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An ideal gas closed system reversible process

  1. Feb 26, 2017 #1
    1. The problem statement, all variables and given/known data
    an Ideal gas at T = 70 C and 1 bar undergoes following reversible
    processes:
    a: Adiabatically compressed to 150 C
    b: then, cooled from 150 to 70 C at constant pressure
    c: finally, expanded isothermally to the original state (T=70 C and P = 1 bar)

    2. Relevant equations
    1-dU=dQ+dW
    2-dU=CvdT
    3-(T2/T1)=(P2/P1)(R/Cp) from TP(1-ϒ)/ϒ=constant
    4-CpdT=dH
    3. The attempt at a solution
    So far for part a) I have done the equation dU=dQ+dW, since it is an adiabatic process I have dQ=0, thus leaving dU=dW. I know that dW=-PdV so I can calculate dU=CvdT to get my change in internal energy. From there I calculated P2 using equation 3.

    For b) knowing it is an Cp problem I used equation 4 having CpdT=dH where my change in temperature is 150 to 70 C

    I am stuck on part c)

    I know it is an isotherm that means that there is no change in temperature thus dU and dH are equal to zero since they are temperature dependent. So dQ=dW. Do I make it so that T is a constant and solve with partial derivatives giving me dT=(V/R)dP+(P/R)dV. With dT=0 we get -PdV=VdP. Using the ideal gas equation to give V=RT/P ==> -(RT/P) dP. Integrating gives me -RTln(P2/P1)

    My problem is I don't know what pressure to use. Do i just use the pressures calculated in my part b going from there back to the original state?
     
  2. jcsd
  3. Feb 27, 2017 #2
    You haven't told us the questions you need to answer!
     
  4. Feb 27, 2017 #3
    Yes.
     
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