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Calculating the work done during an isothermal expansion using integration

  1. Mar 23, 2012 #1
    1. The problem statement, all variables and given/known data
    In calculus, the work done when a gas expands from volume V1 to volume V2 is given by
    W = ∫V2V1 P dV
    Use this expression to show that the work done by n moles of gas at temperature T during an isothermal expansion from volume V1 to V2 is
    W = nRT ln(V2/V1)


    2. Relevant equations
    Q = ΔU + W
    PV = nRT


    3. The attempt at a solution
    W = [VP]V2V1 = PV2 - PV1 = PΔV
    But I think it should be ΔPΔV since this is an isothermal expansion. W = PΔV is for isobaric since P is constant.
    Then I can't even guess where the ln comes from. :frown:
     
  2. jcsd
  3. Mar 23, 2012 #2
    You got P as a constant because you treated it like one when you took the integral.

    But if you look at the ideal gas law you can see that pressure is a function of volume. So then you can put that expression into the integral and n, R, and T are constants, then integrate.
     
  4. Mar 24, 2012 #3
    Hey,

    Unfortunately You have got it wrong.


    See work is defined as
    dW =PdV , where P is external pressure and V is small volume change.

    This comes from the fact that dW=Force * displacement

    dW=(External)Pressure*Area*displacement

    However area * displacement is change in volume so
    dW=PdV

    You have to integrate this expresion to get the value of work.

    Now in isothermal reversible conditions , you have to find work done by system which is a GAS
    In such cases pressure external =pressure of the gas.

    Remember, Ideal gas Equation.?

    How will you integrate PdV now with T being constant.
     
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