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
