Two vessels `A` and `B`, thermally insulated, contain an ideal monoatomic gas. A small tube fitted with a valve 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

5

Explanation

For insulated vessels, no heat how out sides vessless. `U_(1) +U_(2) = U'_(1) +U'_(2)`
`rArr (f)/(2) n_(1) RT_(1) +(f)/(2) n_(2)RT_(2) = (f)/(2) n_(1)RT +(f)/(2)n_(2)RT`
`T = (n_(1)T_(1)+n_(2)T_(2))/(n_(1)+n_(2))`, For `A` vessel
`n_(1) = (P_(1)V_(1))/(RT_(1)) = (2xx10^(5)xx2xx10^(-3))/(((25)/(3))xx100)= (4)/(25)` mole
For `B` vessel
`n_(2) = (P_(2)V_(2))/(RT_(1)) = (4xx10^(5) xx4xx10^(-3))/(25//3xx350) = (480)/(25xx35) = (96)/(175)` mole
Temperature `T = ((4)/(25)xx300+(96)/(175)xx350)/((4)/(25)+(96)/(175) ) = (10500)/(31)K`
From mole conservation
`n_(1) +n_(2) = n'_(1) +n'_(2)`
`(P_(1)V_(1))/(RT_(1)) +(P_(2)V_(2))/(RT_(2)) = (P'V_(1))/(RT) +(P'V_(2))/(RT)`
`P = ((P_(1)V_(1)T_(2)P_(2)V_(2)T_(1))T)/(T_(1)T_(2)(V_(1)+V_(2)))`