Question

A flask contains argon and chlorine in the ratio of 2 : 1 by mass. The temperature of the mixture is 27°C. The ratio of root mean square speed of the molecules of the two gases ($v_{rms}(Ar) / v_{rms}(Cl_2)$) is: (Atomic mass of argon = 40.0 u and molecular mass of chlorine = 70.0 u) ____.

• A. 7/4
• B. $\sqrt{7}/2$
• C. $2/\sqrt{7}$
• D. 7/2

Hint:

In kinetic theory problems, always look for what stays constant. Since $T$ is constant, the lighter molecule will always have a higher $v_{rms}$. Since Argon (40) is lighter than Chlorine (70), its speed must be greater.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The root mean square speed ($v_{rms}$) of a gas molecule depends on the absolute temperature of the gas and its molecular mass.

Step 2: Key Formula or Approach:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \] where $M$ is the molar mass of the gas.

Step 3: Detailed Explanation:
1. Both gases are in the same flask at the same temperature ($T = 27^\circ\text{C} = 300\text{ K}$). 2. Therefore, $v_{rms} \propto \frac{1}{\sqrt{M}}$. 3. The ratio of their speeds is: \[ \frac{v_{rms}(Ar)}{v_{rms}(Cl_2)} = \sqrt{\frac{M(Cl_2)}{M(Ar)}} \] 4. Substitute the given molecular masses ($M(Cl_2) = 70$ and $M(Ar) = 40$): \[ \text{Ratio} = \sqrt{\frac{70}{40}} = \sqrt{\frac{7}{4}} = \frac{\sqrt{7}}{2} \] (Note: The mass ratio 2:1 is irrelevant as $v_{rms}$ depends on molecular mass, not total mass of the sample.)

Step 4: Final Answer:
The ratio of $v_{rms}$ is $\sqrt{7}/2$.