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

$\left(\right.$ Take $\left.\pi=\frac{22}{7}\right)$

<p>The relationship between the root-mean-square (rms) speed ($v_{rms}$) and the average speed ($v_{avg}$) of molecules in a gas can be found using the Maxwell-Boltzmann distribution. The rms speed and average speed are related as follows:</p> $v_{rms} = \sqrt{\frac{3RT}{M}}$ $v_{avg} = \sqrt{\frac{8RT}{\pi M}}$ <p>Where:</p> <ul> <li>$R$ is the ideal gas constant</li> <li>$T$ is the temperature in Kelvin</li> <li>$M$ is the molar mass of the gas</li> <li>$\pi$ is the mathematical constant pi</li> </ul> <p>In this problem, the rms speed of the oxygen molecule is given by:</p> $v_{rms} = \left(1+\frac{5}{x}\right)^{\frac{1}{2}} v_{avg}$ <p>Now, let's divide the expression for $v_{rms}$ by the expression for $v_{avg}$:</p> $$\frac{v_{rms}}{v_{avg}} = \frac{\sqrt{\frac{3RT}{M}}}{\sqrt{\frac{8RT}{\pi M}}} = \left(1+\frac{5}{x}\right)^{\frac{1}{2}}$$ <p>By simplifying the expression, we get:</p> $$\frac{v_{rms}}{v_{avg}} = \frac{\sqrt{3}}{\sqrt{\frac{8}{\pi}}} = \left(1+\frac{5}{x}\right)^{\frac{1}{2}}$$ <p>Square both sides of the equation:</p> $\frac{3}{\frac{8}{\pi}} = 1 + \frac{5}{x}$ <p>Now we will substitute the provided value of $\pi = \frac{22}{7}$:</p> $\frac{3}{\frac{8}{\frac{22}{7}}} = 1 + \frac{5}{x}$ <p>By simplifying the expression, we get:</p> $\frac{3 \cdot \frac{22}{7}}{8} = 1 + \frac{5}{x}$ <p>Now let's solve for $x$:</p> $\frac{66}{56} - 1 = \frac{5}{x}$ $\frac{10}{56} = \frac{5}{x}$ <p>Multiplying both sides by $x$:</p> $\frac{10}{56}x = 5$ <p>Finally, solving for $x$:</p> $x = \frac{5 \cdot 56}{10} = 28$ <p>So, the value of $x$ is $\boxed{28}$.