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.

Question

Engineering

Isai

5 March, 11:00

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.

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Answers (1)

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Amiah Mitchell · Answer 1

(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