FNT12 - FNT#12 Freezing out of modes Damien Martin December...

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FNT #12: Freezing out of modes Damien Martin December 1, 2006 Question 1: See notes on freezing out of modes for an explanation. Question 2: This question shows a graph of molar specific heat c vm against temperature for two gasses. One material is monatomic (Bv) and the other is diatomic (Bc 2 ). There are two curves which are unlabelled. a) Explain which curve represents Bc 2 and its phase in this temperature range We know that the molar specific heat is given by c vm = (# modes/ molecule) 2 R This leads to the conclusion that the lower horizontal line has 3 modes at all temperatures, and the higher line varies from 5 modes per molecule at T =300 K ( c vm 21 J/mol K) to 7 modes per molecule at T =2000 K ( c vm 29 J/mol K). From this, I can conclude that neither are liquids or solids (they would have 6 modes/molecule). The lower line has three modes at all temperatures, and so is a monatomic gas (i.e. Bv). The upper line varies from 5 modes to 7 modes, and this is consistent with a diatomic molecule with its vibrational mode frozen out at low temperatures but activating at high temperatures.
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