The root mean square speed of the molecules of a diatomic gas is $v$. When the temperature is doubled,the molecules dissociate into two atoms. The new root mean square speed of the atom is

The root mean square velocity of the molecules in a sample of helium is $(5/7)^{th}$ that of the molecules in a sample of hydrogen. If the temperature of the hydrogen sample is $0^{\circ}C$,then the temperature of the helium sample is about:

To increase the rms speed of gas molecules by $25 \%$,the percentage increase in absolute temperature of the gas is to be

Temperature of an ideal gas, initially at $27^{\circ} C$, is raised by $6^{\circ} C$. The rms velocity of the gas molecules will

$A$ sample of gas is at $0^{\circ}C$. To what temperature must it be raised in order to double the $r.m.s.$ speed of the molecules? (in $^{\circ}C$)

$A$ gas $A$ has $N$ molecules at temperature $T$ with mass $m$. Another gas $B$ has $2N$ molecules at the same temperature $T$ with mass $2m$. Both are contained in a vessel. If the $rms$ velocity of molecules of gas $B$ is $v$,and the $x$-component of the mean square velocity of molecules of gas $A$ is $w^2$,find the ratio $w^2/v^2$.