especially chemists) use a different one.

    see post#4 here https://www.physicsforums.com/showthread.php?t=516447&highlight=sign+conventions

    Yes

    The pressure is not undefined because any individual part has not got a pressure, it is undefined because every individual part (infinitesimal volume) has a different pressure and you need to take an average.
     
  18. Sep 18, 2012 #17
    Ok, so in this case the molecules each have different pressures (as opposed to a reversible process?), but the overall gas pressure is still equal to the external pressure, correct?

    But if that's the case, wouldn't the expansion stop as soon as it begins, since the overall gas pressure equals the external pressure right at the beginning?

    BiP
     
  19. Sep 18, 2012 #18
    Where did I say that the internal average was equal to the external pressure?

    Incidentally you will learn from the kinetic theory and statistical mechanics that what we call temperature and pressure in classical thermo are averages at the molecular level.

    It is important not to take any model too far and to recognise when the limit has been reached. This applies to any technical discipline.
     
  20. Sep 18, 2012 #19
    I see, I'm just trying to understand the equations. The internal pressure is either equal, or not equal, to the external pressure. If the work done by the system is equal in magnitude to the work done by the surroundings (which you confirmed was true), the magnitude of the force applied by the system must equal the magnitude of the force applied by the surroundings. Dividing by the same area, the pressures must be equal ?

    BiP
     
  21. Sep 18, 2012 #20

    morrobay

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    Would this value for work , external pressure x change in volume be equal to:
    work = nRT ln V2 /V1
     
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