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I Use internal pressure to compute work done by an expanding gas?

  1. Jul 7, 2018 #1
    I would argue that if you are computing the work done by an expanding gas in a frictionless piston, in an irreversible expansion where the inner pressure is greater at all times than the (constant) outer pressure, that you should integrate the internal pressure over volume change, not multiply the volume change by the constant outer pressure.

    However, there are many sources which say that to compute work by the gas, you should just multiply volume change by external pressure. I believe that the work done by the gas is computed by integrating pressure over volume change, but that the work done by the gas plus the piston is computed by multiplying external pressure by volume change, and that the two are different. So I am wondering if all of those sources are subtly wrong, or if I'm missing something.

    Suppose that you have a frictionless piston lying on its side (so that the piston weight doesn't add any extra pressure), held in place by a latch. Inside is gas at 4 atm; external pressure is 1 atm. You release the latch and the piston expands to double its volume (so the new pressure is 2 atm), where it is caught in place by another latch. What is the work done by the gas in this process?

    Here is my argument:
    1) To find work done by a force, you integrate force over distance moved -- one of the first things learned in physics -- so the work done by the gas is found by integrating the (changing) internal pressure over the volume change, because the force exerted is proportional to the internal pressure.
    2) If you consider "the system" to be the gas plus the piston, then the work done by that system is equal to the external pressure times the volume change, because that's the work required to make room for the enlarged system.
    3) The difference between #1 and #2 can be accounted for by the fact that the expanding gas accelerates the piston and gives it kinetic energy. Once the piston locks into place, that kinetic energy is dissipated as heat energy. However, if the gas imparted that energy to the piston, that still counts as work done by the gas, even if it gets dissipated as heat once the piston stops moving.

    It's similar to the question about how if you lift an object 1 meter that weights 10N, you've done 1 Nm of work -- but, a student asks, what if you lift it 1 meter using a force of 50N, haven't you done 50Nm of work? Yes, but most of that that 50 Nm lifting work gets converted into the kinetic energy of the object, and if the object locks into place at the end, that kinetic energy will be dissipated as heat energy.

    (Now, I had assumed that the internal pressure of the gas in the piston could be computed using the ideal gas law. However, if it can't be, I think the internal pressure is obviously greater than the external pressure -- otherwise the piston wouldn't move -- and integrating it will give you a larger value for work done, than you would get by multiplying volume change by external pressure.)

    So first things first, is this correct so far?

    Because there are many sources saying that under the conditions that I've specified, the work done by the gas equals external pressure times volume change, not internal pressure integrated over volume change. (I'm talking about irreversible processes, where the two pressures differ; in reversible processes where they're the same, it doesn't matter.) If I'm right, then I think these sources are in fact wrong:
    http://www.columbia.edu/itc/chemistry/f1403/lectures_ppt/Wk_10_Mon.ppt
    https://facweb.northseattle.edu/hprice/CHEM161/Silberberg PPT to link/ch06_lecture_7e.ppt
    https://chem.libretexts.org/Textbook_Maps/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map:_Physical_Chemistry_(Atkins_et_al.)/Chapter_2:_The_First_Law/Topic_2A:_Internal_Energy
    http://www.quantumstudy.com/chemistry/thermodynamics-4/
    https://quizlet.com/77523508/chemistry-133-chapter-6-thermochemistry-flash-cards/
    http://docfish.com/wp-content/uploa...Energy_First_Law_of_Thermodynamics_HW_KEY.pdf

    (In each of those, you can find the part that I'm disputing, by searching for the phrase "work done by the gas". I found these, and many others, just by googling "pressure volume work" "work done by the gas".)

    On the other hand there are some sources which agree with me:
    https://www.grc.nasa.gov/WWW/k-12/airplane/work2.html
    which says that compute the work done by the gas, you integrate the changing internal pressure, which is clearly less than the external.

    So, which is it?

    To avoid horrible confusion, please be clear what position you're taking:
    1) Is the work done by the gas equal to (A) the external pressure times volume change (and I'm wrong), or is it (B) some larger number that you would get by integrating internal pressure over volume change (and I'm right)?
    2) If your answer to #1 is A, what exactly is wrong with my argument that the internal pressure is greater than the external pressure, and to get work done by the gas, you integrate the force over the distance moved?
    3) If your answer to #1 is A, is the NASA page in fact wrong (the one I think backs me up)?
    4) If your answer to #1 is B, then do you agree that the other sources listed above -- which say to use pressure-times-volume-change -- are in fact wrong?
     
  2. jcsd
  3. Jul 7, 2018 #2
    If it is OK with you, I would like to consider the specific example you posed as a focal problem, in which the gas is at thermodynamic equilibrium at 4 atm. initially, and the latch is suddenly released. Once this happens, the force per unit area exerted on the outside face of the piston is maintained at 1 atm. However, rather than stopping the motion of the piston when it has expanded by a specific amount, say 2x the initial volume, we are going to allow the gas to continue to expand until it again reaches thermodynamic equilibrium. I'm also proposing to initially consider a case where the piston has finite mass (Later, we can consider the case of a massless piston). OK?

    As you indicated, in solving a problem like this, it is important to precisely define the system you are considering. I would like to define the gas alone as the system being considered, and not the combination of gas and piston. With the system defined in this way, the "surroundings" are comprised of the cylinder plus the inside face of the piston. OK?

    I would also like to precisely define the following quantities to work with in our analysis (applicable to both reversible and irreversible expansions):

    ##P_{int}## = the compressive force per unit area exerted by the gas on the inside face of the piston. This force includes not only the "pressure" calculated from the ideal gas law, but also an additional contribution from viscous stresses within the gas (that are present during an irreversible expansion). So, ##P_{int}=P_{ig}+P_{viscous}##. In a reversible expansion, ##P_{int}=P_{ig}##

    ##P_{ext}## = the compressive force per unit area exerted by the inside face of the piston on the gas. From Newton's 3rd law of action reaction, ##P_{ext}=P_{int}##

    ##P_{out}## = the constant compressive force per unit area exerted on the outside face of the piston (1 atm. in our example)

    OK so far? If so, please start thinking about how you would apply Newton's 2nd law force balance o describe the motion of the piston.

    Also, to get a better intuitive feel for the differences between reversible and irreversible expansions, please consider reading my Physics Forums Insights article on this subject: https://www.physicsforums.com/insights/reversible-vs-irreversible-gas-compressionexpansion-work/

    Please get back with me when you are ready to continue.
     
  4. Jul 7, 2018 #3
    Before proceeding, can you clarify: Do you think the answer I got was correct, or do you think my answer was incorrect and you are trying to teach me to look at the problem differently?

    Re-reading through what you wrote in the Stack Exchange comments (https://chemistry.stackexchange.com...iction-on-whether-to-use-internal-or-external , for everyone else new to this party), I think I actually know what you meant now and I'm not sure I need any further explanation, and it sounds like you were agreeing with what I wrote. I was just thrown off by two things:

    1) You were using (and in the post above, are using) "$P_{ext}$" to mean "the compressive force per unit area exerted by the inside face of the piston on the gas". Every other source, and every other sample problem, uses "$P_{ext}$" to mean the external pressure (usually constant, in this case 1 atm). No other source defined any term for "the compressive force per unit area exerted by the inside face of the piston on the gas" since, as you noted, by Newton's 3rd law it's equal to the pressure exerted by the gas on the piston, so every other source just uses $P_{int}$ for this.

    2) You wrote "You can always use $P_{ext}dV$". I had assumed that meant, obviously, that you "could use $P_{ext}dV$" as the correct answer to the question of how much work is done by the gas. But later you wrote that the work done is given by $W(t)=\int{P_{ext}dV}+KE$. I didn't realize you meant "You can always use $P_{ext}dV$ as part of an answer, even though you also need to add some other terms."
     
  5. Jul 7, 2018 #4
    I still don't think that we are on the same page, and I still think you are having some conceptual difficulties. Working the problem together will help clear everything up for you.

    One more thing: When, in other references, they use the word "pressure" instead of "compressive stress" or "compressive force per unit area," they are trying to gloss over the lack of fluid mechanics background on the part of their readers, and are basically trying to slip something past their readers. Students who take thermodynamics for the first time are typically not familiar with fluid dynamic concepts like the stress tensor, viscous stresses, or the fact that the compressive stresses in a deforming fluid depend not only on the amount of deformation (e.g., the volume change) but also on the rate of deformation (e.g., the rate of volume change). When such students hear of the word "pressure," they immediately think "force per unit area," and don't even conceive of the fact that the force per unit area can be a function of the rate of fluid deformation; moreover, they immediately envision the quantity calculated from the ideal gas law (which is valid only at thermodynamic equilibrium). What I've been telling you is that, in a irreversible deformation, the compressive stress (aka compressive force per unit area) is not the value calculated from the ideal gas law, and depends on both the volume change and the instantaneous rate of volume change. It is not really valid to call this entity the pressure (but they do it anyway, because they are trying not to confuse the students and because they want to avoid having to present a significant chunk of a fluid mechanics course to the students). To learn more about this, Google "Newtonian Fluid."

    Now back to our problem and your questions. When I use the term ##P_{ext}##, what I mean is the force per unit area at the interface between our system and its surroundings. This is the only way of correctly calculating the work done by the system on its surroundings. In our case, since we are calling the "gas alone" as our system, the interface with its surroundings (at which work is being done) is the inside face of the piston. It is not the force per unit area on the outside face of the piston, since the piston is not part of our system.

    Now for the Newton's 2nd law force balance on the piston: $$P_{ext}A=P_{out}A+m\frac{dv}{dt}$$ where v is the velocity of the piston and m is the piston mass. Given the definitions of the quantities ##P_{ext}## and ##P_{out}## defined precisely in my post #2, do you agree with this force balance on the piston or not?
     
  6. Jul 8, 2018 #5

    sophiecentaur

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    Hi and welcome to PF
    Just one point. :
    I think your point 1. is a 'false dichotomy', comparing two inappropriate things.
    1. I would say that the work done by the gas in the cylinder would be a function of the internal pressure. The easiest analysis doesn't consider how the reactive force from the piston is generated. It could be against a brake, raising water etc. etc. as well as displacing gas outside. If it's just acting against a gas then what would be the pressure of that gas during the expansion? ~That scenario would require you to know more about the situation, such as how fast the piston would be moving. You would need to know the speed of sound in the experiment.
     
  7. Jul 8, 2018 #6
    Oh OK. So let's use P(out) to refer to the atmospheric pressure around the piston, regardless of what is considered the system.

    So, to save a lot of work here: Let's say I'm not actually interested in computing the answer, the total amount of work done by the gas. I just want to confirm that the answer is not equal to the change in volume times the atmosphere pressure. In other words, in my example where the piston expands from 1 liter to 2 liters (I forgot to specify the initial and final volume in my original post), the "work done by the gas" is not equal to 1 liter*atm, it's some number greater than that.

    Is that a correct statement?

    I'm the kind of learner where I try to synthesize from many different sources, but if it appears that some of those sources are incorrect, I need to get unambiguous confirmation that they're incorrect, before I can move on from that.
     
  8. Jul 8, 2018 #7
    No. If the "system" is regarded as the gas plus the piston, then the interface between the system and the surroundings is the outside face of the piston, and ##P_{ext}=P_{out}##. So, as I said, the answers to some of your questions depends on what one calls the "system." After we determine the work for the case in which the "gas alone" is regarded as the system, we can go back and see how the problem solves when the "gas plus piston" is regarded as the system.
    No, it is not correct.
    I'll help you move on, and I'll help you see why your answer is not correct. I'm the kind of teacher who helps the student work out the correct answer essentially on his own, but, to do this, I need the student's cooperation. When we are done, there will be consensus between the two of us (with me as your unambiguous source) as to what is correct and what is incorrect. Are you willing to work with me for just a couple of posts until I've been able to help you to unambiguously figure it out on your own? If so, please go back to post #4 and confirm or dispute the force balance equation I wrote down for the piston. (This force balance, as you will soon see, is the key to everything).
     
  9. Jul 9, 2018 #8
    I'm not sure why you wrote "No". I said that we should use P(out) represents the atmospheric pressure outside of the piston. That corresponds to what you wrote in your first message in this thread.
     
  10. Jul 9, 2018 #9
    OK. You said it is not a correct statement that the work done by the gas is greater than 1atm*liter. I think I will probably not understand what you are saying unless I first understand exactly what is wrong with the following reasoning:

    1. To find work, you integrate the force exerted over the distance moved.
    2. In the problem I'm describing, at all times the force exerted by the gas on the inner face of the piston is greater than the force exerted by the atmosphere on the outer face of the piston (which is 1 atm).
    3. Therefore, if you integrate force exerted by the gas, over the distance moved as the volume increase 1 liter, the work done by the gas is greater than 1atm*liter.

    Which of these statement(s) do you think is incorrect?
     
  11. Jul 9, 2018 #10
    If I multiply the force balance equation for the piston that I wrote in post #4 by the piston velocity v=dx/dt, I obtain: $$P_{ext}A\frac{dx}{dt}=P_{out}A\frac{dx}{dt}+mv\frac{dv}{dt}$$where x is the distance of the piston from the bottom of the cylinder. If I next integrate this equation between time t = 0 and time t, I obtain the work done by the gas on the piston up until time t:
    $$W(t)=\int_0^t{P_{ext}dV}=P_{out}[V(t)-V(0)]+m\frac{v^2}{2}$$
    So you are correct to say that, at short times, the work done by the gas is greater than ##P_{out}[V(t)-V(0)]##.

    However, the final amount of work attained at the ultimate equilibrium state of the system depends on what happens with
    the piston. There are three basic possibilities:

    1. The piston is not caught in place by another latch, and the piston is simply allowed to re-equilibrate at a new location. During this process, the piston will overshoot the ultimate equilibrium position, then continue on to a final displacement, and then start moving backwards. It will oscillate back and forth about the equilibrium position like in harmonic motion. But, viscous stresses within the gas will cause the motion of the piston to be damped, and, eventually the piston will come to rest at the final thermodynamic equilibrium position (with zero kinetic energy). From our equation, this will be characterized by:
    $$W(\infty)=\int_0^{\infty}{P_{ext}dV}=P_{out}[V(\infty)-V(0)]$$

    2. The piston is not caught in place by another latch, but there is a collar inside the cylinder that prevents the piston from advancing beyond a volume of 2 liters. In this case, if the collision between the piston and the collar is purely elastic, the piston will bounce off the collar and start moving backwards as in Case 1. And, here again, viscous stresses will damp the kinetic energy of the piston (over multiple rebounds from the collar), and, in the end, the kinetic energy of the piston will again be zero. So, in this case also, we will have that
    $$W(\infty)=\int_0^{\infty}{P_{ext}dV}=P_{out}[V(final)-V(0)]$$where, in this case, V(final) = 2 liters.

    3. The piston is caught in place by another latch. In this case, you are correct to say that $$W(\infty)>P_{out}[V(final)-V(0)]$$However, since the kinetic energy of the piston is not known when it reaches the capture latch, there will be no basis for quantifying the actual amount of work done by the gas.

    I think we are both on the same page now.
     
  12. Jul 9, 2018 #11
    OK, so #3 above was the version of the question I was asking all along -- to get confirmation that under the exact conditions I specified, the work done by the gas is greater than atmospheric pressure times change in volume.

    Now, that raises a further question: If the the answer in case #3 is that the work done by the gas is greater than P(Vfinal-V0), then I'm not sure how in case #1 and case #2 it can be equal to P(Vfinal-V0).

    Here is my reasoning:
    1) All three of the above scenarios are identical (with regard to how much work the expanding gas is doing) from time t=0 to the time t=k when the piston has expanded to 2 liters in volume.
    2) In scenario #3, the work done by the piston at time t=k is greater than P(Vfinal-V0).
    3) Therefore in scenarios #1 and #2, the work that has been done by the gas at time t=k is also greater than P(Vfinal-V0).
    4) The work done by the gas in scenarios #1 and #2 after time t=k cannot be negative.
    5) Therefore in scenarios #1 and #2 the total work done by the gas has to also be greater than P(Vfinal-V0).

    The only one I'm unsure about is #4. If the system does work on the gas during some time period, then by convention do we say that "the work done by the gas is negative"?
     
  13. Jul 9, 2018 #12
    Sorry about that. I didn't notice that you were saying that the piston got captured by a second latch.
    Yes. Your assessment regarding Item 4 is the reason. The additional work after time t = k has a net negative value because, in the return strokes, the gas force is in one direction and the piston movement is in the opposite direction. And the net work during the return strokes exceeds the work during the subsequent forward strokes. In addition, the viscous force during both the forward and return strokes is opposite to the direction of movement of the piston.

    As long as the piston movement is damped by viscous forces in the gas (such that the gas causes the piston to slow down and stop), the only possible conclusion from the fundamentally simple work equation I presented is that the overall work done by the gas is ##P_{out}[V(\infty)-V(0)]##
     
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