Question

The total kinetic energy of 1 mole of $N_2$ at ${27}^{0}C$ will be approximately (R = 2 cal / mol/${}^{0}C$):-

• A. 1500 J
• B. 1500 calories
• C. 1500 kilo calories
• D. 1500 erg

Answer 1

The correct option is A 1500 J

According to law of equipartition of energy, a molecule can have energy per degree of freedom.

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here given molecule $N_2$ is diatomic molecule. and we know, there are five degree of freedom of a diatomic molecule ( three translational and two rotational ).

so, total kinetic energy $=\frac{5}{2}nRT$

here, n is number of mole. given, $n=1$

R is universal gas constant i.e., $R=2Cal/mol.K$

and T is temperature in Kelvin. given, T $=27°C=(27+273)K=300K$

now, total kinetic energy$=\frac{5}{2}\times 1\times 2\times 300$

$=5\times 300=1500Cal$

hence, total kinetic energy of $1$ mole of $N_2$ gas at $27°C$ will be $1500Cal.$