Chapter3_A - Chapter 3 - Section A - Mathcad Solutions 3.1...

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a bit of algebra leads to Work c P 1 P 2 P P Pb + d := Work 0.516 J gm = Ans. Alternatively, formal integration leads to Work c P 2 P 1 bln P 2 b + P 1 b + := Work 0.516 J gm = Ans. 3.5 κ ab P + = a 3.9 10 6 atm 1 := b 0.1 10 9 atm 2 := P 1 1 atm := P 2 3000 atm := V1 f t 3 := (assume const.) Combine Eqs. (1.3) and (3.3) for const. T: Work V P 1 P 2 P P + () P d := Work 16.65atm ft 3 = Ans. Chapter 3 - Section A - Mathcad Solutions 3.1 β 1 ρ T ρ d d = κ 1 ρ P ρ d d = P T At constant T, the 2nd equation can be written: d ρ ρ κ dP = ln ρ 2 ρ 1 κ∆ P = κ 44.18 10 6 bar 1 := ρ 2 1.01 ρ 1 = P ln 1.01 κ := 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 Work V 1 V 2 V P d = 21
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P 2 1 bar := T 1 600 K := C P 7 2 R := C V 5 2 R := (a) Constant V: W0 = and UQ = C V T = T 2 T 1 P 2 P 1 := TT 2 T 1 := T 525 K = UC V T := Q and U 10.91 kJ mol = Ans. HC P T := H 15.28 kJ mol = Ans. (b) Constant T: U H = 0 = and QW = Work R T 1 ln P 2 P 1 := Q and Work 10.37 kJ mol = Ans. (c) Adiabatic: Q0 = and UW = C V T = 3.6 β 1.2 10 3 degC 1 := C P 0.84 kJ kg degC := M5 k g := 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 β dT = V 2 V 1 exp β t 2 t 1 () := VV 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. QM C P t 2 t 1 := Q8 4 k J = Ans. H total Q := H total 84kJ = Ans. U total Q Work + := U total 83.99kJ = Ans. 3.8 P 1 8 bar := 22
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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 () := U 41 4.597 10 3 × J mol = H 41 C P T 1 T 4 := H 41 6.436 10 3 × J mol = Q 41 0 J mol := Q 41 0 J mol = W 41 U 41 := W 41 4.597 10 3 × J mol = P 2 3bar := T 2 600K := V 2 RT 2 P 2 := V 2 0.017 m 3 mol = Step 12: Isothermal U 12 0 J mol := U 12 0 J mol = H 12 0 J mol := H 12 0 J mol = γ C P C V := T 2 T 1 P 2 P 1 γ 1 γ := T 2 331.227K = TT 2 T 1 := UC V T := HC 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 1 P 1 := V 1 4.988 10 3 × m 3 mol = 23
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Step 34: Isobaric U 34 C V T 4 T 3 () := U 34 439.997 J mol = H 34 C P T 4 T 3 := H 34 615.996 J mol = Q 34 C P T 4 T 3 := Q 34 615.996 J mol = W 34 R T 4 T 3 := W 34 175.999 J mol = 3.10 For all parts of this problem: T 2 T 1 = and U H = 0 = Also Q Work = and all that remains is to calculate Work. Symbol V is used for total volume in this problem.
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Chapter3_A - Chapter 3 - Section A - Mathcad Solutions 3.1...

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