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Homework Help: Thermo Help

  1. Aug 24, 2010 #1
    1. The problem statement, all variables and given/known data

    1 mole of monatomic ideal gas. P1 = 2 atm V1 = 44.8 L T1 = 1092 K. The path to P2 = 1 atm V2 = 22.4 L and T2 = 273K is P = 6.643E-4 * V^2 + 2/3. Calculate the heat and work.

    2. Relevant equations
    q = nCpΔT
    w = PΔV

    3. The attempt at a solution

    I attempted to directly plug in the data that was available and didn't work. Neither P,V, or T are constants during this state change. I just realized in posting this problem I'm not sure what equation to use since every variable is changing. Suggestions????

    q = -13.5kJ w = -3278 J
    Last edited: Aug 25, 2010
  2. jcsd
  3. Aug 25, 2010 #2


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    You need to use the more general definition of work:

    [tex]W=\int P\,dV[/tex]

    You'll also want to use the first law to solve the problem.
  4. Aug 25, 2010 #3
    I was able to find work by figuring out the area under the curve.

    The solution that I found for q baffles me. This question was the last part of a series of state changes in a cyclic reaction. I found the answer by adding the q and subtracting w from the previous parts of the problem. That shouldn't work like that right?
  5. Aug 25, 2010 #4

    Andrew Mason

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    You are given the change in temperature so you can easily determine [itex]\Delta U[/itex]. You have figured out the work done. So apply the first law to find heat flow:

    [tex]\Delta Q = \Delta U + W[/tex]

  6. Aug 25, 2010 #5
    I suggest memorizing now these two formulas

    dU = dW + dQ (for work done on the environment)
    dW = d(pV)
  7. Aug 26, 2010 #6

    Andrew Mason

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    Except that it should be dQ = dU + dW then, where dW is the work done BY the gas/system (ie. ON the surroundings).

  8. Aug 26, 2010 #7
    Thanks, I did some alegbra incorrectly and stumbled across something that I found very very unusual. Is it safe to assume what I found is a rare coincidence?
  9. Aug 26, 2010 #8

    Andrew Mason

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    I am not sure what you did. Your formula for dq = nCpdT is incorrect, since this is not a constant P process. You determine dQ by finding the work done (by integrating PdV (substituting the given formula for P) and the change in U (using change in temperature) and then applying the first law.

  10. Aug 26, 2010 #9
    Ah, my bad, I got the sign convention wrong. I hate that thing. I should have looked it up. :(

    Anyway, my advice still holds, memorize those formulas and learn how to use them because you'll see more as you go through thermo.

    Oh yeah, and also kalbuskj, mistakes are invaluable. Sometimes you'll learn more than you ever could in getting the problem wrong and figuring out why it is wrong then getting it right to begin with. Don't just work the problem with the goal of getting it right, work it with the goal of finding your misconception, fixing it, and reworking the problem with the right concepts.
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