The pressure and temperature of two different gases are $P$ and $T$,each having the same volume $V$. If they are mixed while keeping the volume and temperature constant,the pressure of the mixture will be:

Two containers of equal volume contain identical gases at pressures $P_1$ and $P_2$ and absolute temperatures $T_1$ and $T_2$ respectively. The vessels are joined and the gas reaches a common pressure $P$ and a common temperature $T$. Then:

$A$ gas mixture consists of $2$ moles of $O_2$ and $4$ moles of $Ar$ at temperature $T$. Neglecting all vibrational modes,the total internal energy of the system is (in $RT$)

$A$ container is divided into two chambers by a partition. The volume of the first chamber is $4.5 \, L$ and the second chamber is $5.5 \, L$. The first chamber contains $3.0 \, \text{moles}$ of gas at a pressure of $2.0 \, \text{atm}$, and the second chamber contains $4.0 \, \text{moles}$ of gas at a pressure of $3.0 \, \text{atm}$. After the partition is removed and the mixture attains equilibrium, the common equilibrium pressure existing in the mixture is $x \times 10^{-1} \, \text{atm}$. The value of $x$ is......... (in $.5$)

$A$ closed vessel contains $0.1$ mole of a monoatomic ideal gas at $200\, K$. If $0.05$ mole of the same gas at $400\, K$ is added to it,the final equilibrium temperature (in $K$) of the gas in the vessel will be closest to:

Three closed vessels $A, B$ and $C$ are at the same temperature $T$ and contain gases which obey the Maxwellian distribution of velocities. Vessel $A$ contains only $O_2$,$B$ only $N_2$ and $C$ a mixture of equal quantities of $O_2$ and $N_2$. If the average speed of the $O_2$ molecules in vessel $A$ is $V_1$,and that of the $N_2$ molecules in vessel $B$ is $V_2$,what is the average speed of the $O_2$ molecules in vessel $C$?