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Homework Help: Ideal gas problem

  1. Jul 26, 2004 #1
    Okay, I've been working on this for a while:

    Three-tenths mole of an ideal gas at 400K is expanded isothermally from a pressure of 5x10^5 N/m^2 to 1.5x10^5N/m^2. If the gas is then heated to 500K at constant volume and then is compressed isothermally back to 5x10^5 N/m^2, and then isobarically back to the original condition, find the net work of the cycle.

    I understand the whole concept of expanding and contracting, but combining the ways that these happen is very confusing to me.

    first i found the volume for the first part using the formula V = nRT/P.
    so for the first part, when its expanded isothermally i used the formula:
    w = nRTln(V2/V1)
    w= (3/10)*(8.314)*(400)*ln(6.65*10^-3/2*10^-3)
    = 1198.68

    same for the second isothermic expansion, except the temperature changed to 500, and therefrore the volume changed and:

    w = -1503.58


    then for the third part, i used the formula for isobaric gas:

    w = P(V2 - V1)
    = 245

    Then for the net work, I added everything together, and got -364.8

    What am I missing here??

    Thanks!
     
  2. jcsd
  3. Jul 26, 2004 #2

    drag

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    Science Advisor

    If you have a set temprtature and pressure, the volume is
    also determined for an equal amount of gas. So, if at the
    end you get the original condition - all of the same parameters,
    what kind of energy difference = work, are you looking for ?
     
  4. Jul 26, 2004 #3
    it just says find the net work. I stated the problem exactly as it is written. Grr...this one is driving me nuts!
     
  5. Jul 26, 2004 #4

    Doc Al

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    Staff: Mentor

    Looks good. (You are defining work as work done by the gas on the environment. So + work means that energy is leaving the gas. That's OK. Just be aware that some define work as work done on the gas.)

    Take a second look at this one. I suspect you are using the wrong final volume.
     
  6. Jul 26, 2004 #5
    daisyi - if you add up all your work terms again, you'll see it will not give you -365. The calculation for all the individual work terms are correct, except for some rounding errors. When I did the calcuations I get: 1201.18 - 1501.47 + 249.42 = -50.87 or -50 after rounding.
     
  7. Jul 27, 2004 #6

    Doc Al

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    Staff: Mentor

    one more thing...

    I never got to this step in my previous post:
    It's an isobaric compression: the work will be negative.

    And yes, as PureEnergy points out, you did add your terms incorrectly. But that's not the real problem. :smile:
     
  8. Jul 27, 2004 #7
    I double checked and it still looks right to me. This is how I did it.

    V = nRT/P

    so, when the gas is heated to 500k with constant volume, the volume is:

    V1 = (3/10)(8.314)(500)/1.5x10^5
    = 8.314x10^-3

    and then when the pressure is compressed back to 5x10^5:

    V2= (3/10)(8.314)(500)/5*10^5
    = 2.49x10^3

    I then plugged that into the equation w = nRTln(v2/v1) to get the 1198.68
     
  9. Jul 27, 2004 #8

    Doc Al

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    Staff: Mentor

    Right.

    No. V1 = (3/10)(8.314)(400)/1.5x10^5 (You know the temp and pressure before it's heated.)

    Right.
     
  10. Jul 27, 2004 #9
    Doc's right, the third step should be negative so the net work is around -550.
     
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