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Entropy of the universe

  1. Apr 9, 2006 #1
    I'm not sure how these Pressure/Volume (PV) diagrams work. We are given this diagram:
    When a gas follows path 123 on the PV diagram in the figure below, 404 J of heat flows into the system and 163 J of work is done.
    and asked the following questions:
    What net work would be done on or by the system if the system followed path 12341?
    What net work would be done on or by the system if the system followed path 14321?
    What is the change in internal energy of the system in the processes described in parts c and d?

    Not looking for a handout, just want to understand how you calculate net work when it doesnt seem I am given much.
  2. jcsd
  3. Apr 9, 2006 #2


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    Work done ON a sample (Energy into sample) = F.dx = -P.A dx = -P.dV ,
    ... with the negative sign for Area pointing outward ;
    ... you have to push inward to decrease the Volume of sample.

    (If the Pressure changes while the compression occurs, use an "avg" P)

    "Net" Work done is just the sum of Work in the individual process steps.
  4. Apr 9, 2006 #3
    F.dx = -P.A dx = -P.dV

    I assume that P and V are pressure and volume, other than that, I'm lost. Units?
  5. Apr 9, 2006 #4


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    IF Pressure is in [N/m^2 = J/m^3], and Volume is in [m^3] , then W is in [J].

    How much Work is done from 1 to 2 ? from 2 to 3 ?
    (... the Area under the curve ...)

    Now, about Heat ... W(in) + Q(in) = Change in Internal Energy = DeltaE .

    Most gases are almost Ideal , so P.V = NkT only depends on Temperature.
    Now, internal Energy also depends only on Temperature ...
    E = 3/2 NkT (if monatomic) ... 5/2 NkT (if 2-d) ... 6/2 NkT (if 3-d)
  6. Apr 9, 2006 #5
    The area under the curve? There are no figures given. I have to go off of the "work" that has already been predetermined such as the 404J of heat and 163 J.

    and if 404 J of heat flows into the system through 123, then how do you determine it for the rest of the system. This is due at midnight, but at this point, I just want to know whats going on for sanitys sake.
  7. Apr 9, 2006 #6
    heres a similar problem:

    A gas expands from I to F, as seen in the figure below.
    Can't do this one either - sure its really simple once you know what you're doing...
  8. Apr 9, 2006 #7

    Andrew Mason

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    There are 2 paths from I to F. In the red path, the gas expands at constant pressure and then the pressure decreases while volume remains constant. In the blue path, the gas pressure is lowered first and then expands.

    Think of work as Force x distance or Pressure x Volume: (F/A) x d x A. If the gas expands a volume dV against an external pressure, the gas does work: W = PdV. If the gas is compressed by external pressure, the work done on it is W = PdV.

    If a gas changes temperature, it undergoes a change in internal energy:[itex]\Delta U = nC_v\Delta T[/itex]

    If there is no change in volume, there can be no work done to or by the gas.

    So in going from I to A, a change in volume of 2 litres against a constant external pressure of 4 atm, the work done is 4 x 2 = 8 atm litres = 8 x 1.01325 10^5 N/m^2 x 10^-3 m^3 = 810.6 Nm = 810.6 J.

    In going from A to F, does it do any work? What is the change in internal energy? Use PV = nRT to determine the temperature change and the change in internal energy.

    Then use the blue path to determine the work done and change in internal energy.

    Last edited: Apr 9, 2006
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