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Work done by an ideal gas

  1. May 22, 2012 #1
    Could someone please prove why the work done by a gas is the area under a [itex] PV [/itex] diagram?

    That is, I know that [itex] dW=P \text{ }dV[/itex], but why is that true physically? I realize that [itex] W=f \cdot d \text{ and } F=P \cdot A [/itex] .Thanks.
     
  2. jcsd
  3. May 22, 2012 #2
    You have more or less done it for yourself!!
    W = P x V......P = F/A and V = A x d
    W = F/A x A/d ....= F x d
     
  4. May 22, 2012 #3
    Well yes, I know that the equation is true, but I was wondering why it's physically true. For example, the area under a velocity-time graph is displacement. Why is the area under the graph the work done?
     
  5. May 22, 2012 #4
    It is because the graph is of P against V so the area under this graph must be the PxV quantity.....which is Fxd = work done.
     
  6. May 22, 2012 #5
    Oh...wow I'm silly. Thanks!
     
  7. May 22, 2012 #6
    you are in good company!!
     
  8. May 22, 2012 #7

    Ken G

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    Gold Member

    Another way to say all this is that, you understand the area under the v vs. t curve is displacement x because you think of v as dx/dt. Thus, to understand why the area under a P vs. V curve is work done, you simply need to think of pressure as dW/dV, i.e., pressure is the amount of work done per unit volume change. In other words, instead of thinking of work as something that comes from pressure, think of pressure as a concept that stems directly from work. That is very much what pressure is-- the concept of the amount of work done per volume change. Indeed, there are situations where the easiest way to calculate pressure P is to first calculate the work W as a function of V and take dW/dV.
     
  9. May 23, 2012 #8
    just remember that this is true under the assumption of quasistatic conditions, which is almost static but notquite.
    think about a situation where the internal pressure is much higher than the external presssure. how do you calculate work then?

    thafeera
     
  10. May 23, 2012 #9
    Another little tip: look at the units of the quantities you are multiplying.
    P x V units are Pa x m^3 = (N/m^2) x m^3 = Nm = energy units
    v x t units are ms^-1 x s = m = distance units
    F x e units N x m = energy units
    V x Q units are volts x charge = V x C = (J/C) x C = J = energy units
     
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