$3$ moles of mono-atomic gas $(\gamma = 5/3)$ is mixed with $1$ mole of a diatomic gas $(\gamma = 7/5)$. The value of $\gamma$ for the mixture will be

$N_1$ 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

The volume $V$ of an enclosure contains a mixture of three gases: $16 \, g$ of oxygen,$28 \, g$ of nitrogen,and $44 \, g$ of carbon dioxide at absolute temperature $T$. Consider $R$ as the universal gas constant. The pressure of the mixture of gases is:

$A$ mixture of carbon dioxide and oxygen has volume $8310 \text{ cm}^3$, temperature $300 \text{ K}$, pressure $100 \text{ kPa}$ and mass $13.2 \text{ g}$. The number of moles of carbon dioxide and oxygen gases in the mixture respectively are . . . . . . . (Assume both carbon dioxide and oxygen gases behave like ideal gases) $[R = 8.31 \text{ J/mol.K}]$

Five moles of helium are mixed with two moles of hydrogen to form a mixture. Take molar mass of helium $M_1=4\ g$ and that of hydrogen $M_2=2\ g$. If the internal energy of the $He$ sample is $100\ J$ and that of the hydrogen sample is $200\ J$,then the internal energy of the mixture is ..... $J$.

The container shown in the figure has two chambers,separated by a partition,with volumes $V_1 = 2.0 \, L$ and $V_2 = 3.0 \, L$. The chambers contain $\mu_1 = 4.0 \, mol$ and $\mu_2 = 5.0 \, mol$ of a gas at pressures $P_1 = 1.00 \, atm$ and $P_2 = 2.00 \, atm$. Calculate the pressure after the partition is removed and the mixture attains equilibrium.