# GRE GR 1777, Problem 009 Solution

Physics GRE GR 1777

9. Classical Mechanics (Kinematics)

Solution)

$\frac{1}{2}Mv^2 = \frac{3}{2}kT$,
where $v$ is the root-mean-square speed of molecules in an ideal gas, $M$ is the mass of the molecule, and $T$ is the temperature of the system.

Since $k = \Big( \frac{n}{N} \Big) R$,
then $\frac{1}{2}Mv^2 = \frac{3}{2} \frac{n}{N} RT$.

Therefore, $v = \sqrt{\frac{3RT}{M}}$.