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Unformatted text preview: Work V 1 V 2 V P â âĄ d = a bit of algebra leads to Work c P 1 P 2 P P P b + â âĄ d â
= Work 0.516 J gm = Ans. Alternatively, formal integration leads to Work c P 2 P 1 b ln P 2 b + P 1 b + â
 â
= Work 0.516 J gm = Ans. 3.5 Îș a b P â
+ = a 3.9 10 6 â
atm 1 â
= b 0.1 10 9 â
atm 2 â
= P 1 1 atm â
= P 2 3000 atm â
= V 1 ft 3 â
= (assume const.) Combine Eqs. (1.3) and (3.3) for const. T: Work V P 1 P 2 P a b P â
+ ( ) P â
â âĄ d â
= Work 16.65atm ft 3 â
= Ans. Chapter 3  Section A  Mathcad Solutions 3.1 ÎČ 1 Ï T Ï d d â
= Îș 1 Ï P Ï d d â
= P T At constant T, the 2nd equation can be written: d Ï Ï Îș dP â
= ln Ï 2 Ï 1 Îș â P â
= Îș 44.18 10 6 â
bar 1 â
= Ï 2 1.01 Ï 1 â
= â P ln 1.01 ( ) Îș = â P 225.2bar = P 2 226.2 bar â
= Ans. 3.4 b 2700 bar â
= c 0.125 cm 3 gm â
= P 1 1 bar â
= P 2 500 bar â
= Since 16 P 2 1 bar â
= T 1 600 K â
= C P 7 2 R â
= C V 5 2 R â
= (a) Constant V: W = and â U Q = C V â T â
= T 2 T 1 P 2 P 1 â
= â T T 2 T 1 = â T 525 K = â U C V â T â
= Q and â U 10.91 kJ mol = Ans. â H C P â T â
= â H 15.28 kJ mol = Ans. (b) Constant T: â U â H = = and Q W = Work R T 1 â
ln P 2 P 1 â
= Q and Work 10.37 kJ mol = Ans. (c) Adiabatic: Q = and â U W = C V â T â
= 3.6 ÎČ 1.2 10 3 â
degC 1 â
= C P 0.84 kJ kg degC â
â
= M 5 kg â
= V 1 1 1590 m 3 kg â
= P 1 bar â
= t 1 0 degC â
= t 2 20 degC â
= With beta independent of T and with P=constant, dV V ÎČ dT â
= V 2 V 1 exp ÎČ t 2 t 1 ( 29 â
â
= â V V 2 V 1 = â V total M â V â
= â V total 7.638 10 5 Ă m 3 = Ans. Work P â V total â
= (Const. P) Work 7.638 joule = Ans. Q M C P â
t 2 t 1 ( 29 â
= Q 84kJ = Ans. â H total Q = â H total 84kJ = Ans. â U total Q Work + = â U total 83.99kJ = Ans. 3.8 P 1 8 bar â
= 17 Step 41: Adiabatic T 4 T 1 P 4 P 1 R C P â
= T 4 378.831K = â U 41 C V T 1 T 4 ( 29 â
= â U 41 4.597 10 3 Ă J mol = â H 41 C P T 1 T 4 ( 29 â
= â H 41 6.436 10 3 Ă J mol = Q 41 J mol = Q 41 J mol = W 41 â U 41 = W 41 4.597 10 3 Ă J mol = P 2 3bar = T 2 600K = V 2 R T 2 â
P 2 = V 2 0.017 m 3 mol = Step 12: Isothermal â U 12 J mol = â U 12 J mol = Îł C P C V = T 2 T 1 P 2 P 1 Îł 1 Îł â
= T 2 331.227K = â T T 2 T 1 = â U C V â T â
= â H C P â T â
= W and â U 5.586 kJ mol = Ans. â H 7.821 kJ mol = Ans. 3.9 P 4 2bar = C P 7 2 R = C V 5 2 R = P 1 10bar = T 1 600K = V 1 R T 1 â
P 1 = V 1 4.988 10 3 Ă m 3 mol = 18 T 4 378.831K = V 4 R T 4 â
P 4 = V 4 0.016 m 3 mol = Step 34: Isobaric â U 34 C V T 4 T 3 ( 29 â
= â U 34 439.997 J mol = â H 34 C P T 4 T 3 ( 29 â
= â H 34 615.996 J mol = Q 34 C P T 4 T 3 ( 29 â
= Q 34 615.996615....
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This note was uploaded on 01/29/2011 for the course CHEM 101 taught by Professor Brown during the Spring '10 term at The University of Akron.
 Spring '10
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