SM_3A - Work V 1 V 2 V P ⌠ ⌠d = a bit of algebra leads...

Info iconThis preview shows pages 1–5. 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

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: Work V 1 V 2 V P ⌠ ⌡ d- = a bit of algebra leads to Work c P 1 P 2 P P P b + ⌠ ⌡ d ⋅ = 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 + ⋅- ⋅ = Work 0.516 J gm = Ans. 3.5 Îș a b P ⋅ + = a 3.9 10 6- ⋅ atm 1- ⋅ = b 0.1- 10 9- ⋅ atm 2- ⋅ = P 1 1 atm ⋅ = P 2 3000 atm ⋅ = V 1 ft 3 ⋅ = (assume const.) Combine Eqs. (1.3) and (3.3) for const. T: Work V P 1 P 2 P a b P ⋅ + ( ) P ⋅ ⌠ ⌡ d ⋅ = Work 16.65atm ft 3 ⋅ = Ans. Chapter 3 - Section A - Mathcad Solutions 3.1 ÎČ 1- ρ T ρ d d ⋅ = Îș 1 ρ P ρ d d ⋅ = P T At constant T, the 2nd equation can be written: d ρ ρ Îș dP ⋅ = ln ρ 2 ρ 1 Îș ∆ P ⋅ = Îș 44.18 10 6- ⋅ bar 1- ⋅ = ρ 2 1.01 ρ 1 ⋅ = ∆ P ln 1.01 ( ) Îș = ∆ P 225.2bar = P 2 226.2 bar ⋅ = Ans. 3.4 b 2700 bar ⋅ = c 0.125 cm 3 gm ⋅ = P 1 1 bar ⋅ = P 2 500 bar ⋅ = Since 16 P 2 1 bar ⋅ = T 1 600 K ⋅ = C P 7 2 R ⋅ = C V 5 2 R ⋅ = (a) Constant V: W = and ∆ U Q = C V ∆ T ⋅ = T 2 T 1 P 2 P 1 ⋅ = ∆ T T 2 T 1- = ∆ T 525- K = ∆ U C V ∆ T ⋅ = Q and ∆ U 10.91- kJ mol = Ans. ∆ H C P ∆ T ⋅ = ∆ H 15.28- kJ mol = Ans. (b) Constant T: ∆ U ∆ H = = and Q W = Work R T 1 ⋅ ln P 2 P 1 ⋅ = Q and Work 10.37- kJ mol = Ans. (c) Adiabatic: Q = and ∆ U W = C V ∆ T ⋅ = 3.6 ÎČ 1.2 10 3- ⋅ degC 1- ⋅ = C P 0.84 kJ kg degC ⋅ ⋅ = M 5 kg ⋅ = V 1 1 1590 m 3 kg ⋅ = P 1 bar ⋅ = t 1 0 degC ⋅ = t 2 20 degC ⋅ = With beta independent of T and with P=constant, dV V ÎČ dT ⋅ = V 2 V 1 exp ÎČ t 2 t 1- ( 29 ⋅ ⋅ = ∆ V V 2 V 1- = ∆ V total M ∆ V ⋅ = ∆ V total 7.638 10 5- × m 3 = Ans. Work P- ∆ V total ⋅ = (Const. P) Work 7.638- joule = Ans. Q M C P ⋅ t 2 t 1- ( 29 ⋅ = Q 84kJ = Ans. ∆ H total Q = ∆ H total 84kJ = Ans. ∆ U total Q Work + = ∆ U total 83.99kJ = Ans. 3.8 P 1 8 bar ⋅ = 17 Step 41: Adiabatic T 4 T 1 P 4 P 1 R C P ⋅ = T 4 378.831K = ∆ U 41 C V T 1 T 4- ( 29 ⋅ = ∆ U 41 4.597 10 3 × J mol = ∆ H 41 C P T 1 T 4- ( 29 ⋅ = ∆ H 41 6.436 10 3 × J mol = Q 41 J mol = Q 41 J mol = W 41 ∆ U 41 = W 41 4.597 10 3 × J mol = P 2 3bar = T 2 600K = V 2 R T 2 ⋅ P 2 = V 2 0.017 m 3 mol = Step 12: Isothermal ∆ U 12 J mol = ∆ U 12 J mol = Îł C P C V = T 2 T 1 P 2 P 1 Îł 1- Îł ⋅ = T 2 331.227K = ∆ T T 2 T 1- = ∆ U C V ∆ T ⋅ = ∆ H C P ∆ T ⋅ = W and ∆ U 5.586- kJ mol = Ans. ∆ 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 ⋅ P 1 = V 1 4.988 10 3- × m 3 mol = 18 T 4 378.831K = V 4 R T 4 ⋅ P 4 = V 4 0.016 m 3 mol = Step 34: Isobaric ∆ U 34 C V T 4 T 3- ( 29 ⋅ = ∆ U 34 439.997- J mol = ∆ H 34 C P T 4 T 3- ( 29 ⋅ = ∆ H 34 615.996- J mol = Q 34 C P T 4 T 3- ( 29 ⋅ = Q 34 615.996615....
View Full Document

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.

Page1 / 45

SM_3A - Work V 1 V 2 V P ⌠ ⌠d = a bit of algebra leads...

This preview shows document pages 1 - 5. Sign up to view the full document.

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