Question

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.
Determine:
a.)the final temperature
b.)change of internal energy
change of enthalpy
change of entropy

Answer 1

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