Entropy Change of Air

Calculate the change in entropy when 10 kg of air is heated at constant volume from a pressure of 101325 N/m² and a temperature of 20^oC to a pressure of 405300 N/m². C_V = 20.934 kJ/kmol.^oC.

Calculations:

Entropy change of an ideal gas,

ΔS = C_p ln (T₂/T₁) - R ln (P₂/P₁)

T₁ = 20^oC = 293 K

P₁ = 101325 N/m²

P₂ = 405300 N/m²

For the constant volume heating process, V₂ = V₁

Therefore, P₂V₁ = RT₂

V₁ = RT₁/P₁ = 293R/101325

Therefore, T₂ =405300 x (293R/101325) /R = 293 x (405300/101325) = 1172 K

C_p = R + C_v = 8.314 + 20.934 = 29.248 kJ/kmol.^oC

ΔS = 29.248 x ln (1172/293) - 8.314 x ln (405300/101325)

= 40.546 - 11.526 = 29.02 kJ/kmol.^oC

kmol of air equivalent to 10 kg of air = 10/29 = 0.345 kmol

Change in entropy of air = 0.345 x 29.02 = 10.007 kJ/^oC