Determine the entropy of an ideal gas with c p =(7/2)R,between the following sta

Question

Determine the entropy of an ideal gas with cp=(7/2)R,between the following states: (a) P1=1bar,T1= 300K, P2= 0.5bar, T2= 500K (b) v1= 0.05m3/mol, T1= 300K ; v2 0.025m3/mol, T2= 500K (c)P1= 1 bar, T1=300K; v2=0.025m3/mol, T2=500K (a) P1=1bar,T1= 300K, P2= 0.5bar, T2= 500K (b) v1= 0.05m3/mol, T1= 300K ; v2 0.025m3/mol, T2= 500K (c)P1= 1 bar, T1=300K; v2=0.025m3/mol, T2=500K

Explanation / Answer

dH = TdS + VdP = nCpdT dS = n*Cp*dT/T - (V/T)*dP dS = n*Cp*dT/T - (nR/P)dP S = n*Cp*ln(Tf/Ti) - nR*ln(Pf/Pi) S/n = Cp*ln(Tf/Ti) - R*ln(Pf/Pi) a) S = (7/2)*(8.314 J/mol-K)*(ln(500K/300K) - (8.314J/mol-K)*ln(0.5 bar / 1 bar) S = 20.6273405 J/mol-K S = 20.6 J/mol/K (b) v1= 0.05m3/mol, T1= 300K ; v2 0.025m3/mol, T2= 500K S/n = Cp*ln(Tf/Ti) - R*ln(Pf/Pi) Pf*Vf/Tf = Pi*Vi/Ti      =>   Pf/Pi = (Vi/Vf)*(Tf/Ti) S/n = Cp*ln(Tf/Ti) - R*ln(Vi*Tf/Vf*Ti) S/n = (7/2)*(8.314 J/mol/K)*ln(500K/300K) - (8.314J/mol/K)*ln(0.05 m3/mol *500K / 0.025 m3/mol *300K) S/n = 4.85468493 S/n = 4.85 J/mol/K (c)P1= 1 bar, T1=300K; v2=0.025m3/mol, T2= 500K P2 = RT/V2 = (8.314 J/mol-K)*(500 K)/(0.025 m3/mol) = 166280 Pa =166280 Pa *(1 bar/ 1e5 Pa) = 1.6628 bar S/n = Cp*ln(Tf/Ti) - R*ln(Pf/Pi) S/n = (7/2)*(8.314 J/mol-K)*ln(500 K / 300K) - 8.314 J/mol-K* ln(1.6628 bar / 1 bar) S/n = 10.6368215 S/n = 10.6 J/mol-K

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