For Boyle's law to hold, the gas should be:

$A$ container $A$ contains an ideal gas at pressure $P$,volume $V$,and temperature $T$. $A$ second container $B$ contains the same gas at pressure $2P$,volume $2V$,and temperature $\frac{T}{2}$. The ratio of the mass of gas in $A$ to that in $B$ is:

$28\, g$ of $N_2$ gas is contained in a flask at a pressure of $10\, atm$ and at a temperature of $57^\circ C$. It is found that due to leakage in the flask,the pressure is reduced to half and the temperature is reduced to $27^\circ C$. The quantity of $N_2$ gas that leaked out is

The given figure represents two isobaric processes for the same mass of an ideal gas. Then:

Absolute zero $(0\, K)$ is that temperature at which

$A$ cylinder of volume $30 \, L$ contains oxygen at an initial gauge pressure of $15 \, atm$ and a temperature of $27^{\circ}C$. After some oxygen is withdrawn from the cylinder,the gauge pressure drops to $11 \, atm$ and its temperature drops to $17^{\circ}C$. The mass of oxygen taken out of the cylinder is approximately ........ $kg$. (Given: $R = 8.31 \, J \, mol^{-1} K^{-1}$,molar mass of $O_2 = 32 \, g/mol$ or $32 \times 10^{-3} \, kg/mol$)