Initially,a gas of diatomic molecules is contained in a cylinder of volume $V_{1}$ at a pressure $P_{1}$ and temperature $250\, K$. Assuming that $25\%$ of the molecules get dissociated,causing a change in the number of moles. The pressure of the resulting gas at temperature $2000\, K$,when contained in a volume $2V_{1}$,is given by $P_{2}$. The ratio $\frac{P_{2}}{P_{1}}$ is:

$\frac{1}{2}$ mole of helium gas is contained in a container at $S.T.P.$ The heat energy needed to double the pressure of the gas,keeping the volume constant (specific heat of the gas $= 3 \; J \; g^{-1} \; K^{-1}$) is ...... $J$.

An ideal gas is placed in a tank at $27^{\circ} C$. The pressure is initially $600 \ kPa$. One fourth of the gas is then released from the tank and thermal equilibrium is established. What will be the pressure if the temperature is $327^{\circ} C$ (in $kPa$)?

Two non-reactive monoatomic ideal gases have their atomic masses in the ratio $3:4$. The ratio of their partial pressures when enclosed in a vessel kept at a constant temperature is $2:3$. The ratio of their densities is

The plot that depicts the behavior of the mean free time $t$ (time between two successive collisions) for the molecules of an ideal gas,as a function of temperature $(T)$,qualitatively,is (Graphs are schematic and not drawn to scale)

$A$ 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 $R = 8.3 \ J \ mol^{-1} \ K^{-1}$).