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

Answer:

a) 0.39795 kJ/K

b) 79.589.37 kJ

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

$\Delta S=mRln\frac{P_1}{P_2}\\\Rightarrow \Delta S=2\times 287.058ln\frac{300}{600}\\\Rightarrow \Delta S=-397.95\ J/K$

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

b)

$W=mRTln\frac{P_1}{P_2}\\\Rightarrow W=2\times 287.058\times 200ln\frac{300}{600}\\\Rightarrow W=-79589.37\ J$

∴ Amount of lost work is 79.589.37 kJ