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Understanding pV Diagrams

  1. Nov 30, 2008 #1

    Part D
    Calculate the work W done by the gas during process 1\rightarrow3\rightarrow6.
    Express your answer in terms of p_0 and V_0.

    Okay so I know that W = p*deltaV, but p is not constant in this case, so what do I do?
  2. jcsd
  3. Dec 1, 2008 #2
    W = p*deltaV only holds for the case of constant p, that is true.

    What is the general form of the work eqn. that holds for all types of paths? (Hint: It also invloves p and V, and it must reduce to the W = p*deltaV, when p is constant)
  4. Dec 1, 2008 #3
    Is it W = p1V1 - p2V2?
  5. Dec 1, 2008 #4
    No, recall that work done is given by W = F.x, hence dW = F.dx, where dx is a infinitesimal displacement in the direction of the force.

    For ideal gases, F = pA, hence dW = pA.dx, or dW = p.dV.

    Work is then found from integrating p.dV, with respect to V from Vfinal to Vinitial, which can be interpreted as the area under the pV graph
  6. Dec 1, 2008 #5
    So pV from Vi to Vf, but isn't p still constant in this case?

    The graph indicates that p is NOT constant.
  7. Dec 1, 2008 #6
    no, p as indicated in this equation of the general case, is a function of V.
  8. Dec 1, 2008 #7
    okay let's say from 1 to 3, pressure is decreasing from 3p to 2p, how would I apply that to pV| from Vi to Vf?

    p(3V) - p(2V)?
  9. Dec 1, 2008 #8
    Integration between two points of a function can be simplified by taking the area under the curve. Because, the graph is linear, it is very easy to find the value of the integral from 1 - 3 - 6, because the area under the curve is a trapezium (or trapezoid).
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