$N$ moles of a polyatomic gas $(f=6)$ must be mixed with two moles of a monoatomic gas so that the mixture behaves as a diatomic gas. The value of $N$ 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:

Two non-reactive monoatomic ideal gases have their atomic masses in the ratio $2: 3$. The ratio of their partial pressures,when enclosed in a vessel kept at a constant temperature,is $4: 3$. The ratio of their densities is:

The molar specific heat of a mixture at constant volume,if one mole of $He$ gas is mixed with three moles of $O_2$ gas,is: (in $R$)

If one mole of an ideal monoatomic gas $\left(\gamma = \frac{5}{3}\right)$ is mixed with one mole of a diatomic gas $\left(\gamma = \frac{7}{5}\right)$,the value of $\gamma$ for the mixture is:

Two containers of equal volume contain identical gases at pressures $P_1$ and $P_2$ and absolute temperatures $T_1$ and $T_2$ respectively. The vessels are joined and the gas reaches a common pressure $P$ and a common temperature $T$. Then: