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Calculating work done in a thermodynamic process

  1. Dec 3, 2011 #1
    1. The problem statement, all variables and given/known data

    The figure shows a thermodynamic process followed by 130 mg of helium. How much work is done on the gas during each of the three segments?

    [tex]p_1 = 1.08 atm, p_2 = 5.41 atm, p_3=1.08 atm[/tex]

    [tex]T_1=133 C, T_2=T_3=1760 C[/tex]

    [tex]V_1=V_2=1000 cm^3, V_3=5000 cm^3[/tex]

    These are answers to previous parts of the question that I've been able to work out and have been marked correct by Mastering Physics. I've also attached a picture of the figure for reference, but 1→2 is isochoric, 2→3 is isothermal and 3→1 is isobaric.


    2. Relevant equations

    [tex]-\int_{V_i}^{V^f} p dV[/tex]

    W=0, isochoric process

    [tex]W= -p \Delta V[/tex] for an isobaric process

    [tex]W= -p_iV_i ln(\frac {V_f} {V_i})[/tex] for an isothermal process.


    3. The attempt at a solution

    [tex]W_1→2 = 0[/tex] because it's an isochoric process. Hooray, easy!


    [tex]W_2→3 =-p_2V_2 ln(\frac {V_3} {V_2})=-5.41*.001*ln(\frac {.005} {.001}) = -0.0087[/tex]

    [tex]W_3→1 = -p \Delta V=-1.08*(.001-.005)=0.00432[/tex]

    So, my answers to the work from 2→3 and 3→1 are not right. Because I'm redoing the problems for practice, I know what the correct answers are, but I've been unable to back solve my way into them.

    Any help as to where I'm going awry would be greatly appreciated!
     

    Attached Files:

  2. jcsd
  3. Dec 4, 2011 #2
    not sure if you could do this because there is a curve, but all you really need to do with a graph is calculate the area inclosed by the triangle to get the work.

    These questions always messed me up because i forgot the graph stuff
     
  4. Dec 4, 2011 #3

    I like Serena

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    Welcome to PF, flygirl! :smile:

    Your work and calculations looks fine!

    However, you left out the units and their powers of 10, and you should do something with them.

    Btw, when you use temperature and mass instead of pressure and volume, you'll get slightly different values.
    So your answer is good, but calculating it in a different manner might give you the expected result.
     
  5. Dec 5, 2011 #4
    Thanks for the welcome Serena! I've lurked for a long time and finally decided to ask a question of my own!

    Well, it's good to know my calculations are right and I'm not at all surprised to find out that my units are messing me up. I'm bollocks with units! I've been trying to get better about using them!

    Thanks for your help!
     
  6. Dec 6, 2011 #5

    I like Serena

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    You're welcome! ;)

    So... is your question answered?
     
  7. Dec 6, 2011 #6
    Since you mentioned that I'd get different results if I used temp and mass, I did the calculations again using those numbers and got the right answer!

    Then I went back and tried to figure out what I was doing wrong above

    [tex]W_2→3 =-p_2V_2 ln(\frac {V_3} {V_2})=-5.41*.001*ln(\frac {.005} {.001}) = -0.0087[/tex]

    It looks like for the pressure and volume I should have been using my *final* values, rather than the values at point 2, like I was. I redid the math with the final values and FINALLY got the right answer!

    You definitely helped lead me down the right path. Thank you!
     
  8. Dec 6, 2011 #7

    I like Serena

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    Well, for an ideal gas and an isothermal process you should have (with the appropriate units like Kelvin):
    [tex]p_2V_2=p_3V_3=nRT_2=nRT_3[/tex]
    but as you can see, with the values given these equations are only approximately true, leading to subtly different results.
     
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