Question

(4) 1500 erg.
The total kinetic energy of 1 mole of N, at 27° C will be approximately :-
(1) 1500 J
(2) 1500 Calories
(3) 1500 kilo Calories​

Answer 1

Answer:

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

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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 =25nRT 

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=25×1×2×300

=5×300=1500Cal

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