test1s - Chemistry 2740 Spring 2011 Test 1 Solutions 1....

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Chemistry 2740 Spring 2011 Test 1 Solutions 1. Intensive: temperature, pressure Extensive: mass, number of moles, volume, internal energy, enthalpy, entropy 2. (a) mass of sample and temperature change (b) to obtain the calorimeter’s heat capacity (c) the specific (or molar) internal energy change 3. Some possible examples: Heat flowing spontaneously from a cold to a hot body Gases unmixing Water cooling to 0 C and forming ice in a room at 20 C (or any temperature above 0 C) by spontaneously transferring heat from a glass of water to the room Conversion of heat into work with perfect efficiency 4. (a) We first need to convert the molar entropy change to a per residue basis: Δ S = 52 J K - 1 mol - 1 6 . 022 142 0 × 10 23 mol - 1 = 8 . 6 × 10 - 23 J / K = k B ln Ω f - k B ln Ω i = k B ln Ω f - k B ln 1 = k B ln Ω f ln Ω f = Δ S k B = 8 . 6 × 10 - 23 J / K 1 . 380 650 3 × 10 - 23 J / K = 6 . 3 Ω = e 6 . 3 = 520 conformations per residue (b) k B ln Ω
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This note was uploaded on 03/03/2012 for the course CHEM 2740 taught by Professor Roussel during the Fall '07 term at Lethbridge College.

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test1s - Chemistry 2740 Spring 2011 Test 1 Solutions 1....

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