For enthalpy 2 going in the heater, the result is -0.15 kj/mol air. How was this obtained? ||!PDF Elementary Principles of Chemica X PDF *Elementary Principles of Chemic X +58°FSunnyFile | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDFDrive%20).pdfDrawT Read aloud500(---)T1000(-)652 APPENDIX B Physical Property TablesQ SearchA0← →0.000734Ĥ 187822933021Û 1791216927650.001726 0.0024910.005112179820512316259416531888212723690.001446 0.001628 0.001893 0.0022460.002000672 of 6950.000000272325290.003882U.VIVZTABLE B.9 Specific Enthalpies of Selected Gases:U.S. Customary UnitsHOPTABLE B.8 Specific Enthalpies of Selected Gases: SI UnitsĤ (kJ/mol)Reference state: Gas, Pref= 1 atm, Tref = 25°CCOCO₂ H₂OAir 0₂-0.72 -0.730.00N₂-0.73H₂-0.720.000-0.73-0.92 -0.84250.00 0.000.000.000.001002.192.242.192.162.192.902.542005.155.315.135.065.167.086.013008.178.478.127.968.17 11.589.5713.2317.01400 11.24 11.72 11.15 10.89 11.25 16.35500 14.37 15.03 14.24 13.83 14.38 21.34600 17.55 18.41 17.39 16.81 17.57 26.53 20.91700 20.80 21.86 20.59 19.81 20.82 31.88 24.92800 24.10 25.35 23.86 22.85 24.13 37.36 29.05900 27.46 28.89 27.19 25.93 27.49 42.94 33.321000 30.86 32.47 30.5629.0430.91 48.60 37.691100 34.31 36.07 33.99 32.19 34.37 54.33 42.181200 37.81 39.70 37.46 35.39 37.87 60.14 46.781300 41.34 43.38 40.97 38.62 41.40 65.98 51.471400 44.89 47.07 44.51 41.90 44.95 71.89 56.251500 48.45 50.77 48.06 45.22 48.51 77.84 61.09V.VIIT32483439294630910.006112 0.007000285725910.002668V.VIZU310527950.003106CVIJO361032240.007722332429710.003536VUITT377133500.008418352631310.003953{"ⓇJ6350DENGSign in(0)+07:53 AM5/22/2023 ||!PDF Elementary Principles of Chemica X PDF *Elementary Principles of Chemic X +70°FSunnyFile | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDFDrive%20).pdfDrawT Read aloudExample 8.3-5SolutionEnergy Balance on a Gas PreheaterA stream containing 10% CH4 and 90% air by volume is to be heated from 20°C to 300°C. Calculatethe required rate of heat input in kilowatts if the flow rate of the gas is 2.00 × 10³ liters (STP)/min.Q Search441 of 695Basis: Given Flow RateAssume ideal-gas behavior.2000 L (STP)/min, 20°Cn(mol/min)0.100 mol CH4/mol0.900 mol air/molCDn =2000 L (STP)minHEATER(010)Q(kW)Recall that specifying the flow rate in liters (STP)/min does not imply that the feed gas is at standardtemperature and pressure; it is simply an alternative way of giving the molar flow rate.8.3 Changes in Temperature 4211 mol22.4L (STP)n(mol/min), 300°C0.100 mol CH₂/mol0.900 mol air/mol= 89.3 mol/minThe energy balance with kinetic and potential energy changes and shaft work omitted is Q = AH. Thetask is to evaluate AH = [n¡Â¡- ΣnĤ₁. Since each species has only one inlet condition and one outletoutincondition in the process, two rows are sufficient for the enthalpy table.References: CH4(g, 20°C, 1 atm), air(g, 25°C, 1 atm)ninĤinnoutSubstance (mol/min) (kJ/mol) (mol/min)Ĥ out(kJ/mol){"ƠⓇJ6360DENGSign in(0)+O11:34 AM5/22/2023

Question

For enthalpy 2 going in the heater, the result is -0.15 kj/mol air.  How was this obtained? ||!PDF Elementary Principles of Chemica X PDF *Elementary Principles of Chemic X +58°FSunnyFile | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDFDrive%20).pdfDrawT Read aloud500(---)T1000(-)652 APPENDIX B Physical Property TablesQ SearchA0← →0.000734Ĥ 187822933021Û 1791216927650.001726 0.0024910.005112179820512316259416531888212723690.001446 0.001628 0.001893 0.0022460.002000672 of 6950.000000272325290.003882U.VIVZTABLE B.9 Specific Enthalpies of Selected Gases:U.S. Customary UnitsHOPTABLE B.8 Specific Enthalpies of Selected Gases: SI UnitsĤ (kJ/mol)Reference state: Gas, Pref= 1 atm, Tref = 25°CCOCO₂ H₂OAir 0₂-0.72 -0.730.00N₂-0.73H₂-0.720.000-0.73-0.92 -0.84250.00 0.000.000.000.001002.192.242.192.162.192.902.542005.155.315.135.065.167.086.013008.178.478.127.968.17 11.589.5713.2317.01400 11.24 11.72 11.15 10.89 11.25 16.35500 14.37 15.03 14.24 13.83 14.38 21.34600 17.55 18.41 17.39 16.81 17.57 26.53 20.91700 20.80 21.86 20.59 19.81 20.82 31.88 24.92800 24.10 25.35 23.86 22.85 24.13 37.36 29.05900 27.46 28.89 27.19 25.93 27.49 42.94 33.321000 30.86 32.47 30.5629.0430.91 48.60 37.691100 34.31 36.07 33.99 32.19 34.37 54.33 42.181200 37.81 39.70 37.46 35.39 37.87 60.14 46.781300 41.34 43.38 40.97 38.62 41.40 65.98 51.471400 44.89 47.07 44.51 41.90 44.95 71.89 56.251500 48.45 50.77 48.06 45.22 48.51 77.84 61.09V.VIIT32483439294630910.006112 0.007000285725910.002668V.VIZU310527950.003106CVIJO361032240.007722332429710.003536VUITT377133500.008418352631310.003953{&#34;ⓇJ6350DENGSign in(0)+07:53 AM5/22/2023 ||!PDF Elementary Principles of Chemica X PDF *Elementary Principles of Chemic X +70°FSunnyFile | E:/Elementary%20Principles%20of%20Chemical%20Processes,%204th%20Edition%20(%20PDFDrive%20).pdfDrawT Read aloudExample 8.3-5SolutionEnergy Balance on a Gas PreheaterA stream containing 10% CH4 and 90% air by volume is to be heated from 20°C to 300°C. Calculatethe required rate of heat input in kilowatts if the flow rate of the gas is 2.00 × 10³ liters (STP)/min.Q Search441 of 695Basis: Given Flow RateAssume ideal-gas behavior.2000 L (STP)/min, 20°Cn(mol/min)0.100 mol CH4/mol0.900 mol air/molCDn =2000 L (STP)minHEATER(010)Q(kW)Recall that specifying the flow rate in liters (STP)/min does not imply that the feed gas is at standardtemperature and pressure; it is simply an alternative way of giving the molar flow rate.8.3 Changes in Temperature 4211 mol22.4L (STP)n(mol/min), 300°C0.100 mol CH₂/mol0.900 mol air/mol= 89.3 mol/minThe energy balance with kinetic and potential energy changes and shaft work omitted is Q = AH. Thetask is to evaluate AH = [n¡Â¡- ΣnĤ₁. Since each species has only one inlet condition and one outletoutincondition in the process, two rows are sufficient for the enthalpy table.References: CH4(g, 20°C, 1 atm), air(g, 25°C, 1 atm)ninĤinnoutSubstance (mol/min) (kJ/mol) (mol/min)Ĥ out(kJ/mol){&#34;ƠⓇJ6360DENGSign in(0)+O11:34 AM5/22/2023

Accepted Answer

Specific enthalpy of the air is to be determined.