When one mole of monatomic gas is mixed with one mole of a diatomic gas,then the equivalent value of $\gamma$ for the mixture will be (vibration mode neglected).

If $4$ moles of an ideal monoatomic gas at temperature $400 \,K$ is mixed with $2$ moles of another ideal monoatomic gas at temperature $700 \,K$, the temperature of the mixture is:

$A$ mixture of ideal gases consists of $7 \, kg$ of nitrogen and $11 \, kg$ of $CO_2$. Then (Take $\gamma$ for nitrogen and $CO_2$ as $1.4$ and $1.3$ respectively):

$A$ 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:

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:

Three perfect gases at absolute temperatures $T_1, T_2$ and $T_3$ are mixed. The masses of molecules are $m_1, m_2$ and $m_3$ and the number of molecules are $n_1, n_2$ and $n_3$ respectively. Assuming no loss of energy,the final temperature of the mixture is