Lec5 - Problem from Last Time Calculate the change in...

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1 1 Calculate the change in entropy of the freezer (surroundings), S surroundings , when 0.444 mol of water freezes at 0 o C (273 K). The temperature of the freezer is –15 o C (258 K) For ice, H fusion = 6.01 kJ/mol S = q/T A) + 10.3 J/K B) + 23.3 J/K C) + 9.77 J/K D) + 22.0 J/K Problem from Last Time 2 Freezer (the surroundings) S freezer = q freezer /T freezer The freezer absorbs energy at its temperature. (258 K) so q freezer > 0. q freezer = q water S freezer = q freezer /T freezer = (+0.444 mol x 6010 J/mol)/258 K = +10.3 J/K
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2 3 The Universe (total) S universe = S water + S freezer S universe = 9.8 J/K + 10.3 J/K = + 0.5 J/K So the total entropy change is positive, the total entropy increases, and water freezes spontaneously at 0 0 C because energy is transferred to surroundings at a lower temperature. T q S 4 Gibbs Free Energy S universe = S surroundings + S system J Willard Gibbs, a Yale professor regarded as the father of Thermodynamics, wanted to express these ideas in terms of the system only! So he rearranged equations and invented a new concept that is named after him. 1839 - 1903
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3 5 S universe = S surroundings + S system S surroundings = q surroundings /T = H system / T S universe = H system / T + S system Multiply through by T to obtain  S universe = H system T S system How did he do this??? 6  S universe = H system -T S system Right-hand side only depends on the system. So Gibbs define a new function
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Lec5 - Problem from Last Time Calculate the change in...

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