$A$ mixture of ideal gases consists of $7 \, kg$ of nitrogen and $11 \, kg$ of $CO_2$. Then (Take $\gamma$ for nitrogen and $CO_2$ as $1.4$ and $1.3$ respectively):

One mole of a gas with $\gamma = 7/5$ is mixed with one mole of a gas with $\gamma = 4/3$. What will be the $\gamma$ of the mixture?

$1 \, \text{mole}$ of a gas having $\gamma = \frac{7}{5}$ is mixed with $1 \, \text{mole}$ of a gas having $\gamma = \frac{4}{3}$. What will be the $\gamma$ for the mixture?

$1 \text{ mole}$ of a gas with $\gamma = 7/5$ is mixed with $1 \text{ mole}$ of a gas with $\gamma = 5/3$. The value of $\gamma$ for the resulting mixture is:

$N_1$ molecules of a gas at temperature $T_1$ are mixed with $N_2$ molecules at temperature $T_2$. The resulting temperature of the mixture gas 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 $.5$)