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Propane gas at 1 bar and 35°C is compressed to a final state of 135 bar...

Propane gas at 1 bar and 35°C is compressed to a final state of 135 bar and 195°C. Estimate the molar volume of the propane in the final state and the enthalpy and entropy changes for the process. In its initial state, propane may be assumed an ideal gas.

Expert Solution

Calculate the reduced properties of propane

Reduced temperature = given temperature /critical temperature

Tr = T/Tc

= (195+273.15)/369.8

= 1.266

Similarly reduced pressure

Pr = P/Pc

= 135/42.48 = 3.178

Accentric factor w = 0.152

Lee Kesler equation

At the calculated reduced properties

Z0 = 0.6141

Z1 = 0.1636

Z = Z0 + wZ1

= 0.6141 + 0.152*0.1636

= 0.639

Molar volume

V = ZRT/P

= 0.639*83.126 cm3-bar/mol·K*(195+273.15)K/135 bar

= 184.2 cm3/mol

Calculate the enthalpy at critical temperature

HR0 = - 2.496 RTc

= - 2.496*8.314*369.8 = - 7.674 x 10^3 J/mol

HR1 = - 0.586 RTc

= - 0.586 x 8.314 x 369.8 = - 1.802 x 10^3 J/mol

HR = HR0 + w*HR1

= - 7.674 x 10^3 + 0.152 * (- 1.802 x 10^3)

= - 7.948 x 10^3 J/mol

Enthalpy change

H = HR + R Cp dT

H = HR + R (1.213 + 28.785*10^-3 T - 8.824*10^-6 T^2)dT

H = - 7.948 x 10^3 + 8.314 (1.213 + 28.785*10^-3 T - 8.824*10^-6 T^2)dT

H = 6734.9 J/mol

Calculate the entropy

SR0 = - 1.463 R

= - 1.463*8.314 = - 12.163 J/mol-K

SR1 = - 0.717 R

= - 0.717 x 8.314 = - 5.961 J/mol-K

SR = SR0 + w*SR1

= - 12.163 + 0.152 * (- 5.961)

= - 13.069 J/mol-K

Entropy change

S = R (1.213 + 28.785*10^-3 T - 8.824*10^-6 T^2)dT - ln (P/P0) + SR

S = 8.314 (1.213 + 28.785*10^-3 T - 8.824*10^-6 T^2)dT + ln (135/1) - 13.069

S = - 15.9 J/mol-K

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