Question

9. ICO4] One lbmol of carbon at 77-F and 1 is then brought to 77 F and 1 atm, the conditions of the surroundings. Assuming the reversible work for this process is during the prooest hennd i atm is burned steadily with 1 bmal of oygen at ne come suatunmiye formed the is complete, 51.03 Btu b. 169,300 Btu C 48.9 Btu d. 169,680 Btu e(fill in the answer on your bubble sheet and provide supporting calculations below Answer.

Answer 1

Solution: (d) 169,680 Btu Carbon is burned steadily withpure oxygen. The reversible workassociated with this process is to be determined Assumptions: 1. Combustion is complete. 2. Steady-flow conditions exist during combustion. 3. Oxygen and the combustion gases are ideal gases. 4. Changes in kinetic and potential energies are negligible Properties: The Gibbs function of formation at 77°F and 1 atm is 0 for C and O2, and -169,680 Btu/lbmol for CO2 The enthalpy of formation is 0 for C and O2, and -169,300 Btu/lbmol for CO2 The absolute entropy is 1.36 Btu/lbmol-R for C, 49.00 Btu/lbmol-R for O2, and 51.07 Btu/lbmol-R for CO2 Analysis: C + O2 → CO2 The combustion equation is The C, O2, and CO2 are at 77°F and 1 atm, which is the standard reference state and also the state of the surroundings. Therefore, the reversible work in this case is simply the difference between the Gibbs function of formation of the reactants and that of the products 0 +No.gjo = (-1 lbmol)(-169.680 Btu/lbmol) 169,680 Btu

Another Method: We could also determine the reversible work without involving the Gibbs function by using Substituting the enthalpy of formation and absolute entropy values, we obtain Wey-(1 lbmol C)[O (537 R) (1.36 Btu/lbmol R)] R)(49.00 Btu/lbmol R)] + (1 lbmol 02)[0 (537 -(1 lbmo1 CO-)[-169,300 Btu/lbmol-(537 R)(51.07 Btu/lbmol . R)] - 169,680 Btu