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A problem from thermodynamics -- Freezing of water at 273 K and 1 atm
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[QUOTE="Andrew Mason, post: 5457397, member: 15795"] ##\Delta U## includes potential energy. In order to form ice, the water molecules lose potential energy and rotational kinetic energy. Average[I] translational [/I]kinetic energy does not change (i.e. temperature) but average potential energy per molecule decreases[I] and there is loss of rotational kinetic energy[/I]. The process is exothermic (heat flows out of the water to form ice). Since ##\Delta Q < 0##, ##\Delta S = \int dQ/T = \Delta Q/T < 0##. Since P and T are constant, ##\Delta G = \Delta Q - T\Delta S = T\Delta S - T\Delta S = 0## I am not sure about w = 0, however. Ice takes up more volume than liquid water (which is why ice floats). So there is a small amount of work done on the surroundings. [I][later edits are in italics][/I] AM [/QUOTE]
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A problem from thermodynamics -- Freezing of water at 273 K and 1 atm
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