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Work done on a gas

  1. May 2, 2009 #1
    1. The problem statement, all variables and given/known data
    A 26-L sample of an ideal gas with (gamma)= 1.67 is at 250 K and 50 kPa . The gas is compressed adiabatically until its pressure triples, then cooled at constant volume back to 250 K, and finally allowed to expand isothermally to its original state.



    2. Relevant equations
    I've called the three states a,b, and c, so the subscripts are W_(from-to)

    W_ab= (P_a*V_a - P_b*V_b)/(gamma-1)

    P_b=(50 kPa)(3)=150 kPa

    P_a(V_a)^gamma=P_b(V_b)^gamma

    W_bc= 0

    W_ca= nRT*ln(V_a/V_c)

    pV=nRT
    V_c=V_b


    3. The attempt at a solution

    The first process is adiabatic, so I used the first equation to find the work. I first needed to find the V_b, so I used the second equation: (50 kPa)(26 L)^1.67=(150 kPa)(V_b)^1.67
    I found that V_b= 13 L (based on another part of the question that I solved correctly I know that they rounded it to 13).
    Then W_ab= [(50)(26 )-(150)(13)]/(1.67-1) = -970 J

    The second process was isochoric, so W= 0.

    The third process is isothermal, so I used:
    W_ca= p_a(V_a)ln(V_a/V_c) = (50)(26) * ln (26/13) = 901 J

    So then the total work done is: W= 901-970 = -69 J

    can someone please tell me where my mistake is? Thanks!
     
    Last edited: May 2, 2009
  2. jcsd
  3. May 2, 2009 #2
    Check the sign. Since this is work done on the system.

    Edit: I did remove the part about kPa * L = J when I quickly realized I was wrong.
     
    Last edited: May 2, 2009
  4. May 2, 2009 #3

    diazona

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    Homework Helper

    Actually kPa * L = J is true...

    It seems like your math is correct, but I wonder about the rounding. You're not supposed to round intermediate values like the 13L; rounding is only applied when you write out the final answer. In this case, the difference between 13L and the unrounded value is large enough that it makes a noticeable (well, more than just "noticeable") difference in the answer.
     
  5. May 3, 2009 #4
    Yeah, I tried answering with both the rounded and not rounded intermediate and I noticed the difference too, but it still returned both answers as incorrect. Usually when just the sign is wrong it'll say so as well... I've been going off of an example in my book that is pretty much the same problem with different values and somehow I keep getting it wrong. In any case its good to know that its not just me making some very obvious mistake, thanks!
     
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