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An ideal gas having a mass of 2 kg at 465K and 415 kPa expands in a reversible adiabatic process to 138 kPa. The ideal gas constant is 242 J/kg. K and k = 1.4. Determine

a. Final temperature
b. Change in internal energy
c. The work
d. Cp
e. Cv

Non-flow isometric

a. T2 : PVT

T2 / T1 = (P2 / P1) k-1/k

= 465 (138/415) 0.4/1.4

= 339.5 K

b. change in internal energy = mCv (T2 – T1)

Cp / Cv = k = 1.4
Cp = Cv (1.4)
Cp – Cv = R
1.4 Cv – Cv = 0.242 kJ / kg K
Cv = 0.605 kJ / kg K

change in internal energy = 2(0.605) (339.5 – 465)
change in U = -151.85 kJ = 1.51.85 kJ (decrease)

c. Wnf = P2V2 – P1V1 / 1 – k = mR (T2 – T1) / 1 – k

Q = Wnf + change in U
Wnf – Au = (-151.85)
Wnf = 151.85 kJ

d. Cp = 1.4 (0.605) = 0.847 kJ / kg K
e. Cv = 0.605 kJ / kg K


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