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Entropy of gas at constant pressure and volume

  1. Nov 25, 2009 #1
    1. The problem statement, all variables and given/known data

    1m^3 of air is heated reversibly at constant pressure from 288K to 573K. Then it is cooled reversibly at constant volume back to the initial T. Initial P is 103kPa Calculate overall change in entropy.
    Cp=1.02
    Cv=0.702

    2. Relevant equations

    dS=Cp x ln(T2/T1)-R x ln(P2/P1)


    3. The attempt at a solution

    I have found P2 to be 911kPa but when I put all the data into the above equation I end up with the wrong answer (0.076kJ/K). Please help!

    I have found
     
  2. jcsd
  3. Nov 25, 2009 #2
    tell me how u got 911kPa..
    its wrong from thr only...
     
  4. Nov 25, 2009 #3
    I used T2/T1 = (P2/P2)^(n-1/n)

    and n = Cp/Cv
     
  5. Nov 25, 2009 #4
    for 1st proces which is isobaric....
    V1/T1=V3/T3... V1= 1m^3 so one can find V3.....3 is the intermediate stage
    and then for isochoric process....
    P3/T3=P2/T2.....
    n bcoz T2= T1=288 and P3=P1.....
    n T3=573
    so jus find value n den see...
    u r goin in wrong direction...dats actually for adiabatic processes
     
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