Please provide the detailed solution not the final answer on

Please provide the detailed solution!!! not the final answer only!! Please

Please provide the detailed solution!!!

Process: 1

According to 1^st law of thermodynamics

Q = W + U

W = pdv : Since V = 0 , W = 0

Therefore, Q = 0 +240 = 240 KJ

Process: 2

Q = W + U

Q = pdv + U

-50 = 1000*(0.81-0.27) + U (Substitute pressure in kpa 1 bar = 100 KPa)

U = -590 kJ

Process: 4

Q = W + U

-100 = 300 * (0.81-0.27) + U

U = -262 kJ

Process: 3

In a cyclic process, the net heat transfer is equal to network transfer

Q = W

Q1 + Q2 + Q3 +Q 4 = W1 +W2+W3+W4

240-50+Q3-100 = 0+(1000*(0.81-0.27))+0+(300*(0.81-0.27))

Q3 = 612 KJ

Q = W + U

774 = 0 + U

U = 774 KJ

Cant able to find KE and PE