GRE GR 1777, Problem 009 Solution

Physics GRE GR 1777
9. Classical Mechanics (Kinematics)
\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}}.
The answer is (D).

Leave a Reply