In the figure,the volume of container $X$ is twice the volume of container $Y$. Both contain an ideal gas. The temperature of $X$ is $200 \, K$ and the temperature of $Y$ is $400 \, K$. If the mass of the gas in $X$ is $m$,what will be the mass of the gas in $Y$?

The lowest pressure (the best vacuum) that can be created in the laboratory at $27^{\circ}C$ is $10^{-11} \; mm$ of $Hg$. At this pressure,the number of ideal gas molecules per $cm^{3}$ will be:

One mole of an ideal gas undergoes a process in which pressure $P$ varies with volume $V$ as $P = 3 - g \left(\frac{V}{V_0}\right)^2$, where $V_0$ and $g$ are constants. The maximum temperature attainable by the ideal gas during this process is ($All$ quantities are in $SI$ units and $R$ is the gas constant).

An ideal gas at pressure $P$ and temperature $T$ is enclosed in a vessel of volume $V$. Some gas leaks through a hole from the vessel and the pressure of the enclosed gas falls to $P'$. Assuming that the temperature of the gas remains constant during the leakage, the number of moles of the gas that have leaked is

$A$ container holds $13 \, g$ of gas at a temperature of $27^{\circ}C$. If the temperature of the container is raised to $52^{\circ}C$ while keeping the pressure constant,the amount of gas that must be released is ..... $g$.

$A$ long cylindrical pipe of radius $20 \,cm$ is closed at its upper end and has an airtight piston of negligible mass as shown. When a $50 \,kg$ mass is attached to the piston,it moves down. If the air in the enclosure is cooled from temperature $T$ to $T-\Delta T$,the piston moves back to its original position. Then $\Delta T / T$ is close to (Assuming air to be an ideal gas,$g=10 \,m/s^2$,atmospheric pressure is $10^5 \,Pa$)