Answer to Question #180094 in Physics for Jason

Question

A mixture of two ideal gases, denoted individually by A and B, is in thermal equilibrium at a temperature of 600 K. Molecule A is characterized by four times less mass than the mass of molecule B. Also, the mean square velocity of gas molecule A is 400 m / s. Determine the mean square velocity of gas molecules B. (Answer: V_B = 200m / s)

Answer 1

Since the mixture of two ideal gases is at thermal equilibrium, mean kinetic energy of molecule of each gas is $\frac{i}{2} k T$, hence $\frac{m_A v_A^2}{2} = \frac{i}{2} k T$ and $\frac{m_B v_B^2}{2} = \frac{i}{2} k T$, from where $m_A v_A^2 = m_B v_B^2$, and $v_B = \sqrt{\frac{m_A}{m_B}} v_A = \frac{1}{2} v_A = 200\frac{m}{s}$.