Thermodynamics filled in class notes_Part_58

Thermodynamics filled in class notes_Part_58 - 123 4.5. VAN...

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Unformatted text preview: 123 4.5. VAN DER WAALS GAS 2 1 Now 1 w3 = 1 w2 + 2 w3 = v2 1 w3 v1 = RT1 = RT1 ln = 200 v2 − b v1 − b J kg J −138095 , kg −138.095 (4.227) (4.228) +a P a m6 kg 2 −140408 = P dv since 2 → 3 is at constant volume. So 1 1 , − v2 v1 3 3 0.0585 m − 0.001 m kg kg (300 K ) ln m3 m3 0.598 kg − 0.001 kg J kg K + 150 = 2 1 P dv = RT a − 2 dv, v−b v v2 v2 dv dv , −a 2 v1 v v1 v − b = = 3 2 P dv + 1 0.0585 + 2313 m3 kg J kg 1 0.598 − m3 kg (4.229) , (4.230) , (4.231) (4.232) kJ . kg (4.233) The gas is compressed, so the work is negative. Since u is a state property: T3 u3 − u1 = 1 1 − v1 v3 cv (T )dT + a T1 . (4.234) Now cv = = J kg K J 290 kg K 350 J kg K 2 J + 0.2 kg K 2 (T − (300 K )), (4.235) T. + 0.2 (4.236) so T3 u3 − u1 290 = T1 = = J kg K J 290 kg K 290 + 150 = = = J kg K J kg J , 291687 kg kJ 292 . kg J kg K 2 (T3 − T1 ) + 0.1 T J kg K 2 dT + a 1 0.598 + 91000 m3 kg J kg 1 0.0585 m3 kg − 2313 J kg − 1 1 − v1 v3 2 2 T3 − T1 + a ((1000 K ) − (300 K )) + 0.1 P a m6 kg 2 203000 + 0.2 J kg K 2 , 1 1 − v1 v3 (4.237) , (4.238) (1000 K )2 − (300 K )2 , (4.239) , (4.240) (4.241) (4.242) CC BY-NC-ND. 18 November 2011, J. M. Powers. 124 CHAPTER 4. MATHEMATICAL FOUNDATIONS OF THERMODYNAMICS Now from the first law u3 − u1 1 q3 = 1 q3 − 1 w3 , = u3 − u1 + 1 w3 , kJ kJ − 138 = 292 kg kg 1 q3 1 q3 = 154 (4.243) (4.244) , (4.245) kJ . kg (4.246) The heat transfer is positive as heat was added to the system. Now find the entropy change. Manipulate the Gibbs equation: T ds = ds = ds = ds = ds = s3 − s1 = = = du + P dv, 1 P du + dv, T T a P 1 cv (T )dT + 2 dv + dv, T v T 1 a RT 1 cv (T )dT + 2 dv + − T v T v−b cv (T ) R dT + dv, T v−b T3 cv (T ) v3 − b dT + R ln , T v1 − b T1 1000 290 kgJ K J + 0.2 T kg K 2 300 290 J kg K J + 200 kg K = = J kg K J 21 , kg K = 349 0.021 1000 K 300 K ln 0.0585 ln 0.598 + 140 + 0.2 m3 kg m3 kg J kg K (4.247) (4.248) (4.249) a v2 dv, (4.250) (4.251) (4.252) dT + R ln v3 − b , v1 − b J kg K 2 − 0.001 − 0.001 − 468 (4.253) ((1000 K ) − (300 K )) m3 kg m3 kg J kg K , (4.254) , (4.255) (4.256) kJ . kg K (4.257) Is the second law satisfied for each portion of the process? First look at 1 → 2 u2 − u1 1 q2 = = 1 q2 − 1 w2 , u2 − u1 + 1 w2 , = 1 q2 T2 cv (T )dT + a T1 (4.258) (4.259) 1 1 − v1 v2 CC BY-NC-ND. 18 November 2011, J. M. Powers. + RT1 ln v2 − b v1 − b +a 1 1 − v2 v1 . (4.260) ...
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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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