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Find the value of Cv and Cp for nitrogen. (Given R = 8.3 J mol-1 K-1 ; also, for a diatomic gas, Cv = \(\frac{5}{2}\) R.

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Nitrogen is known to exist as a diatomic molecule. Hence

Cv(for nitrogen) = \(\frac{5R}{2}\) 

= \(\frac{5}{2}\) × 8.3 J mol-1 K-1

= 20.75 J mol-1 K-1

From Mayer’s relation, Cp = Cv + R

Cp(for nitrogen) = (20.75 + 8.3) J mol-1 K-1

= 29.05 K mol-1 K -1

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