Initially for container A,p0V0=n0RT0 For container B,p0V0=n0RT0 ∴ n0=
p0V0
RT0
Total number of moles =n0+n0=2n0
Since, even on heating the total number of moles is conserved Hence, n1+n2=2n0 ...(i) If p be the common pressure, then For containerA,pV0=n1R2T0∴n1=
pV0
2RT0
For containerA,pV0=n2RT0∴n2=
pV0
RT
Substituting the value of n0,n1 and n2 in Eq. (i) we get
pV0
2RT0
+
pV0
RT0
=
2.p0V0
RT0
⇒p=
4
3
p0 Number of moles in container A (at temperature 2T0) =n1=