One mole of an ideal diatomic gas is taken through the cycle as shown in the figure.
$1 \rightarrow 2$: isochoric process
$2 \rightarrow 3$: straight line on $P-V$ diagram
$3 \rightarrow 1$: isobaric process
The average molecular speed of the gas in the states $1, 2$ and $3$ are in the ratio

$A$ box contains $N$ molecules of a perfect gas at temperature ${T_1}$ and pressure ${P_1}$. The number of molecules in the box is doubled while keeping the total kinetic energy of the gas the same as before. If the new pressure is ${P_2}$ and temperature is ${T_2}$,then:

An oxygen cylinder of volume $30 \ L$ has $18.20 \ \text{moles}$ of oxygen. After some oxygen is withdrawn from the cylinder, its gauge pressure drops to $11 \ \text{atm}$ at a temperature of $27^{\circ} \text{C}$. The mass of the oxygen withdrawn from the cylinder is nearly equal to (Given: $R = \frac{100}{12} \ \text{J mol}^{-1} \text{K}^{-1}$, molecular mass of $O_2 = 32 \ \text{g/mol}$, $1 \ \text{atm} = 1.01 \times 10^5 \ \text{Pa}$) (in $\text{kg}$)

$A$ jar contains a gas and a few drops of water at $T \, K.$ The pressure in the jar is $830 \, mm$ of mercury. The temperature of the jar is reduced by $1\%.$ The saturated vapour pressure of water at the two temperatures are $30 \, mm$ and $25 \, mm$ of mercury. Then the new pressure in the jar will be ..... $mm$ of $Hg$.

$A$ thermally insulated vessel contains an ideal gas of molecular mass $M$ and ratio of specific heats $\gamma$. It is moving with speed $v$ and is suddenly brought to rest. Assuming no heat is lost to the surroundings,its temperature increases by

For a diatomic gas,the change in internal energy for a unit change in temperature at constant pressure and constant volume is $U_1$ and $U_2$ respectively. The ratio $U_1 : U_2$ is