Two vessels of equal volume contain the same gas. Their pressures and temperatures are $P_1, T_1$ and $P_2, T_2$ respectively. When they are connected,the common pressure and temperature are $P$ and $T$. Then $P/T$ is equal to:

$A$ gaseous mixture consists of $2$ moles of oxygen and $4$ moles of argon at an absolute temperature $T$. Neglecting all vibrational modes,the total internal energy of the mixture of the gases is (in $RT$)

$A$ mixture of two non-reactive ideal gases is enclosed in a vessel consisting of one mole of a monatomic gas '$A$' and 'n' moles of a diatomic gas '$B$' at a temperature '$T$'. If the adiabatic constant of the gaseous mixture is $\frac{13}{9}$,then the value of 'n' is:

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 $Pa$)

$A$ gas mixture consists of $3$ moles of oxygen and $5$ moles of argon at temperature $T$. Assuming the gases to be ideal and the oxygen bond to be rigid,the total internal energy (in units of $RT$) of the mixture is

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?