Question

The total energy of 1 mole of an ideal monatomic gas at $27{}^{\circ }C$ is ............ $\text{cal}$.

Answer 1

As we know,
$KE$ per mole $=\frac{3}{2}RT=\frac{3}{2}\times 2\times 300=900cal$

Where R= 1.99 cal K${}^{-1}$ mol${}^{-1}$

The total energy of 1 mole of an ideal monatomic gas at 27 ${}^{\circ }$C is900 cal.