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Homework Help: Determining work done by rapidly expanding gas

  1. Apr 12, 2014 #1
    1. The problem statement, all variables and given/known data

    A gas initially occupying a volume V undergoes a sudden explosion, causing its volume to triple to 3V. Assuming that the pressure of the environment (atmospheric pressure po) remains constant, find an expression for the work done by the gas.
    Hint: Please notice that an explosion is a rapid and irreversible process. Because this process happens far from equilibrium, the pressure of the gas (p) is not well defined and the expression dw = - pdV cannot be applied to the gas undergoing the explosion.

    2. Relevant equations

    dU = TdS-pdV
    dU = dQ+ dW

    3. The attempt at a solution

    I'm not really sure how to even begin. I was trying to relate this to the Joule free expansion, but I don't think that is correct. I think entropy will come into account.
  2. jcsd
  3. Apr 12, 2014 #2
    I'm tempted to say not really enough information. The statement "a sudden explosion" doesn't really explain whether there's additional energy added via the explosion. But the volume of gas is expanding into a background at constant pressure. Thus the volume change works against the background gas at atmospheric pressure, pushing it back. The pressure inside the gas volume is then irrelevant, only the gas pressure it's working against. Try that.
  4. Apr 13, 2014 #3
    I think "a sudden explosion" just means the gas expands rapidly and we will ignore any energy needed to create the explosion. So, I should find the work that the volume change does on the constant background pressure?

    I'm unsure of how to do this.
  5. Apr 13, 2014 #4
    A sudden explosion only stands for "fast process" - in my opinion. Since the heat transfer is usually a fairly slow process this only gives you a hint that your system will not exchange any heat with the environment (surrounding gas).

    And yes, entropy will come into account. Firstly I suggest you find out what changes and what doesn't. So the questions you should answer: dp=?, dV=? and dT=?

    After that, how does one calculate the entropy of ideal gas?

    EDIT: Oh, and everything I just wrote may be completely wrong. :) It's been a while since I had Thermodynamics.
  6. Apr 13, 2014 #5
    Fast process means dQ = 0. There is no time for heat to move in or out of the system.
  7. Apr 13, 2014 #6
    See my Physics Forums Blog in my PF personal area. Qraal was correct in post #2.

  8. Apr 13, 2014 #7

    Ken G

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    Gold Member

    Entropy is a red herring because it's not reversible, and dQ does not equal TdS, but as has been said, it does equal 0. So you have dU = dW, so you could answer the question if you knew dU. Unfortunately, you don't, so there no better way than to just calculate the work on the surrounding gas. The surrounding gas does have a well-defined momentum flux (i.e., pressure) at the surface of the expanding gas, because the particles bouncing off the expanding wall don't know it's coming until it gets there. That means the problem is just like you had a momentum flux of tennis balls coming at you, and you smack them all with one swing of a moving racket. It doesn't matter how fast you swing the racket, it only matters the distance the racket moves, because if you swing it faster you do proportionally more work but over proportionally less time. So you just integrate PdV, where P is the external pressure so is constant. That's a trivial integral, yes?
  9. Apr 14, 2014 #8
    yeah Ken's right. But now the P in the integral is not the pressure of the gas, but the pressure it is pushing against: ##\int P_0 dV##
  10. Apr 16, 2014 #9
    Ken G, I think I understand what you're saying. So we know dQ = 0 because it is a fast process and does not have time to exchange heat (as said by skrat)? And the only thing we can actually calculate is the work done on the surroundings gas? Which is the integral of the constant surrounding pressure, but with what limits of integration? V to 3V? V to 3V is the change of volume of the internal gas rather than the outside gas. Why would we use these limits?

    Sorry, I'm having so much trouble understanding this. (haha)
  11. Apr 16, 2014 #10

    Ken G

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    Imagine the outside gas has volume Vo originally. What is it's volume ultimately? It is Vo - 2V, because it has lost volume 2V in the expansion. So actually, the "real" limits of integration are Vo to Vo - 2V, but if you work out that integral, it would be the same as using V to 3V with the opposite sign (work done on, rather than work done by, the usual care you must take to get the sign of work right-- I usually just impose the sign logically). The key point is that for doing PdV integrals, it is not the initial and final V that matters, it is the change in V that matters.
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