$A$ container of fixed volume contains a gas at $27^{\circ} C$. To double the pressure of the gas,the temperature of the gas should be raised to . . . . . . ${ }^{\circ} {C}$.

An air bubble of volume $2.9 \ cm^3$ rises from the bottom of a swimming pool $5 \ m$ deep. At the bottom of the pool, the water temperature is $17^{\circ} C$. The volume of the bubble when it reaches the surface, where the water temperature is $27^{\circ} C$, is . . . . . . $cm^3$. ($g = 10 \ m/s^2$, density of water $\rho = 10^3 \ kg/m^3$, and $1 \ atm = 10^5 \ Pa$)

The mass of oxygen gas occupying a volume of $11.2 \text{ litres}$ at a temperature $27^{\circ}C$ and a pressure of $760 \text{ mm}$ of mercury in kilograms is $[$ Molecular weight of oxygen $= 32]$

Two different masses $m$ and $3m$ of an ideal gas are heated separately in a vessel of constant volume. The pressure $P$ versus absolute temperature $T$ graphs for these two cases are shown in the figure. The ratio of the slopes of curves $B$ to $A$ is

You have two containers of equal volume. One is full of helium gas. The other holds an equal mass of nitrogen gas. Both gases have the same pressure. How does the temperature of the helium compare to the temperature of the nitrogen?

We write the relation for Boyle's law in the form $PV = C$ when the temperature remains constant. In this relation,the magnitude of $C$ depends upon