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Finding the work done by a gas under quasi-static, isothermal expansion

  1. Mar 27, 2010 #1
    1. So, an ideal gas is initially at 293K and 200kPa, and has a volume of .004m3
    It undergoes a quasi-static, isotheral expansion until its pressure is reduced to 100kPa
    What is the Work done by the gas

    2.I know that for isothermal conditions W=nRTln(Pf/Pi)

    3. my first attempt was making n=1 and Pf=200kPa and Pi=100kPa
    i got W to be 1685w.

    My first problem with this is that I don't actually know what n is, and i'm not even sure what Pi and Pf are, can't find it in my lecture notes anywhere.

    My next step was to isntead of guessing n=1, make PV=nRt because I have P and V, but i'm not sure which P to use.

    I know this dosen't seem like a difficult problem, but i won't be getting the answers until next month, so any guidance will be appreciated
  2. jcsd
  3. Mar 27, 2010 #2


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    Both P and V change during the expansion. You don't know the final volume, but you do know the initial volume. So, use the initial pressure as well.
  4. Mar 28, 2010 #3

    Andrew Mason

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    It should be Pi/Pf.

    dW = PdV. If you substitute P = nRT/V (where T is constant since it is isothermal) you get:

    [tex]\int_{V_i}^{V_f} PdV = nRT\int_{V_i}^{V_f} dV/V = nRT\ln (\frac{V_f}{V_i})[/tex]

    Since, by the ideal gas law, PiVi/Ti = PfVf/Tf, Vf/Vi = Pi/Pf (where Ti and Tf are equal)

    You have to find n. Use the ideal gas law:


    You know Pi, Vi and Ti so you can find n. n does not change.

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