Question

The total kinetic energy of 1 mole of N2 at 27 Celsius will be approximately

Answer 1

according to law of equipartition of energy, a molecule can have $\frac{1}{2}nRT$ energy per degree of freedom.

here given molecule N2 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 = 2 Cal/mol.K

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

now, total kinetic energy = 5/2 × 1 × 2 × 300

= 5 × 300 = 1500 Cal

hence, total kinetic energy of 1 mole of N2 gas at 27°C will be 1500 Cal.

Answer 2

Explanation:

Ans is 1500 cal

For explanation see image