One gram mole of an ideal gas $A$ with the ratio of constant pressure and constant volume specific heats $\gamma_{A} = 5/3$ is mixed with $n$ gram moles of another ideal gas $B$ with $\gamma_{B} = 7/5$. If the $\gamma$ for the mixture is $19/13$,then what will be the value of $n$?

When one mole of an ideal monatomic gas $(\gamma = 5/3)$ is mixed with one mole of a diatomic gas $(\gamma = 7/5)$, what is the value of $\gamma$ for the mixture?

In a container of volume $16.62 \ m^3$ at $0 \ ^{\circ}C$ temperature,$2 \ moles$ of oxygen,$5 \ moles$ of nitrogen,and $3 \ moles$ of hydrogen are present. Then,the pressure in the container is (Universal gas constant $R = 8.31 \ J \ mol^{-1} \ K^{-1}$) (in $Pa$)

Two ideal gases at temperatures $T_1$ and $T_2$ are mixed. There is no loss of energy. If the number of moles of the two gases are $n_1$ and $n_2$ respectively,then the temperature of the mixture will be:

$A$ closed vessel contains $0.1$ mole of a monoatomic ideal gas at $200\, K$. If $0.05$ mole of the same gas at $400\, K$ is added to it,the final equilibrium temperature (in $K$) of the gas in the vessel will be closest to:

$A$ gas mixture contains $n_1$ moles of a monoatomic gas and $n_2$ moles of a gas of rigid diatomic molecules. Each molecule in the monoatomic and diatomic gas has $3$ and $5$ degrees of freedom respectively. If the adiabatic exponent $\left(\frac{C_p}{C_V}\right)$ for this gas mixture is $1.5$, then the ratio $\frac{n_1}{n_2}$ will be