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Thermodynamics question

  1. May 24, 2006 #1
    Can someone give us a hand in explaining a little about this question??

    Gas in a cylinder expands and does 150 joulesof work moving a piston. The Internal energy of the gas changes from 310J to 100J during the expansion.
    During this process the change in heat is best described as:


    (a): 60J of heat is released by the gas


    (b): 360J of heat is released by the gas


    (c): 60J of heat is added to the gas


    (d): 210J of heat is released by the gas


    (e): 210J of heat is added to the gas



    Any hints would be much appreciated
     
  2. jcsd
  3. May 24, 2006 #2

    dav2008

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    Gold Member

    Do you have any work or thoughts on the problem? Do you know of any relationships that can relate the heat input to the work done and the change in internal energy?
     
    Last edited: May 24, 2006
  4. May 24, 2006 #3
    Use the 1st law of thermo dynamics:
    dU = dQ + dW
    You are given the change in internal energy (dU), and the work done (dW)
     
  5. May 24, 2006 #4

    dav2008

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    Personally I think it makes more sense to write it as [tex]\Delta U=Q-W_{by}[/tex] where [tex]\Delta U[/tex] is the change in internal energy of a gas, [tex]Q[/tex] is the heat added to the system, and [tex]W_{by}[/tex] is the work done by the system.

    That way you have defined all your variables with respect to the gas. A positive W means the gas is doing work (expanding). A negative W means work is being done on the gas (contracting). A positive Q means heat is being put into the gas; a negative Q means heat is being taken out of the gas. A positive change in U means that the internal energy of the gas is increasing; a negative change means its decreasing.

    It's easy to realize what's happening if you look at it as: [tex]Q=\Delta U+W_{by}[/tex]. Any heat added to the gas can either go into changing the internal energy of the gas or towards the gas expanding and thus doing work.
     
    Last edited: May 24, 2006
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