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Thermodynamics - expansion, compression, work

  1. Oct 17, 2016 #1
    Hi everyone!
    I am confusing myself with this topic and I would appreciate some insight into this. You see, one takes the area of a PV graph to obtain the work done on the system (for compression) or by the system (for expansion). However, I am learning that in the irreversible path, compression takes the area above the PV curve while expansion is below it. For representational purposes something like the graph on this paper:http://www.csrc.ac.cn/~suncp/papers/PR/PRE08.pdf

    a1) Does this mean that the amount of work done in compression is always greater than expansion (because of the "extra" area? I don't find this intuitive because I am thinking that for compression, the external pressure (outside of the system) approaches that of the gas(boundary) plus the atmospheric pressure and, aided by the gravitational force, this reduces the amount of Pexternal necessary to apply unto the system to achieve compression.
    a2) I realized that perhaps the reason for a1 may be due to the starting Pext which changes depending on compression (high P to low P, or left to right) or expansion (low P to high P, or right to left). But how does this reflect on the work (i.e. area below the graph?) Please help! I think I am going in circles! =S

    On another note, one can determine what the maximum or minimum Pexternal and work can be used to achieve compression or expansion in the reversible process. Compression corresponds to minimum Pext and work that the surroundings has to do to compress the system, whereas expansion corresponds to maximum Pext and work that the system can perform on the surroundings to achieve the expansion.

    b) This approach to infinitesimal changes sort of makes sense if I follow my reasoning on (a), but i am not sure if this is correct at all. In addition, why is w(irreversible)>w(reversible) in compression, and vice versa for expansion? Does it have to do with the "extra" or "less" area acquired in the irreversible process in comparison to the closer approximation ot the PV curve by the infinitesimal changes in the reversible process?

    Finally (as in my last question for this thread), when calculating the reversible work for some expansion or compression one has to make the following integration:
    dw=-∫PdV, where P is Pexternal but for an ideal gas P=nRT/V,
    So, dw = -nRT∫ (1/V) dV => w= -nRTln(V2-V1)

    c) When is it possible to substitute P1V1 for nRT in the above result, such that w= -P1V1ln(V2-V1)? I would simply retain the previous solution; and I assume that for an ideal gas, both solutions should give the same answer?

    I hope that someone could take the time to read through my (rather long and pesky) entry and provide some helpful insight... it would certainly help me get rid of many doubts and correct some of my reasoning mistakes as well. Thanks! =D
     
  2. jcsd
  3. Oct 19, 2016 #2

    Simon Bridge

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    Welcome to PF;
    your link takes too long to load - so I'm not very clear what you are referring to.

    Meantime... here's a pic showing a reversible and irreversible process:
    rev_irr_comp_isotherm.gif ... it's fairly typical of the kind of thing encountered in courses.

    When we compress something we usually think in terms of reducing the volume ... though we could also mean "increasing the pressure".
    There is no work to increase the pressure at constant volume, but there is work needed to reduce the volume at constant pressure.
    In this particular process, there is more work needed during the irreversible compression process than you get out during the expansion.
    You can swap this over by reversing the direction of the arrows on the diagram.

    To work out how the diagram produces work you need to refer to the exact way that the process happened.
    On the diagram it does have a lot to do with the extra areas.

    The following notes may help:
    https://ocw.mit.edu/ans7870/16/16.unified/thermoF03/mud/T11mud03.html
     
  4. Oct 19, 2016 #3
    Thank you for your response Simon Bridge, I understand a bit more about what is going on after reading more from your link. I think that I was making the procedure more complicated than it needs to be to explain thermo concepts! Thank you once again for giving me such insight! =)
     
  5. Oct 19, 2016 #4
    Hi SpinzTronics,

    Welcome to Physics Forums.

    In rapid irreversible expansions and compressions, a gas behaves in a different way mechanically from when it is expanded or compressed very gradually (reversible). In rapid deformations, the viscous nature of the gas comes into play in addition to is usual (elastic-type) P-V behavior. For very gradual deformations, the viscous behavior is negligible, and the P-V behavior dominates.

    A good conceptual model of the behavior of a gas in expansion or compression is provided if you imagine that, unbeknownst to you, the gas has been replaced by a spring and damper (i.e., viscous element) in parallel. This spring-damper combination will exhibit mechanical behavior qualitatively similar to the gas for both (slow) reversible and (rapid) irreversible deformations, and, if the parameters for the spring and damper are chosen properly, can even closely mimic the behavior of the actual gas quantitatively.

    Specifically too address the kind of questions you raised in your post, I wrote a Physics Forums Insights article that explores more fully the similarity between a gas and a spring-damper combination during irreversible expansion or compression, and interprets the behavior quantitatively in terms of the viscous behavior of the gas. Here is the link:
    https://www.physicsforums.com/insights/reversible-vs-irreversible-gas-compressionexpansion-work/

    If you have any questions, please feel free to ask.

    Chet
     
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