University:
Amherst College
Course:
CHEM 161 | Chemical Principles
Academic year:

2009
Views:

230

Pages:

5
Author:

Brennen Smith

The Thermodynamics of Ideal Gases 28. 88.0 g N₂O × (1 mol N₂O / 44.02 g N₂O) = 2.00 mol N₂O At constant pressure, qp = ΔH ΔH = nCpΔT = (2.00 mol)(38.70 J mol⁻¹C⁻¹)(55°C - 165°C) ΔH = -8510 J = -8.51 kJ = qp w = -PΔV = -nRΔT = -(2.00 mol)(8.3145 J mol⁻¹K⁻¹)(-110.K) = 1830 J = 1.83 kJ ΔE = q + w = -8.51 kJ + 1.83 kJ = -6.68 kJ 29.Pathway I: Step 1: (5.00 mol, 3.00 atm, 15.0 L) → (5.00 mol, 3.00 atm, 55.0 L) w = -PΔV = -(3.00 atm)(55.0 - 15.0 L) = -120.L atm w = -120.L atm × (101.3 J / L.atm) × (1 kJ / 1000 J) = -12.2 kJ ΔH = qp = nCpΔT = nCp(Δ(PV) / nR) = (CpΔ(PV)) / R; For an ideal monatomic gas: Cp = (5/2)R ΔH = q = 300. L atm × (101.3 J / L.atm) × (1 kJ / 1000 J) = 30.4 kJ ΔE = q + w = 30.4 kJ - 12.2 kJ = 18.2 kJ Step 2: (5.00 mol, 3.00 atm, 55.0L) → (5.00 mol, 6.00 atm, 20.0 L) ΔE = nCvΔT = n((3/2)R)(Δ(PV) / nR) = (3/2)ΔPV ΔE = (3/2)(120. - 165)L atm = -67.5 L atm (Carry extra significant figure.) ΔE = -67.5 L atm × (101.3 J / L.atm) × (1 kJ / 1000 J) = -6.8 kJ Pathway II: Step 3: (5.00 mol, 3.00 atm, 15.0 L) → (5.00 mol, 6.00 atm, 15.0 L) ΔE = q = (3/2)Δ(PV) = (3/2)(90.0 - 45.0) L.atm = 67.5 L.atm ΔE = q = 67.5 L.atm × (101.3 J / L.atm) × (1 kJ / 1000 J) = 6.84 kJ w = -PΔV = 0 since ΔV = 0 ΔH = ΔE + Δ(PV) = 67.5 L.atm + 45.0 L.atm = 112.5 L.atm = 11.40 kJ Step 4: (5.00 mol, 6.00 atm, 15.0 L) → (5.00 mol, 6.00 atm, 20.0 L) ΔH = q = (5/2)Δ(PV) = (5/2)(120.0 - 90.0) L.atm = 75 L.atm ΔH = q = 75 L.atm × (101.3 J / L.atm) × (1 kJ / 1000 J) = 7.6 kJ Summary: Pathway Step 1 Step 2 Total ΔH 30.4 kJ -11 kJ 19 kJ w -12.2 kJ 21.3 kJ 9.1 kJ ΔE 18.2 kJ -6.8 kJ 11.4 kJ 30. (continued) Pathway II: w = -PΔV = -(6.00 atm)(20.0 - 15.0)L = -30.L.atm w = -30.L atm × (101.3 J / L.atm) × (1 kJ / 1000 J) = -3.0 kJ ΔE = q + w = 7.6 kJ - 3.0 kJ = 4.6 kJ Summary: Pathway Step 1 Step 2 Step 3 Total q 6.84 kJ 7.6 kJ 101.3 kJ 115.7 kJ w 0 -3.0 kJ -40.5 kJ -35.4 kJ ΔE 6.84 kJ 4.6 kJ 60.8 kJ 60.8 kJ ΔH 11.40 kJ 12.7 kJ 101.3 kJ 101.3 kJ

Chemistry 12 Assignment 3 - Solutions Fall , 2009

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