A piston-cylinder assembly contains 5 kg of air, initially at 2.0 bar, 30 C. The air undergoes a process to a state where the pressure is 2 bar, during which the pressure-volume relationship is pV1.5 = constant. Assume ideal gas behavior for the air. Determine the work and heat transfer, in kJ. Note molar mass of air is 28.96 g/mole

Question

Engineering

Wyatt Sutton

24 May, 07:30

A piston-cylinder assembly contains 5 kg of air, initially at 2.0 bar, 30 C. The air undergoes a process to a state where the pressure is 2 bar, during which the pressure-volume relationship is pV1.5 = constant. Assume ideal gas behavior for the air. Determine the work and heat transfer, in kJ. Note molar mass of air is 28.96 g/mole

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Melvin Case · Answer 1

Answer: wor done is 145. 06kJ

Heat transfer is 135.53kJ

Explanation:

No of moles of air = mass/molar mass = 5000g/28gmol^-1 = 172.65mol

P1 = 2bar = 2*101300 = 202600pa

T1 = 30° + 273k = 303k

P2 = p1 = 202600pa

V2 = ? T2 = ?

Using pV = nRT

R = 8.314 PA m^3 mol^-1 k^-1

V1 = (172.65*8.314*303) / 202600

V1 = 2.146m^3

For second state, 1.5pv = const = P1V1

V2 = (202600*2.146) / (1.5*202600)

V2 = 1.43m^3

Volume change = 2.146 - 1.43 = 0.715m^3

Word done = pressure * volume change

W = 202600*0.716 = 145061.6J

= 145.061kJ

Using V1/T1 = V2/T2

T2 = V2T1/V1

= (1.43*303) / 2.146 = 201.9k

For internal energy U

U = nCv * (T2 - T1)

*CV is the heat capacity at const. vol approximately 0.718J mol^-1 k^-1

U = 172.65*0.718 * (201.9-303)

U = - 12532.6J = - 12.532kJ

The - ve means the system lost internal energy.

Q = U+W = total heat energy of system

Q = - 12.532+145.061 = 132.52 kJ