The container shown in Fig. 9(EP).6 has two chambers, separated by a partition, of volumes `V_(1) = 2.0 litre and V_(2) = 3.0 litre`. The chambers contains `mu_(1) = 4.0 and mu_(2) = 5.0 "moles"` of a gas at pressures `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

Here, `V_(1) = 2.0` litre, `V_(2) = 3.0` litre, `mu_(1) = 4.0` mole , `mu_(2) = 5.0` mole
`p_(1) = 1.00 atm , p_(2) = 2.00` atm
After the partition is removed and the mixture attains equilibrium, `p=?`
we can show that `p=(p_(1)V_(1)+p_(2)V_(2))/(V_(1)+V_(2)) = (1.00 xx 2.0 + 2.00 xx 3.0)/(2.0 + 3.0) = 8.0/5.0 = 1.6` atmosphere.