Thermodynamics filled in class notes_Part_71

Thermodynamics filled in class notes_Part_71 - 4.11...

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Unformatted text preview: 4.11. EQUILIBRIUM IN A TWO-COMPONENT SYSTEM 149 The molar concentrations are a little smaller than for the isochoric case, mainly because the volume is larger at equilibrium. Now consider the heat transfer. One knows for the isobaric process that Q = H 2 − H 1 . The initial enthalpy is given by H 1 = n N 2 h N 2 = (1 kmole ) parenleftbigg 2 . 05848 × 10 5 kJ kmole parenrightbigg = 2 . 05848 × 10 5 kJ. (4.515) The enthalpy at the final state is H 2 = n N 2 h N 2 + n N h N , (4.516) = (0 . 882147 kmole ) parenleftbigg 2 . 05848 × 10 5 kJ kmole parenrightbigg + (0 . 235706 kmole ) parenleftbigg 5 . 9727 × 10 5 kJ kmole parenrightbigg , (4.517) = 3 . 22368 × 10 5 kJ. (4.518) So 1 Q 2 = H 2 − H 1 = 3 . 22389 × 10 5 kJ − 2 . 05848 × 10 5 kJ = 1 . 16520 × 10 5 kJ. (4.519) Heat needed to be added to keep the system at the constant temperature. This is because the nitrogen dissociation process is endothermic . Relative to the isochoric process, more heat had to be added to maintain the temperature. This to counter the cooling effect of the expansion.maintain the temperature....
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This note was uploaded on 11/26/2011 for the course EGN 3381 taught by Professor Park-sou during the Fall '11 term at FSU.

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Thermodynamics filled in class notes_Part_71 - 4.11...

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