The speed of an oxygen molecule in a gas is $u$. If the temperature is doubled and the molecule dissociates into two atoms,the speed will be:

The temperature of an ideal gas is increased from $140 \,K$ to $560 \,K$. If the r.m.s. speed of gas molecules is $v$ at $140 \,K$, then at $560 \,K$, the r.m.s. speed becomes:

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 speed distribution for a sample of $N$ gas particles is shown below. $P(v) = 0$ for $v > 2 v_0$. How many particles have speeds between $1.2 v_0$ and $1.8 v_0$ (in $N$)?

Column-$I$ represents the formula for ${v_{rms}}$ and Column-$II$ represents the corresponding condition (phenomena). Match them correctly:

Column-$I$	Column-$II$
$(a)$ ${v_{rms}} = \sqrt {\frac{3P}{\rho}}$	$(i)$ For $1 \text{ mole ideal gas}$
$(b)$ ${v_{rms}} = \sqrt {\frac{3RT}{M_0}}$	$(ii)$ For one molecule of gas
$(c)$ ${v_{rms}} = \sqrt {\frac{3{k_B}T}{m}}$	$(iii)$ On the basis of kinetic theory

The molecular mass of a gas having $r.m.s.$ speed four times as that of another gas having molecular mass $32$ is