The container shown in Fig. 13.6 has two chambers, separated by a partition, of volumes V1 = 2.0 litre and V2= 3.0 litre. The chambers contain μ1 = 4.0and 5.0 μ2 = moles of a gas at pressures p1 = 1.00 atm and p2 = 2.00 atm. Calculate the pressure after the partition is removed and the mixture attains equilibrium.
Given:
V1= 2.0litre, V2=3.0litre
When the partition is removed the gases in both the partition mix without any loss of energy and acquire a common pressure P after mixing.
Total volume of gas after equilibrium = 
From ideal gas equation, product of pressure p and volume V is
Where
n=no. of moles of gas
….(i)
….(ii)
Also, from kinetic theory of gases,
..(0)
…..(iii)
….(iv)
Where,
n=no. of moles of gas
E=internal kinetic energy of gas
Initially 
is present
Total initial energy
Using eqn.(iii) and (iv), we get
After removing the partition,
Total moles=
Total volume =
Using eqn-(0) we have final product of pressure p and volume V equal to
Common pressure p is given by
Putting the values of 
we get
Therefore, total final pressure after the partition is removed is 1.6atm.
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