$\bar{v}=\sqrt{\frac{8 R T}{\pi m}}=\sqrt{\frac{2.5 R T}{m}}$

and $v_{p}=\sqrt{\frac{2 R T}{m}}$

From these expressions, we can see that

$v_{p}<\bar{v}$

Again, $v_{r m s}=v_{p} \frac{\sqrt{3}}{2}$

and average kinetic energy of a gas molecule

$E_{k}=\frac{1}{2} m v_{rm s}^{2}$

$E_{k}=\frac{1}{2} m\left(\sqrt{\frac{3}{2}} v_{y}\right)^{2}=\frac{1}{2} m \times \frac{3}{2} v_{p}^{2}=\frac{3}{4} m v_{p}^{2}$