In given figure, an adiabatic cylindrical tube of volume 2V₀ is divided in two equal parts by a frictionless adiabatic separator. An ideal gas in left side of a tube having pressure P₁ and temperature T₁ where as in the right side having pressure P₂ and temperature T₂. Cₚ/Cᵥ = γ is the same for both the gases. The separator is slid slowly and is released at a position where it can stay in equilibrium. Find ; (a) the final volumes of the two parts, (b) the heat given to the gas in the left part and (c) the final common pressure of the gases. Which of the following gives the correct final result?

Question

G1 Physics · Accepted Answer

Both gases change quasi-statically and adiabatically, so PV raised to γ remains constant on each side. At equilibrium the final pressures are equal and Vₗ+Vᵣ=2V₀. Solving gives Vₗ and Vᵣ proportional to p₁ raised to (1/γ) and p₂ raised to (1/γ), respectively. Because the system is adiabatic, Qₗ=0. Therefore option A.