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Thermodynamics: Calculating the work done

  1. Jun 16, 2013 #1
    1. The problem statement, all variables and given/known data
    One mole of a certain ideal gas is contained under a weightless piston of a vertical cylinder at a temperature ##T##. The space over the piston opens into the atmosphere. What work has to be performed in order to increase isothermally the gas volume under the piston ##n## times by slowly raising the piston? The friction of piston against the cylinder walls is negligibly small.


    2. Relevant equations



    3. The attempt at a solution
    Work done in isothermal process is
    [tex]nRT\ln\frac{V_2}{V_1}[/tex]
    In the given question, ##n=1 \text{mol}## and ##V_2=nV_1##. Hence work done is:
    [tex]RT\ln n[/tex]
    but this is wrong. :confused:
     
  2. jcsd
  3. Jun 16, 2013 #2

    I like Serena

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    Hey Pranav!

    You have calculated the work done on the gas, but you have omitted the minus sign. :surprised
    So if you (slowly) extract an amount of energy equal to ##RT\ln n## from the gas, the process will be isothermal.
    This is not the total amount of work.

    Note that it is a bit hard to force a change and still getting energy back instead of putting it into it. :wink:
     
  4. Jun 16, 2013 #3
    Where is the atmosphere in your equations?
     
  5. Jun 16, 2013 #4
    Thanks voko and ILS for the replies! :smile:

    Okay, so there will be some work done on the atmosphere too. The change in volume of atmosphere is ##(n-1)V_1##. The work done on the atmosphere is ##PV_1(n-1)## where P is the initial pressure. Also ##PV_1=RT##, hence net work done is ##(n-1)RT-RT\ln n=RT(n-1-\ln n)##, correct?
     
  6. Jun 16, 2013 #5
    This looks good to me.
     
  7. Jun 16, 2013 #6

    I like Serena

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    Yep. Looks good.
     
  8. Jun 16, 2013 #7
    Thanks!
     
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