Two ideal gases $A$ and $B$ having the same temperature $T$,same pressure $P$,and same volume $V$,are mixed together. If the temperature of the mixture is kept constant and the volume occupied by the mixture is reduced to $\frac{V}{2}$,then the pressure of the mixture will become:

One mole of monoatomic gas and three moles of diatomic gas are put together in a container. The molar specific heat (in $J\,K^{-1}\,mol^{-1}$) at constant volume is $(R = 8.3\,J\,K^{-1}\,mol^{-1})$.

$A$ closed container of volume $0.02 \ m^3$ contains a mixture of neon and argon gases at a temperature of $27^{\circ}C$ and a pressure of $1 \times 10^5 \ N/m^2$. The total mass of the mixture is $28 \ g$. If the atomic weights of neon and argon are $20$ and $40$ respectively,find the mass of each gas in the container in $g$,assuming they behave as ideal gases. $(R = 8.314 \ J/mol \cdot K)$

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:

If two moles of diatomic gas and one mole of monoatomic gas are mixed,then the ratio of specific heats $\gamma$ for the mixture is:

Four moles of hydrogen,two moles of helium and one mole of water vapour form an ideal gas mixture. $[C_v$ for hydrogen $= \frac{5}{2} R, C_v$ for helium $= \frac{3}{2} R, C_v$ for water vapour $= 3 R]$. What is the molar specific heat at constant pressure of the mixture?