Question

At 300 K, the value of \(u_{{rms}}^2\) of an ideal gas is \(x \, {m}^2/{s}^2\). What is the \(u_{{av}}^2\) (in \({m}^2/{s}^2\)) of this gas at the same temperature? 
 

• A. \(\frac{8x}{3\pi}\)
• B. \(\frac{3\pi}{8x}\)
• C. \(\frac{2x}{3\pi}\)
• D. \(\frac{3\pi}{2x}\)

Hint:

When converting from \(u_{{rms}}\) to \(u_{{av}}\) or vice versa, remember that these velocity measures are related through the distribution of molecular speeds in a gas, with the root mean square being greater than the average speed.

Answer 1

The Correct Option is A

The relationship between the root mean square velocity \(u_{{rms}}\) and the average velocity \(u_{{av}}\) of an ideal gas is given by: \[ u_{{av}} = \frac{u_{{rms}}}{\sqrt{\frac{8}{3\pi}}} \] Given \(u_{{rms}}^2 = x\), we find \(u_{{av}}^2\) as follows: \[ u_{{av}}^2 = \left(\frac{u_{{rms}}}{\sqrt{\frac{8}{3\pi}}}\right)^2 = \left(\frac{\sqrt{x}}{\sqrt{\frac{8}{3\pi}}}\right)^2 = \frac{x}{\frac{8}{3\pi}} = \frac{3\pi x}{8} \]