When the temperature of a gas in a closed container is increased by $1 ^\circ C$,its pressure increases by $0.4 \%$. What is the initial temperature of the gas?

The temperature of a gas having $2.0 \times 10^{25}$ molecules per cubic meter at $1.38 \text{ atm}$ (Given,$k = 1.38 \times 10^{-23} \text{ J K}^{-1}$) is: (in $\text{ K}$)

The pressure of a given mass of gas at $20^{\circ}C$ is $P$. At constant volume,at what temperature in $^{\circ}C$ will the pressure of the gas become $2P$?

The expansion of an ideal gas of mass $m$ at a constant pressure $P$ is given by the straight line $D$. Then the expansion of the same ideal gas of mass $2m$ at a pressure $P/2$ is given by the straight line

When $2 \, g$ of a gas are introduced into an evacuated flask kept at $25 \, ^\circ C$,the pressure is found to be $1 \, atm$. If $3 \, g$ of another gas is added to the same flask,the pressure becomes $1.5 \, atm$. The ratio of the molecular weights of these gases will be:

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