Question

Equal volumes of two gases are kept in different containers having densities in the ratio $1 : 16$. They exert equal pressures on the wall of their respective containers. Then the ratio of their r.m.s. velocities is

• A. $16 : 1$
• B. $1 : 8$
• C. $4 : 1$
• D. $1 : 12$

Hint:

Since velocity is inversely proportional to the square root of density, the less dense gas must move faster to generate the exact same pressure. Because its density is 16 times smaller, its velocity must be $\sqrt{16} = 4$ times larger, instantly pointing you to a ratio starting with the larger number.

Answer 1

The Correct Option is C

Step 1: Understand the question.
Two gases have densities in the ratio $1:16$ and exert equal pressures. We find the ratio of their root mean square (rms) speeds.
Step 2: Recall the pressure-density relation.
From kinetic theory, the pressure of a gas is \[ P = \frac{1}{3}\rho\,v_{rms}^2, \] where $\rho$ is density and $v_{rms}$ is the rms speed.
Step 3: Solve for the rms speed.
Rearranging, \[ v_{rms} = \sqrt{\frac{3P}{\rho}}. \] So at equal pressure, $v_{rms}$ is inversely proportional to the square root of density.
Step 4: Write the ratio.
\[ \frac{v_1}{v_2} = \sqrt{\frac{\rho_2}{\rho_1}}. \]
Step 5: Put in the density ratio.
Here $\frac{\rho_1}{\rho_2} = \frac{1}{16}$, so $\frac{\rho_2}{\rho_1} = 16$. Then \[ \frac{v_1}{v_2} = \sqrt{16} = 4. \]
Step 6: State the answer.
The ratio of rms speeds is $4:1$. \[ \boxed{v_1 : v_2 = 4 : 1} \]