A cylindrical tube of volume V having adiabatic walls contains an ideal gas. The internal energy of this ideal gas is given by 1.5 nRT . A fixed diathermic wall divides the tube into two equal parts. Initially, the pressure and temperature are p₁, T₁ on the left side and p₂, T₂ on the right side. The system is allowed sufficient time for the temperatures on both sides to become equal. Find the final pressures on the

Step-by-step solution

Let the volume of each part be V₀ = V 2 . The number of moles of gas in the left part is: n₁ = p₁ V₀ R T₁ The number of moles of gas in the right part is: n₂ = p₂ V₀ R T₂ Since the outer walls are adiabatic and the total volume is constant, the total internal energy of the system remains conserved. Let the final common temperature be T_f . U_ initial = U_ final n₁ C_v T₁ + n₂ C_v T₂ = (n₁ + n₂) C_v T_f Solving for T_f : T_f = n₁ T₁ + n₂ T₂ n₁ + n₂ Substituting the expressions for n₁ and n₂ : n₁ T₁ + n₂ T₂ = p₁ V₀ R