Question

Air in a closed piston cylinder device is initially at 800 K and at 250 kPa.
The air undergoes a process until it is at 2200 K and at 750 kPa.

What is the change in the air's specific entropy during this process (kJ/kgK)?

Answer 1

Entropy of air (or any ideal gas) depends upon both its temperature and pressure.

s = f(p,T)

Using Table A-1 (Molar mass, gas constant and critical point properties);

R = 0.287 kJ/kg.K

Using Table A-17 (Ideal gas properties of air);

• at T₁ = 800 K;

s₁^o = 2.71787 kJ/kg.K

• at T₂ = 2200 K;

s₂^o = 3.9191 kJ/kg.K

The change in specific entropy is given as;

Δs = (s₂^o - s₁^o) - R ln(P₂/P₁)

=> Δs = (3.9191 - 2.71787) - (0.287) ln(750/250)

=> Δs = 0.88593 kJ/kg.K

Answer:

Δs = 0.886 kJ/kg.K