$3\, \text{moles}$ of an ideal gas at a temperature of $27^{\circ}\, \text{C}$ are mixed with $2\, \text{moles}$ of an ideal gas at a temperature of $227^{\circ}\, \text{C}$. Determine the equilibrium temperature $(^{\circ}\, \text{C})$ of the mixture, assuming no loss of energy.

$A$ mixture of ideal gas containing $5$ moles of monatomic gas and $1$ mole of rigid diatomic gas is initially at pressure $P_0$,volume $V_0$ and temperature $T_0$. If the gas mixture is adiabatically compressed to a volume $V_0 / 4$,then the correct statement$(s)$ is/are:
(Given $2^{1.2}=2.3$; $2^{3.2}=9.2$; $R$ is gas constant)
$(1)$ The final pressure of the gas mixture after compression is in between $9 P_0$ and $10 P_0$.
$(2)$ The average kinetic energy of the gas mixture after compression is in between $18 RT_0$ and $19 RT_0$.
$(3)$ The work $|W|$ done during the process is $13 RT_0$.
$(4)$ Adiabatic constant of the gas mixture is $1.6$.

Two ideal polyatomic gases at temperatures $T_{1}$ and $T_{2}$ are mixed such that there is no loss of energy. If $F_{1}$ and $F_{2}$,$m_{1}$ and $m_{2}$,$n_{1}$ and $n_{2}$ are the degrees of freedom,masses,and number of molecules of the first and second gas respectively,then the temperature of the mixture of these two gases is:

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.

Two moles of helium are mixed with $n$ moles of hydrogen. If $\frac{C_P}{C_V} = \frac{3}{2}$ for the mixture,then the value of $n$ is:

Considering the gases to be ideal,the value of $\gamma = \frac{C_P}{C_V}$ for a gaseous mixture consisting of $3$ moles of carbon dioxide and $2$ moles of oxygen will be $(\gamma_{O_2} = 1.4, \gamma_{CO_2} = 1.3)$.