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Basic Thermodynamics Q

  1. Apr 18, 2010 #1
    1. The problem statement, all variables and given/known data

    http://imgur.com/44ay2.png" 44ay2.png

    2. Relevant equations

    Below.

    3. The attempt at a solution

    A.
    [itex]W=Fd[/itex] 2000N x 0.001m = 2J

    B.
    This is where I am stuck. Can I assume that the pressure doesn't change when it is pushed in because it says that it is pushed in "suddenly"?
    If so then I get: [itex]W = \int p \cdot dV = p(V_f - V_i)[/itex] 105Pa x (0.01m^2 x 0.001m) = 1.05E-3 J

    C
    Is this just the difference between the two above i.e. 2 - 1.05E-3 = 1.99895J?

    I get the feeling this is wrong since I didn't use the information that its 1 litre and 300K anywhere here...
     
    Last edited by a moderator: Apr 25, 2017
  2. jcsd
  3. Apr 18, 2010 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    a) The system consists of the cylinder of air and the piston. So the work done on the system is 2000 N x .001 m = 2 Joules, as you found. Note: if the piston had mass, this would not be the work done on the gas. This is because the piston gains kinetic energy so some of the added energy would go into the piston rather than the gas. But since the piston is massless, this is ignored.

    b) The question does not ask how much work has been done on the gas. It asks how much heat has been added to the gas. Has there been any heat flow from the surroundings into the gas (to the instant before the piston is stopped)?

    c) Is there is any heat flow out of the gas? So where does all the energy go? (use the first law).

    AM
     
  4. Apr 18, 2010 #3
    Thanks for the reply.
    None from the surrounding, but my reasoning is that of the work done by pushing the piston, some will go into compressing the gas and the rest will go to internal energy and therefore increase the temperature. So actually the answer would be 2 - 1.05E-3 = 1.99895J ?
    Nope. So it's just the total amount - ie. 2J?
     
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