JEE Advance - Chemistry (1981 - No. 12)

The question deals with the calculation of the ratio of the root mean square velocity to the average velocity of gas molecules. To solve it, we need equations for both these velocities.

The root mean square velocity, $$v_{rms}$$, is given by the formula:

$$ v_{rms} = \sqrt{\frac{3RT}{M}} $$

where:

• $$R$$ is the universal gas constant,


• $$T$$ is the temperature,


• $$M$$ is the molar mass of the gas.

The average velocity, $$v_{avg}$$, can be calculated using the formula:

$$ v_{avg} = \sqrt{\frac{8RT}{\pi M}} $$

To find the ratio of the root mean square velocity to the average velocity, we take the ratio of these two expressions:

$$ \frac{v_{rms}}{v_{avg}} = \frac{\sqrt{\frac{3RT}{M}}}{\sqrt{\frac{8RT}{\pi M}}} $$

Simplifying this, we can cancel out $$RT/M$$ from the numerator and denominator since they are present in both terms:

$$ \frac{v_{rms}}{v_{avg}} = \frac{\sqrt{3}}{\sqrt{\frac{8}{\pi}}} = \sqrt{\frac{3\pi}{8}} $$

To solve $$ \sqrt{\frac{3\pi}{8}} $$ numerically, we know that $$\pi \approx 3.14159$$. Plugging in the values, we get:

$$ \sqrt{\frac{3 \times 3.14159}{8}} \approx \sqrt{1.178} \approx 1.085 $$

So, the ratio of $$v_{rms}$$ to $$v_{avg}$$ is approximately 1.085, which is nearest to Option A, 1.086:1.

Therefore, the correct answer is Option A, 1.086:1.