As shown in the figure, a cylindrical tube possesses adiabatic walls and is equipped with an adiabatic separator. An external mechanism can be used to slide the separator within the tube. An ideal gas ( = 1.5 ) is introduced into both sides at equal pressures and temperatures, with the separator initially in equilibrium at the middle. Subsequently, the separator is shifted to a position such that it divides the tube

Step-by-step solution

Let the initial volume of each part of the tube be V₀ . The total volume of the tube is 2V₀ . The separator is shifted such that it divides the tube in the ratio 1 : 3 . Let the final volumes of the two parts be V₁ and V₂ . V₁ = 1 4 2V₀ = V₀ 2 V₂ = 3 4 2V₀ = 3V₀ 2 Since the walls and the separator are adiabatic, the gas in both parts undergoes a reversible adiabatic process. For an adiabatic process, TV^ - 1 = constant . For the first part: T₁ V₁^ - 1 = T₀ V₀^ - 1 T₁ = T₀ ( V₀ V₁ )^ - 1 = T₀ ( V₀ V₀ / 2 )^ 1.5 - 1