The container shown in figure has two chambers separated by a partition of volumes `V_(1) =2.0 L` and `V_(2) =3.0 L`. The chambers contain `mu_(1) 4.0` and `mu_(2)=5.0` mole of a gas at pressure `p_(1) =1.00` atm and `P_(2) =2.00` atm. Calculate the pressure after the partition is removed and the mixture attains equilibrium.

Explanation

Consider the diagram
`"Given"" "V_(1) 2.0 L, V_(2) =3.0 L`
`mu_(1) 4.0 mol, mu_(2) =5.0 mol`
`p_(1) =100 atm p_(2) 2.00 atm`
For chamber `1,p_(1) ,V_(1) =mu_(1) RT_(1)`
For chamber `2, p_(2), V_(2) =mu_(2) RT_(2)`
When the parition is removed the gasses get mixed wihtout any loss of energy . the mixture now attains a common equilibrium pressure and the total volume of the system is sum of the volume of individual chamber `V_(1)` and `V_(2)`.
`"So"" " mu =mu_(1)+mu_(2), V =V_(1)+V_(2)`
From kinetic theory of gasses,
`"For l mole"" " pV =2/3 mu_(1)E_(1) " "[underset("kinetic energy")(E="translational"]]`
`"For " mu_(1) " moles"" "p_(1)V_(1)=2/3 mu_(2) E_(1)`
`"For "mu_(2) " moles"" "P_(2)V_(2) =2/3 mu_(2) E_(2)`
`"Total energy is "" "(mu_(1)E_(1)+mu_(2)E_(2)) =3/2 (P_(1)V_(1)+P_(2)V_(2))`
`"From the above relation"" "pV=2/3 E_("Total") =2/3 muE_("per mole")`
`p(V_(1)+V_(2)) =2/3 xx3/2(P_(1)V_(1)+p_(2)V_(2))`
`p=(p_(1)V_(1)+p_(2)V_(2))/(V_(1)+V_(2))`
`=((1.00 xx 2.0 + 2.00xx3.0)/(2.0+3.0))"atm"`
`=(8.0)/(5.0)=1.60 " atm"`