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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/m^{2} and a temperature of 20^{o}C to a pressure of 405300 N/m^{2}. C_{V} = 20.934 kJ/kmol.^{o}C.

Calculations:

Entropy change of an ideal gas,

DS =
C_{p} ln (T_{2}/T_{1}) - R ln (P_{2}/P_{1})

T_{1} = 20^{o}C = 293 K

P_{1} = 101325 N/m^{2}

P_{2} = 405300 N/m^{2}

For the constant volume heating process, V_{2} = V_{1}

Therefore, P_{2}V_{1} = RT_{2}

V_{1} = RT_{1}/P_{1} = 293R/101325

Therefore, T_{2} =405300 x (293R/101325) /R = 293 x (405300/101325) = 1172 K

C_{p} = R + C_{v} = 8.314 + 20.934 = 29.248 kJ/kmol.^{o}C

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

= 40.546 - 11.526 = 29.02 kJ/kmol.^{o}C

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/^{o}C

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Last Modified on: 11-Sep-2014

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