Question

The total $KE$ of an ideal monoatomic gas at ${27}^{\circ }C$ is :

• A. 900 $cal$
• B. 1800 $cal$
• C. 300 $cal$
• D. None

Answer 1

The correct option is A 900 $cal$

Total kinetic energy at temperature $T$ is

$KE=\frac{3}{2}RT$

Here, $R=2cal/mol/K$ and $T={27}^{o}C+273=300K$

$KE=\frac{3}{2}\times 2\times 300=900cal$

So, the correct option is $A$