Question

The total kinetic energy of 1 mole of $N_2$ at ${27}^{o}C$ will be approximately:

• A. $1500J$
• B. $1500$ calorie
• C. $1500$ kilo calorie
• D. $1500$ erg

Answer 1

The correct option is B $1500$ calorie

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

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 = $\frac{5nRT}{2}$

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 = $5/2\times 1\times 2\times 300$

=$5\times 300=1500Cal$