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Q and W for Van-der-Waals Gases

  1. Oct 26, 2015 #1
    1. The problem statement, all variables and given/known data
    I have the following task:

    43921_1.PNG

    2. Relevant equations


    3. The attempt at a solution

    I already managed to calculate Delta Um, but how do I calculate Q und W. Can I use the equations for the isothermic expansion for ideal gases, even if this are Van-der-Waals Gases?
     
  2. jcsd
  3. Oct 26, 2015 #2

    TSny

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    Hello. Welcome to PF!

    What is the most basic formula you know for calculating the work done by a system for any quasi-static process? Hint: This formula was probably presented when you first introduced the concept of work in thermodynamics.
     
  4. Oct 26, 2015 #3
    e701045752be3b1a02f4255078f9754e.png
    This one?
     
  5. Oct 26, 2015 #4
    Yes. They obviously want you to assume that the expansion is reversible. Your equation is correct if W represents the work done by the surroundings on the system.

    Chet
     
  6. Oct 26, 2015 #5
    Ok, thx a lot!

    therefore, Q = -W?
     
  7. Oct 26, 2015 #6
    No. ΔUm is not equal to zero.

    Chet
     
  8. Oct 26, 2015 #7
  9. Oct 26, 2015 #8
    You don't need to ask me this. Why don't you solve the same the same problem using the ideal gas law and see how the numbers compare? Incidentally, why did you need wolframalpha to do the integration for you? Why didn't you do the integration yourself?

    Chet
     
  10. Oct 27, 2015 #9
    I calculated it by myself, but I often use wolfram alpha to check my results, and in this case, the result was the same, so it was easier just to paste the wolfram alpha link :)

    Well, since delta U should be =0 for ideal gases, this small change seems possible?
     
  11. Oct 27, 2015 #10
    Sure. Wouldn't you have expected that?

    Chet
     
    Last edited: Oct 27, 2015
  12. Oct 27, 2015 #11
    I wasn't sure,

    but now it makes sense.

    Thanks a lot!
     
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