100 kg of refrigerant-134a at 200 kPa iscontained in a piston-cylinder device whose volume is 12.322 m3. The piston is now moved until the volume is one-half its original size. This is done such that thepressure of the refrigerant-134a does not change. Determine (a) the final temperatureand (b) the change in the specific internal energy.

(a) Final temperature is 151.2 K

(b) Change in the specific internal energy is -30.798 kJ/kg

Explanation:

(a) P1 = P2 = 200 kPa

V1 = 12.322 m^3

V2 = 1/2 × V1 = 1/2 × 12.322 = 6.161 m^3

mass of refrigerant-134a = 100 kg

MW of refrigerant-134a (C2H2F4) = (12×2) + (2×1) + (19×4) = 24 + 2 +76 = 102 g/mol

number of moles of refrigerant-134a (n) = mass/MW = 100/102 = 0.98 kgmol

R = 8.314 kJ/kgmol.K

Final temperature (T2) = P2V2/nR = 200×6.16/0.98×8.314 = 151.2 K

(b) Change in specific internal energy = change in internal energy (∆U)/mass (m)

∆U = Cv(T2 - T1)

Cv = 20.785 kJ/kgmol.K

T1 = P1V1/nR = 200×12.322/0.98×8.314 = 302.4 K

∆U = 20.785(151.2 - 302.4) = 20.785 × -151.2 = -3142.7 kJ/kgmol × 0.98 kgmol = -3079.8 kJ

Change in specific internal energy = -3079.8 kJ/100 kg = -30.798 kJ/kg

answer:

6.3 is the answer

explanation:

because it is the ex