Question

If the vapour behaves ideally at 1000 $K$, Then average translational kinetic energy of a molecule is $x\times {10}^{-20}J/molecule$.
value of $x$ is....(as nearest integer)

Answer 1

As we know,
Average $KE=\left(3/2\right)kT$
$=\left(3/2\right)\times 1.38\times {10}^{-23}\times 1000$
$=2.07\times {10}^{-20}J/molecule$
So, value of $x$ is 2.