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Isothermic Work done by a gas

  1. Jan 20, 2006 #1
    Work done by a gas; Isothermic situation

    A gas expands isothermically from [tex] p_1 , V_1 [/tex] to [tex] p_2 , V_2 [/tex] . What is the work done by the gas if [tex] p_1 = 8 atm[/tex] [tex] V_1 = .1m^3[/tex] [tex] p_2 = 1 atm [/tex][tex] V_2 = .8m^3[/tex]?
    I solved the problem using the integral P*dv but my teacher posted the solutions this evening and drew a linear relationship between the two and solved that way. The problem as you see above is how it was given to us verbatim. Should I have assumed that there was a linear relationship between the two? This is Physics with Calc so it isn't like we can't do the integration. Here was my solution
    [tex] W=pdv=\int_ {V_1}^{V_2}\frac{nRTdV}{V} [/tex]
    [tex]nRT=p_1V_1[/tex]
    [tex]nRT\ln\frac{V_2}{V_1}[/tex]
    [tex]p_1V_1\ln\frac{V_2}{V_1} [/tex]
    Then I just plugged in my values and got an answer. I guess my real question is why should I have assumed there was a linear relationship here??!?!?
     
    Last edited: Jan 20, 2006
  2. jcsd
  3. Jan 20, 2006 #2

    Doc Al

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    Staff: Mentor

    Your solution looks correct to me. I don't see how the correct answer could be obtained by assuming a linear relationship (between P and V, I presume) where one doesn't exist. Did he get the same answer as you?
     
  4. Jan 20, 2006 #3
    Thanks for your input Doc! Just got a reply from my professor. He said he made a mistake when writing the solutions guide and verified my answer.
     
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