Question

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

• A. $1500J$
• B. $15633cal$
• C. $1500kcal$
• D. $1500erg$

Answer 1

The correct option is B $15633cal$

For $n$ mole of any gas the total kinetic energy is given as $E=\frac{3}{2}nRT$

Where $R$ is gas constant having value $8.31J/mole-K$ or $8.31\times 4.18cal/mole-K=34.74\text{Cal per mole per Kelvin}$

$T$ is temperature in Kelvin which is $T=27+273=300K$

So putting all values we get $E=1.5\times 1\times 34.74\times 300=15633Calorie$