1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Isothermal exapansion, work done

  1. Feb 3, 2013 #1
    1. The problem statement, all variables and given/known data
    Derive an expression for the amount of work done when one mole of an ideal gas expands isothermally at a temperature T from an initial volume V1 to a final volume V2.


    2. Relevant equations
    PV = nRT
    W = V2V1Pdv

    3. The attempt at a solution
    I'm not really sure how to go about this, I know if it's isothermal the temperature doesn't change so Volume would be inversely proportional to Pressure. But other than that i don't know what to do, to be honest i'm not really exactly what they want me to write. Any help would be much appreciated :)
     
  2. jcsd
  3. Feb 3, 2013 #2
    The question seems to be incomplete. It is isothermal expansion , fine. But its isothermal reversible or isothermal irreversible process , its just not mentioned.

    I ask you a question : Do you want to calculate maximum work done by the gas , or not in an isothermal process. If former , consider the process reversible , else irreversible.

    Oh , they do not concern you about external pressure. Consider your process reversible then.

    Hint :

    W = V2V1Pdv ....(i)

    PV=nRT => P=nRT/V

    Now replace this value of P in (i) and carry on your integration.
     
  4. Feb 3, 2013 #3
    Thanks for the Hint thats makes sense, my final answer (upon using that hint) would be W = RT * Ln(V). I think that's the answer but i'll try and confirm it with my lecturer tommorow if i can. Thanks for your help.
     
  5. Feb 3, 2013 #4
    No you should get W=RT*ln(V2/V1)

    See your integral again. Be careful while doing integration.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook