Question

Equal volumes of two gases, having their densities in the ratio of $1 : 16$ exert equal pressures on the walls of two containers. The ratio of their rms speeds ($C_1 : C_2$) is

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

Hint:

Whenever pressure is constant, remember the handy shortcut $C \propto \frac{1}{\sqrt{\rho}}$. Because Gas 2 is $16$ times denser than Gas 1, its molecules are heavier and move slower. Taking the square root of $16$ immediately reveals that Gas 1 must move $4$ times faster than Gas 2.

Answer 1

The Correct Option is B

Step 1: Picture the two gases.
Two gases sit in equal volumes and push with equal pressure on their container walls. Their densities are in the ratio $\rho_1 : \rho_2 = 1 : 16$. We want the ratio of their rms speeds.
Step 2: Recall the pressure formula from kinetic theory.
Kinetic theory gives the pressure of an ideal gas as $P = \dfrac{1}{3}\rho C^2$, where $C$ is the rms speed and $\rho$ is the density.
Step 3: Solve for the rms speed.
Rearranging, $C = \sqrt{\dfrac{3P}{\rho}}$. The factor of $3$ and the pressure $P$ are common to both gases here.
Step 4: Use the equal-pressure condition.
Since $P_1 = P_2$, the only thing that changes $C$ is the density. So $C \propto \dfrac{1}{\sqrt{\rho}}$ - lighter (less dense) gas molecules move faster.
Step 5: Build the ratio.
Therefore $\dfrac{C_1}{C_2} = \sqrt{\dfrac{\rho_2}{\rho_1}}$. The denser gas ends up with the smaller speed.
Step 6: Put in the numbers.
$\dfrac{C_1}{C_2} = \sqrt{\dfrac{16}{1}} = \dfrac{4}{1}$. So the first (lighter) gas moves four times as fast.
\[ \boxed{C_1 : C_2 = 4 : 1\ \text{(option 2)}} \]