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Change of Entropy Calculation - How to do this without Final Pressure Volume?

  1. Oct 20, 2012 #1
    1. The problem statement, all variables and given/known data

    I am having trouble working out the change in Entropy. The question is as follows:

    A mass of 1 kg of air is initially at 4.8 bar and 150 degC and it is enclosed in a rigid container. The air is heated until the temperature is 200 degC.
    Calculate the change in entropy and also estimate the heat transferred by using a finite difference approximation for the equation: Q = ∫Tds.

    For air, R = 0.287 kJ/kgK and Cv = 0.718 kJ/kgK

    Summary:

    m = 1 kg
    p1 = 4.8 Bar
    T1 = 150 DegC
    T2 = 200 DegC

    Answers: [0.08 kj/K , ~ 35.9 kJ]

    2. Relevant equations

    s2-s1 = Cv*Ln(T2/T1) + R*Ln(v2/v1)
    s2-s1 = Cp*Ln(T2/T1) - R*Ln(p2/p1)

    (p1*v1)/t1 = (p2*v2)/t2

    3. The attempt at a solution

    I calculated initial volume by using: PV=MRT and rearranging.

    V1 = 89.6875x10^-6

    Next I'm not sure as you don't have an final pressure to work out final velocity.
    Without either I can't see how I can use the s2-s1 equations.

    I am thinking this may have something to do with the "it is enclosed in a rigid container". Does this mean constant volume or something?

    Thanks in advance.

    Steve
     
  2. jcsd
  3. Oct 20, 2012 #2
    Yes. It means exactly that.
     
  4. Oct 21, 2012 #3
    Thanks for confirming that, I realised I was tripping up with the temperature. I was using degrees C rather than absolute.
     
  5. Oct 21, 2012 #4
    How come it gives values for mass an initial pressure if you don't actually end up using them?

    For the next part I did:

    ((T1*ds) + (T2*ds))/2

    This gave the correct answer but I'm concerned I'm not doing it correctly as I never used the mass or pressure values
     
  6. Oct 21, 2012 #5
    You actually did use the mass, but apparently you don't even know you used it. If the mass were 2 kg, would your answer have been the same? Entropy is a so-called extensive property, which means it is proportional to the amount of mass.

    As far as the pressure is concerned, in real life you often have information that isn't needed. You didn't need the pressure because you were able to assume an ideal gas, but, if the gas were not ideal, the internal energy would be a function not only of temperature but also of the pressure. In a way, you used the 4.8 bars to ascertain that it was valid to treat the air as an ideal gas.
     
  7. Oct 22, 2012 #6
    Ok thank you for help, I appreciate it!
     
  8. Oct 22, 2012 #7
    Ok thank you for help, I appreciate it!
     
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