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5 pounds of an ideal gas with R=38.7 and K=1.668 have 300Btu of heat added during a reversible constant pressure change of state, initial temperature=80°F.
a.)the final temperature
b.)change of internal energy
change of enthalpy
change of entropy

  • Thermodynamics - ,

    a) Since it is a constant pressure heating, dQ = M Cp dT = [K*M*R/((K-1)]*dT

    Use that to determine the temperature changle dT, and add that to 80 F.

    You need to specify the units of R. They should be Btu/lbm*degF

    b) dU = M Cv dT = [M/(K-1)]dT
    c) dH = Cp*dT = dQ
    d) dS = Integral of dQ/T
    = Integral of M Cp dT/T
    = M Cp ln(T2/T1)
    The T's must be in degrees Rankine

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