Two monconducting containers having volume `V_(1)` and `V_(2)` contain monoatomic and dimatomic gases respectivley. They are connected as shown in figure. Pressure and temperature in the two containers are `P_(1), T_(1)` and `P_(2), T_(2)` respectively. Initially stop cock is closed, if the stop coc is opened find the final pressure and temperature.

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 be become 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`
`implies T= ((3P _(1)V_(1)+5P _(2) V_(2))T_(1) T_(2))/(3 P _(1) V _(1) T_(1) + 5P _(2)V_(2)T _(1))`
`P = ((3 P _(1) V _(1)+ 5P _(1) V _(2))/(3P _(1) V _(1)T_(2)+ 5P _(2)V_(2) T_(1)))((P _(1)V_(1) T_(2) + P _(2) V_(2)T_(2))/(V _(1) + V_(2)))`