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Problem about adiabatic expansion of gases

  1. Sep 20, 2011 #1
    Hello everyone. I would like to know if I did this correctly.

    1. The problem statement, all variables and given/known data

    Calculate the heat, internal energy and work of 1 mol of hydrogen, which undergoes a reversible adiabatic expansion from a volume of 5.25 m^3 at 300 K to a volume of 25.5 m^3



    3. The attempt at a solution

    The first thing that I have to do is to find the final temperature using the following equation:

    (Vf/Vi)^1-y * Ti = Tf

    y = gamma

    y = Cp/Cv

    Because it's HYDROGEN (diatomic gas):

    y = 7/2/5/2 = 1.4

    (25.5 m^3/5.25 m^3)^1.4 * 300 K = Tf

    Tf = 2741.9 K

    so...

    a) Heat

    becuase it's an ADIABATIC expansion

    Q = 0 J

    b) Internal energy:

    E = n Cv (Tf - Ti)

    Cv = 5/2 * R

    E = 1 mol * (5/2) * 8.31 J/mol K * ( 2741.9 K - 300 K)

    E= 50730.47 J

    c) Work

    I've seen the equation in two different ways:

    E = Q + W

    AND...

    E = Q - W

    Anyway, I'll use the second one (because is the one I have to use according to the worksheet)...

    W = Q - E

    W = 0 J - 50730.47 J

    W = - 50730.47 J

    Is it ok?

    Please check it out, because the worksheet doesn't have the answers and I want to know if I did it correctly.

    Thanks in advance!
     
  2. jcsd
  3. Sep 21, 2011 #2

    rude man

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    " (25.5 m^3/5.25 m^3)^1.4 * 300 K = Tf "

    Do you see an error here? You should be suspicious since you have computed an increase in E while at the same time having done positive work W ( = integral pdV). The 1st law doesn't like that if, as you correctly state, Q = 0 ... in the same vein, computing W as negative can't be right either ...

    Tip - I would always use U = Q - W rather than U = Q + W, this defines W as the work done BY the system. This goes back to the early days of thermodynamics when steam engines were the fad and W was of course always positive. (Most people use U rather than E).

    I would also add that the formulas for Cv are approximate - you can use the ideal gas law to get a precise answer without reverting to Cv.
     
  4. Sep 28, 2011 #3
    Oh thank you. I see it now.

    It was 1 - y and I forgot about doing this:

    1 - 1.4 = - 0.4

    So it plug into the equation the -0.4 instead of the 1.4 It get 113.6 K (final temperature)


    So dT would be: 113.6 K - 300 K = - 186.4 K

    That means that the Internal Energy is NEGATIVE

    Therefore

    U = q - w

    q = 0

    U = - w

    w = - U

    w = - (- 3873.4 J )

    w = 3873.4 J

    Is it ok now?
     
  5. Sep 28, 2011 #4

    rude man

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    Much better!
    Except the internal energy U is not negative; the CHANGE in U is negative.
     
  6. Sep 28, 2011 #5
    Thank you again! I can't believe that I skipped that.

    So, the final correct answer would be that:

    * q = 0

    * U = -3873.4 J

    * w = 3873.4 J

    U negative and w positive
     
  7. Sep 28, 2011 #6

    rude man

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    It certainly would be. Except I haven't checked your algebra. But all your equations look right now. And Q = 0, ΔU < 0 and W = - ΔU are all most definitely correct.
     
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