One mole of a monoatomic gas is mixed with three moles of a diatomic gas. The molecular specific heat of mixture of constant volume is α24RJ/molK; then the value of will be________.

(Assume that the given diatomic gas has no vibrational mode.)

2 Views|Posted 6 months ago
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1 Answer
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6 months ago

 Cv=α2R4J/molk

As, Cv (mix) = 1*32R+3*52R4=9R4=α2R4

α=3

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Physics NCERT Exemplar Solutions Class 12th Chapter Nine 2025

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