The rms speed of an oxygen molecule in a vessel at a 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: (Take $\pi=\frac{22}{7}$)

Given below are two statements:
Statement $I$: The average momentum of a molecule in a sample of an ideal gas depends on temperature.
Statement $II$: The rms speed of oxygen molecules in a gas is $v$. If the temperature is doubled and the oxygen molecules dissociate into oxygen atoms,the rms speed will become $2v$.
In the light of the above statements,choose the correct answer from the options given below.

If the rms speed of the molecules of a diatomic gas at a temperature of $322 \ K$ is $2000 \ m \ s^{-1}$,then the gas is

When the temperature of an ideal gas is increased from $27^{\circ} C$ to $127^{\circ} C$,calculate the percentage increase in its $v_{\text{rms}}$. (in $\%$)

The rms speed of a gas molecule is $V$ at pressure $P$. If the pressure is increased by two times,then the rms speed of the gas molecule at the same temperature will be

If the $rms$ speeds of helium and oxygen are equal,then the ratio of the temperatures of helium and oxygen is