Two vessels `A` and `B`, thermally insulated, contain an ideal monoatomic gas. A small tube fitted with a value connects these vessels. Initially the vessel `A` has `2` litres of gas at `300K` and `2 xx 10^(5)Nm^(-2)` pressure while vessel `B` has `4` litres of gas at `350 K` and `4 xx 10^(5)Nm^(-2)` pressure. The value is now opened and the system reaches equilibrium in pressure and temperature. Calculate the new pressure and temperature . `(R = (25)/(3)J//mol-K)`

Answer

`P = (10)/(3)xx10^(5) N//m^(2), T = (10500)/(31) K ~~ 338.71 K`

Explanation

From `PV = nRT`
`n = (PV)/(RT) n_(A) = (2 xx 10^(5) xx2xx 10^(-3))/((25)/(3) xx 300) = (4)/(25)`,
`n_(B) = (4 xx 10^(5) xx4 xx 10^(-3))/((25)/(3)xx350) = (96)/(175)`
`U_(1) +U_(2) = U_(mix)`
`rArr n_(A) C_(V)T_(1) +n_(B)C_(V)T_(2)=(n_(A) +n_(B)) C_(V)T`
`rArr T = (n_(A)T_(1)+n_(B)T_(2))/(n_(A)+n_(B))`

Putting values
`T = (10500)/(31) K ~~ 338.71 K`
`P = ((n_(A)+n_(B))RT)/(V_(A)+V_(B))`
Putting values
`P = (10)/(3) xx 10^(5) N//m^(2)`