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Thermodynamics in action

  1. Jan 13, 2005 #1
    Okay, i found this problem to be pretty tough, maybe someone can lend me a hand.

    Theres a container with a piston, and theres a gas on the bottom of the piston.

    State 1
    [itex]A=1.2 x 10^{-2}m^{2}[/itex]
    [itex]V_1=1.5 x 10^{-3}m^{3}[/itex]
    [itex]P_1=1.02 x 10^{5}Pa[/itex]

    State 2
    Theres a 150kg weight ontop to the piston, wish causes the piston to go down.

    State 3
    The cylinder is brought in contact with boiling water raising temperature to [itex]100^{\circ{}}C[/itex]
    The weight goes up, the gas pushes up the piston, and the piston pushes up the weight.

    State 4
    The block is removed, and the gas expands to fill the container, which is a cylinder.

    All I need is how to get this started I think I can get it going from there.

    The truth is I dont even know how to start this, looked at it for like 30 minutes, maybe I missed something.

    But any help would be appreciated.

    Last edited: Jan 13, 2005
  2. jcsd
  3. Jan 13, 2005 #2


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    You need to understand the fact that any thermodynamical transformations that take place in the gas (probably assumed ideal) are assumed quasistatical and reversible.So u can apply the law of Mendeleev and Clapeyron.

    You didn't say what the problem was asking you... :tongue2:

  4. Jan 13, 2005 #3
    Hmm, does not seem I learned those laws, onless my book and teacher are calling them by different names. O, and I asked to get this started, for example finding pressure on state 2. From what I see, is that the 150 kg weight does work on the gas? I might be wrong. But if so it does 1470 J, so I got Work. Work is [itex]\Delta W= P \Delta V[/itex]. I can see that pressure goes up, and volume goes down. But I have no clue as in how to find either of those.
    Last edited: Jan 13, 2005
  5. Jan 13, 2005 #4
    O and thank you for replying. :smile:
  6. Jan 13, 2005 #5


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    That "A" is it the surface of the piston??If so,u can compute the pressure the piston is exeritng on the gas...Temperature remains constant,so you'd have to apply
    [tex]P_{1}V_{1}=P_{2}V_{2} [/tex]

  7. Jan 13, 2005 #6
    Hmm, I didnt think of that. I think I can take it from here. Thank you very much. :smile:
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