Two ideal gases at absolute temperatures $T_1$ and $T_2$ are mixed. There is no loss of energy. The masses of the molecules are $m_1$ and $m_2$ and the number of molecules in the gases are $n_1$ and $n_2$ respectively. The temperature of the mixture will be:

$A$ vessel of volume $20 \ L$ contains a mixture of hydrogen and helium at a temperature of $27^{\circ} C$ and pressure $2 \ atm$. The mass of the mixture is $5 \ g$. Assuming the gases to be ideal,the ratio of the mass of hydrogen to that of helium in the given mixture will be:

$A$ 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 . . . . . . $kg/m^3$.

$A$ 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 $.5$)

One mole of an ideal monoatomic gas $(\gamma = 5/3)$ is mixed with one mole of a diatomic gas $(\gamma = 7/5).$ What is $\gamma$ for the mixture? $\gamma$ denotes the ratio of specific heat at constant pressure to that at constant volume.

Two gases have temperature $T$,pressure $P$,and volume $V$ each. If they are mixed in a container of volume $V$ at the same temperature $T$,what will be the new pressure and mass of the mixture?