Thermodynamics filled in class notes_Part_71

Thermodynamics filled in class notes_Part_71 - 4.11...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

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.

Page1 / 2

Thermodynamics filled in class notes_Part_71 - 4.11...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online