Entropy Change of Air

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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 20oC to a pressure of 405300 N/m2. CV = 20.934 kJ/kmol.oC.

Calculations:

Entropy change of an ideal gas,

ΔS = Cp ln (T2/T1) - R ln (P2/P1)

T1 = 20oC = 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

Cp = R + Cv = 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

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Last Modified on: 01-May-2024

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