Question

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

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