Posted by **Janus** on Saturday, March 6, 2010 at 11:53am.

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

- Thermodynamics -
**drwls**, Saturday, March 6, 2010 at 12:34pm
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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