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Calculate kinetic energy of 5 moles of N...

Calculate kinetic energy of 5 moles of Nitrogen at `27^@C`.

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A. Kinetic energy `=3/2 nRT`
where, n = 5 moles, `R = 8.314 "mol"^(-1) k^(-1)`
`T = 27^(@) C + 273 = 300 K`
Kinetic energy
`K_(k) = 3/2 xx 5 "mol" xx 8.314 "J mol"^(-1) K^(-1) xx 300 K = 18706.50 J`
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