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Q.Find the total kinetic energy of one mole of nitrogen at 27°C.

Rajasthan RbseRajasthan Board Senior Secondary Part-I Examination 2017Subjective· 2mImportance★★★★★
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Nitrogen (diatomic, 5 degrees of freedom) has total molar kinetic energy (5/2)RT; at 300 K this comes to about 6235 J.

By the law of equipartition of energy, each degree of freedom of a gas molecule contributes (1/2)RT of energy per mole (R = 8.314 J/mol·K). A nitrogen molecule (N2) is diatomic, and at ordinary temperatures it has 5 degrees of freedom: 3 translational (motion along x, y, z) and 2 rotational (about the two axes perpendicular to the bond axis). So its total (translational + rotational) kinetic energy per mole is:

KE_total = (5/2) R T

First convert the temperature to kelvin: T = 27°C + 273 = 300 K.

KE_total = (5/2)(8.314 J/mol·K)(300 K)

= (5/2)(2494.2 J)

= 6235.5 J …

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