Torentical containers each of volume V₀ are joined by a small pipe. The containers contain identical gases at temperature T₀ and pressure P₀ . One container is heated to temperature 2 T₀ while maintaining the other at the same temperature. The common pressure of the gas is P and n is the number of moles of gas in container at temperature 2 T₀

Step-by-step solution

Initially for container A, p₀ V₀=n₀ R T₀ For container B, p₀ V₀=n₀ R T₀ n₀= p₀ V₀ R T₀ Total number of moles =n₀+n₀=2 n₀ Since, even on heating the total number of moles is conserved Hence, n₁+n₂=2 n₀ ...(i) If p be the common pressure, then For container A, p V₀=n₁ R 2 T₀ n₁= p V₀ 2 R T₀ For container A, p V₀=n₂ R T₀ n₂= p V₀ R T₀ Substituting the value of n₀, n₁ and n₂ in Eq. (i) we get p V₀ 2 R T₀ + p V₀ R T₀ = 2 p₀ V₀ R T₀ p= 4 3 p₀ Number of moles in container A (at temperature 2 T₀ ) aligned & =n₁= P V₀ 2 R T