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Calculating entropy for an adiabatic system

  1. Nov 7, 2017 #1
    1. The problem statement, all variables and given/known data

    1. A container of 1.5 Kg of gas is at a temperature and pressure of 293 K and 1 bar respectively. The gas is adiabatically compressed until its temperature and pressure are 450 K, 4.49 bars. Adiabatic processes are processes with no heat transfer. The properties of this gas are cv = 10.3 KJ/(Kg K) and R = 4.158 KJ/(Kg K). Neglect kinetic and potential energy terms.


    2. Relevant equations

    a) Use the first law to determine the work into the system.

    b) Calculate the entropy production for this process.

    c) Is this a reversible process?

    3. The attempt at a solution

    I solved A by saying that the work done = the change in internal energy. So Work done = P1V1 - P2V2 / Y-1 where Y = Cp/Cv = 1.403.
    Since P1V1 = NRT1 and P2V2 = nRT2 , I calculated Work done as nRT1-nRT2 / 1.403 - 1 = -2427.79 kJ. The negative sign representing work being done on the system.

    I am having trouble calculating B. My understanding would be that the change in entropy for an adiabatic process where volume is constant would be calculated as Cv ln (T2/T1), However I am finding conflicting ideas online, where people are actually calculating it as Cp ln (T2/T1) - R ln (P2/P1). When I calculate it that way, I get a negative number of -61.71 J/K for entropy production.

    Can anyone explain if that is the right way to calculate it, or if I should be doing it the way I originalyl thought was correct (Entropy production = Cv ln (T2/T1)?


    Thanks!
     
  2. jcsd
  3. Nov 7, 2017 #2
    You calculated the work incorrectly. What is the equation for the change in internal energy of an ideal gas?
     
  4. Nov 7, 2017 #3
    the change in internal energy of an ideal gas would be Δ U = n Cv Δ T . . . since there is no heat transfer for this, should I be saying that work actually equals this equation?
     
  5. Nov 7, 2017 #4
    the change in internal energy of an ideal gas would be Δ U = n Cv Δ T . . . since there is no heat transfer for this, should I be saying that work actually equals this equation?
     
  6. Nov 7, 2017 #5
    Sure
     
  7. Nov 7, 2017 #6
    Okay, thanks.

    My other question is in regards to part B. Can you please help me understand which equation I should be using in the two that I listed above to solve for entropy?
     
  8. Nov 7, 2017 #7
    You should be using the one that the other people are using, not the one you wanted to use. The one they are using is, of course, per unit mass, so you have to multiply by the mass of gas. The reason their equation is correct is that, in addition to the temperature changing, the pressure and volume also changed. The answer should come out positive, so maybe you made a mistake in arithmetic. Please show your work.

    Chet
     
  9. Nov 7, 2017 #8

    Hi Chet,

    Thanks for your response. You are very helpful.

    Here is my math:
    Cp - Cv = R
    Cp = R + Cv
    Cp = 4.158 KJ/Kg*K + 10.3 KJ/Kg*k = 14.458 KJ/Kg*k

    Entropy production calculation:
    Delta S = M*Cp*ln(T2/T1) - M*R*ln (P2/P1)
    Delta S = (1.5 Kg)*(14.458 KJ/Kg*K)*ln(450K/293K) - (1.5 Kg)*(4.158 KJ/Kg*K)*ln(449000 pascals/100000 pascals)
    Delta S = 9.305 - 9.948 KJ/K
    Delta S = -0.643 KJ/K
     
  10. Nov 7, 2017 #9
    For the 2nd term, I get 9.367, not 9.948. So, for the change in entropy, I get ##\Delta S=-0.0617##. This is pretty close to zero, so I would call it reversible.

    That being the case, the result you got for the work using the equation that applies only for an adiabatic reversible change (i.e., in your original post) should come very close to matching the result you got using the relationship for the internal energy change. Does it?
     
  11. Nov 7, 2017 #10
    Hi Chet,

    Sorry about that. I did mess up the arithmetic! I also got the same answer as you for the second term. I didn't know that it was OK to consider an entropy change that is very close to zero as zero, thus making it reversible. Thank you for the info.

    When I calculated the work done by using the internal energy change (the way you suggested) I get 2425.65 KJ, as opposed to the 2429.79 KJ in my original attempt.
     
  12. Nov 7, 2017 #11
    Looks good. Nice job.
     
  13. Nov 7, 2017 #12
    Thank you so much for your help!
     
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