$N$ molecules each of mass $m$ of gas $A$ and $2N$ molecules each of mass $2m$ of gas $B$ are contained in the same vessel at temperature $T$. The mean square of the velocity of molecules of gas $B$ is $v^2$ and the mean square of $x$ component of the velocity of molecules of gas $A$ is $w^2$. The ratio $\frac{w^2}{v^2}$ is

$A$ thermally isolated cylindrical closed vessel of height $8 \ m$ is kept vertically. It is divided into two equal parts by a diathermic (perfect thermal conductor) frictionless partition of mass $8.3 \ kg$. Thus,the partition is held initially at a distance of $4 \ m$ from the top. Each of the two parts of the vessel contains $0.1 \ mol$ of an ideal gas at temperature $300 \ K$. The partition is now released and moves without any gas leaking from one part of the vessel to the other. When equilibrium is reached,the distance of the partition from the top (in $m$) will be. . . . . . (take the acceleration due to gravity $g = 10 \ m/s^2$ and the universal gas constant $R = 8.3 \ J \ mol^{-1} \ K^{-1}$).

An insulated container contains $4$ moles of an ideal diatomic gas at temperature $T$. Heat $Q$ is supplied to the gas,causing $2$ moles of the gas to dissociate into atoms. If the temperature of the gas remains constant,then:

$2.8 \ g$ of nitrogen gas $(N_2)$ is in a vessel at a temperature of $127^{\circ} C$. The amount of heat energy required to increase the rms speed of the nitrogen molecules by $41.4 \%$ is $(R = 8.31 \ J \ mol^{-1} \ K^{-1})$. (in $J$)

$CO_2$ $(O-C-O)$ is a triatomic gas. The mean kinetic energy of one gram of the gas will be (where $N$ is Avogadro's number,$k$ is Boltzmann's constant,and the molecular weight of $CO_2 = 44$).

The mass of a hydrogen atom is $3.32 \times 10^{-27} \, kg$. If $10^{23}$ atoms per second strike a wall of area $2 \, cm^2$ at an angle of $45^{\circ}$ with the normal to the wall and reflect elastically with a speed of $10^3 \, m/s$,the pressure exerted on the wall will be: $(N/m^2)$