Physical Chemistry 2 Quiz
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STATISTICAL THERMODYNAMICS
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The vibrational partition function for a diatomic molecule is qvib=11−e−θv/Tq_{vib} = \frac{1}{1-e^{-\theta_v/T}} where θv=hν/k\theta_v = h\nu/k is the vibrational temperature. For CO with θv=3103 K\theta_v = 3103 \text{ K}, at what temperature does qvib=1.5q_{vib} = 1.5?
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