Question

The rms speed of oxygen molecule in a vessel at particular temperature is \((1+\frac{5}{x})^{\frac{1}{2}} v\), where \(v\) is the average speed of the molecule. The value of x will be : (Take \(π = \frac{22}{7}\))

• A. 4
• B. 8
• C. 27
• D. 28

Hint:

Remember that the relationship between rms speed and average speed for molecules can involve constants such as \( \pi \), and care should be taken when substituting numerical values.

Answer 1

The Correct Option is D

The relationship between rms speed \( v_\text{rms} \), average speed \( v \), and \( x \) is given as: \[ v_\text{rms} = \left( 1 + \frac{5}{x} \right)^{\frac{1}{2}} v. \]

From the kinetic theory of gases, the rms speed \( v_\text{rms} \) and average speed \( v \) are related as: \[ v_\text{rms} = \sqrt{\frac{3k_BT}{m}}, \quad v = \sqrt{\frac{8k_BT}{\pi m}}. \]

Taking the ratio: \[ \frac{v_\text{rms}}{v} = \sqrt{\frac{3}{8/\pi}} = \sqrt{\frac{3\pi}{8}}. \]

Substitute \( \pi = \frac{22}{7} \): \[ \frac{v_\text{rms}}{v} = \sqrt{\frac{3 \times \frac{22}{7}}{8}} = \sqrt{\frac{66}{56}} = \sqrt{\frac{33}{28}}. \]

Equating this to the given relation: \[ \sqrt{\frac{33}{28}} = \left( 1 + \frac{5}{x} \right)^{\frac{1}{2}}. \]

Square both sides: \[ \frac{33}{28} = 1 + \frac{5}{x}. \]

Simplify: \[ \frac{33}{28} - 1 = \frac{5}{x}. \]

\[ \frac{5}{28} = \frac{5}{x}. \]

Solve for \( x \): \[ x = 28. \]