Question

Two kilograms of air in a piston-cylinder assembly undergoes an isothermal process from an initial state of 200K, 300kPa to 600kPa. a) How much entropy is generated in the process, in kJ/K. b) Calculate the amount of lost work.

Answer 1

a) 0.39795 kJ/K

b) 79.589.37 kJ

Step-by-step explanation:

m = Mass of air = 2 kg

Temperature = 200 K

P₁ = Initial pressure = 300 kPa

P₂ = Final pressure = 600 kPa

R = mass-specific gas constant for air = 287.058 J/kgK

a) For isentropic process

`\Delta S=mRln(P_1)/(P_2)\\\Rightarrow \Delta S=2* 287.058ln(300)/(600)\\\Rightarrow \Delta S=-397.95\ J/K`

∴ Entropy is generated in the process is 0.39795 kJ/K

b)

`W=mRTln(P_1)/(P_2)\\\Rightarrow W=2* 287.058* 200ln(300)/(600)\\\Rightarrow W=-79589.37\ J`

∴ Amount of lost work is 79.589.37 kJ