Determine the work done and change in internal energy.

Given We Assume R= 0.287 KJ/KgK for air ,m = 1 kg,

Initial temp T1 = 156⁰C = 156+273= 429⁰K,

Initial pressure P1= 1 bar = 100KPa ,

Final Volume V2= 0.28$m^3$

Applying characteristic gas equation,

P1V1 = mRT1

100 X V1= 1 X 0.287 X 429

V1 =1.231 $m^3$

For Isothermal process,

P1V1=P2V2

100 X 1.231 = P2 X 0.28

P2 = 439.64 KPa

Work done for Isothermal process

dW= P1V1loge(V2/V1)

= 100 X 1.231 X loge(0.28/1.23)

= - 182.138 KJ Minus sign indicates that work is supplied during compression. Change in Internal energy = dU = mCv(T2-T1) = 0

(As T1 = T2 For isothermal process)