Find: (i) Change in internal energy (ii) Work done by air Assume Cp = 1.005 KJ/kg K and Cv = 0.718 KJ/kg K

Given data: Initial pressure = P1 = 14 Bar = 14 x 105 N/m2

Final pressure = P2 = 1.4 Bar = 1.4 x 105 N/m2

Initial volume = V1 = 0.6 $m^3$

Cp = 1.005 kJ/kgK,

Cv = 0.718 kJ/kgK and

Process is $PV^{1.25} = C$

Air mass m = 1 kg

R = Cp-Cv = 0.287 kJ/kgK

For Polytropic process, we have

[V1/ V2] n = [P2/ P1] or [V1/ V2] = [P2/ P1] 1/n

[ 0.6/ V2 ] = [1.4/ 14] 1/1.25

[ 0.6/ V2 ] = [0.1]0.8

V2 = 0.158 $m^3$

Also from following two general gas equations for initial & final conditions

T1 = P1 V1 / m R = 1400 × 0.6/0.287 = 2926 K

T2 = P2 V2 / m R= 140 × 0.158 /0.287 = 77K

Now change in internal energy

dU= m Cv[T2 – T1] = 1×0.718 [ 2926.8 – 77] = 2045 KJ

Workdone = 1400 × 0.6 – 140 × 0.58 / 1.25-1= 327152 kJ