$A$ vessel has a capacity of $3 \ L$. If it contains a mixture of $6 \ g$ of $O_2$,$8 \ g$ of $N_2$,and $5 \ g$ of $CO_2$,what will be the pressure of the vessel at a temperature of $27^{\circ}C$? (Given $R = 8.31 \ J/mol \cdot K$)

$A$ container holds $8 \, g$ of $O_2$ and $7 \, g$ of $N_2$ at a pressure of $10 \, atm$. If $O_2$ is removed,the pressure becomes ... $atm$.

One mole of ideal monoatomic gas $(\gamma = 5/3)$ is mixed with one mole of diatomic gas $(\gamma = 7/5)$. What is $\gamma$ for the mixture? $\gamma$ denotes the ratio of specific heat at constant pressure to that at constant volume.

Three identical containers contain three different gases. The masses of the molecules are $m_1, m_2$,and $m_3$,and the number of molecules in the containers are $N_1, N_2$,and $N_3$,respectively. The pressures of the gases in the containers are $P_1, P_2$,and $P_3$,respectively. If these three gases are mixed and filled into a single container of the same volume,what will be the pressure of the mixture?

The pressure and temperature of two different gases are $P$ and $T$,each having the same volume $V$. If they are mixed while keeping the volume and temperature constant,the pressure of the mixture will be:

$N$ 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: