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Chem288_2005_PS5_solutions

# Chem288_2005_PS5_solutions - Chemistry 288 Problem Set 5...

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Chemistry 288 Problem Set 5 Solutions Spring 2005, Marohn A USEFUL THERMODYNAMIC RELATION We start with Z = summationdisplay ASN e ( - ε ) (1) where ASN stands for “all states and numbers of parti- cles,” as discussed in class. Taking partial derivatives, and applying the chain rule of differentiation very care- fully, we find ∂τ log( Z ) = 1 Z summationdisplay ASN parenleftbigg - 1 τ 2 parenrightbigg ( - ε ) e ( - ε ) (2) and ∂μ log( Z ) = 1 Z summationdisplay ASN parenleftbigg N τ parenrightbigg e ( - ε ) (3) Adding these together as follows, the terms with the chemical potential cancel, and we get τμ ∂μ log Z + τ 2 ∂τ log Z = 1 Z summationdisplay ASN ε e ( - ε ) = U (4) The last equality follows be definition of the average en- ergy. DEATH BY CARBON MONOXIDE POISONING Part a Use the ideal gas relation (derived in class) for the activity λ e μ/τ λ = n n Q (5) where n Q = parenleftbigg 2 π ¯ h 2 parenrightbigg 3 / 2 (6) is the “quantum concentration” for the gas, m is the mass of one gas molecule, ¯ h is Planck’s constant, τ is the temperature in fundamental temperature units, and n in the gas per-molecule concentration. Calculate the O 2 and CO quantum concentrations. Work in SI units: ¯ h = 1 . 05 × 10 - 34 J/s τ = 310 K × 1 . 38 × 10 - 23 J/K = 4 . 28 × 10 - 21 J m a m (O 2 ) = 32 × 1 . 67 × 10 - 27 kgr bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright m H = 5 . 34 × 10 - 26 kgr m b m (CO) = 28 × 1 . 67 × 10 - 27 kgr bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright m H = 4 . 70 × 10 - 26 kgr Compute: n a Q n Q (O 2 ) = 1 . 89 × 10 32 particle/m 3 = 3 . 14 × 10 5 mol/ n b Q n Q (CO) = 1 . 55 × 10 32 particle/m 3 = 2 . 57 × 10 5 mol/ Let λ a = 1 . 0 × 10 - 5 be the O 2 activity, n a the O 2 concentration, and let λ b = 1 . 0 × 10 - 7 be the CO activity, n b the CO concentration. Compute the concentrations using eq 5: n a = λ a n a Q = 10 - 5 × 3 . 14 × 10 5 mol/ = 10 - 5 × 1 . 89 × 10 32 particle/m 3 = 1 . 89 × 10 27 particle/m 3 = 3 . 14 mol/ n b = λ b n b Q = 10 - 7 × 2 . 57 × 10 5 mol/ = 10 - 7 × 1 . 55 × 10 32 particle/m 3 = 1 . 55 × 10 25 particle/m 3 = 25 . 7 mmol/ Compute partial pressures. Convert concentration to pressure using the ideal gas law: p = (as derived in class). Compute (use that J = N m): p a = n a τ = 1 . 89 × 10 27 m - 3 × 4 . 28 × 10 - 21 N m = 8 . 09 × 10 6 N / m 2 = 80 . 1 atm = 6 . 08 × 10 4 Torr p b = n a τ = 1 . 55 × 10 25 m - 3 × 4 . 28 × 10 - 21 N m = 6 . 63 × 10 4 N / m 2 = 0 . 656 atm = 499 Torr

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Chem288_2005_PS5_solutions - Chemistry 288 Problem Set 5...

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