Question

The total kinetic energy of $1$ mole of $N_2$ at $27$C will be approximately

• A. 3739.662 J
• B. 1500 calorie
• C. 1500 kilo calorie
• D. 1500 erg.

Answer 1

The correct option is A 3739.662 J

The kinetic enrgy of one mole is given by:

KE=$\frac{3}{2}{K}_{B}T$

The kinetic enrgy of 1 mole of $N_2$ atoms is:

KE=$\frac{3}{2}{K}_{B}T$ where $N$ is Avogadro's number,$K_B$ is Boltzmann's constant and $T$ is temperature

KE=$\frac{3}{2}\times (6.022\times {10}^{23})\times (1.38\times {10}^{-23})\times 300$

$=3739.662J$