As shown in the figure, a cylindrical tube with adiabatic walls is equipped with a diathermic separator. An external mechanism allows the separator to be slid inside the tube. An ideal gas is introduced into both compartments at equal pressures and equal temperatures, with the separator initially in equilibrium at the midpoint. The separator is subsequently slid to a location where it partitions the tube in the ratio

Step-by-step solution

Let the initial volume of each compartment be V₀ . The total volume of the cylindrical tube is 2V₀ . Initially, both compartments contain an ideal gas at the same pressure P₀ and temperature T₀ . According to the ideal gas law, the number of moles in each compartment is equal: n₁ = n₂ = n = P₀ V₀ R T₀ The separator is then slid to partition the tube in a 1:3 ratio. The new volumes of the two compartments are: V₁ = 1 1+3 2V₀ = V₀ 2 V₂ = 3 1+3 2V₀ = 3V₀ 2 Since the separator is diathermic, it allows heat exchange bet