$A$ closed container of volume $0.02 \ m^3$ contains a mixture of neon and argon gases at a temperature of $27^{\circ}C$ and a pressure of $1 \times 10^5 \ N/m^2$. The total mass of the mixture is $28 \ g$. If the atomic weights of neon and argon are $20$ and $40$ respectively,find the mass of each gas in the container in $g$,assuming they behave as ideal gases. $(R = 8.314 \ J/mol \cdot K)$

Two gases $A$ and $B$ having the same pressure $P$, volume $V$, and absolute temperature $T$ are mixed. If the mixture has the volume $V$ and temperature $T$, then the pressure of the mixture is:

Five moles of helium are mixed with two moles of hydrogen to form a mixture. Take molar mass of helium $M_1 = 4 \ g/mol$ and that of hydrogen $M_2 = 2 \ g/mol$. The equivalent molar mass of the mixture is:

$A$ 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 $.5$)

$A$ mixture of $7 \, g$ of nitrogen and $11 \, g$ of carbon dioxide is kept in a container at a temperature of $T = 290 \, K$. If the pressure of the mixture is $1 \, atm$ $(1.01 \times 10^5 \, N/m^2)$,then the density of the mixture is approximately . . . . . . $kg/m^3$.

Three identical containers contain three different gases. The masses of the molecules are $m_1, m_2$,and $m_3$,and the number of molecules in the containers are $N_1, N_2$,and $N_3$,respectively. The pressures of the gases in the containers are $P_1, P_2$,and $P_3$,respectively. If these three gases are mixed and filled into a single container of the same volume,what will be the pressure of the mixture?