Two adiabatic containers have volumes `V_(1)` and `V_(2)` respectively. The first container has monoatomic gas at pressure `p_(1)` and temperature `T_(1)`. The second container has another monoatomic gas at pressure `p_(2)` and temperature `T_(2)`. When the two containers are connected by a narrow tube, the final temperature and pressure of the gases in the containers are P and T respectively. Then

Explanation

`n_(1) = (P_(1)V_(1))/(RT_(1)) n_(2) = (P_(2)V_(2))/(RT_(2))`
`n = n_(1) +n_(2)` (number of moles are conserved)
Finally pressure in both parts & temperature of the both the gases will becomes equal.
`(P(V_(1)+V_(2)))/(RT) = (P_(1)V_(1))/(RT_(1)) +(P_(2)V_(2))/(RT_(2))`
From energy conservation
`(3)/(2)n_(1)RT_(1)+(5)/(2) n_(2)RT_(2) = (3)/(2)n_(1)RT + (5)/(2) n_(2)RT`
`rArr T = ((3P_(1)V_(1)+5P_(2)V_(2))T_(1)T_(2))/(3P_(1)V_(1)T_(2)+5P_(1)V_(2)T_(1)) rArrP =((3P_(1)V_(1)+5P_(2)V_(2))/(3P_(1)V_(1)T_(2)+5P_(1)V_(2)T_(1)))((P_(1)V_(1)T_(2)+P_(2)V_(2)T_(1))/(V_(1)+V_(2)))`