Hint: The total no. of molecules will remain the same.
Consider the diagram,
Given,
V1=2.0 L, V2=30 L
μ1=40 mol, μ2=50 mol
p1=1.00 atm, p2=2.00 atm
Step 1: Find the relation between the pressure, volume and the total energy of the gas.
For chamber 1, p1V1=μ1RT1
For chamber 2, p2V2=μ2RT2
When the partition is removed the gasses get mixed without any loss of energy. The mixture now attains a common equilibrium pressure and the total volume of the system is the sum of the volume of individual chambers V1 and V2.
So, μ=μ1+μ2,V=V1+V2
From the kinetic theory of gases,
For 1 mole, pV=23E [E=translational kinetic energy]
For μ1 moles, p1V1=23μ1E1
For μ2 moles, p2V2=23μ2E2
The total energy is, (μ1E1+μ2E2)=32(p1V1+p2V2)
Step 2: Find the final pressure of the chamber.
From the above relation,
pV=23Etotal=23μEper mole
p(V1+V2)=23×32(p1V1+p2V2)
p=p1V1+p2V2V1+V2
=(1.00×2.0+2.00×302.0+3.0)atm
=8.05.0=160 atm