An adiabatic vessel of total volume V is partitioned into two equal parts by a fixed, conducting separator. The left part holds one mole of an ideal gas ( U = 1.5 nRT ) and the right part contains two moles of the same gas. Initially, the pressure on each side is p . The system is allowed to rest for sufficient time until a steady state is achieved. Find the final common temperature reached by the gases.

Step-by-step solution

Let the initial temperatures of the left and right parts be T_L and T_R respectively. The volume of each part is V 2 . For the left part: p ( V 2 ) = (1) R T_L T_L = pV 2R For the right part: p ( V 2 ) = (2) R T_R T_R = pV 4R The internal energy of the gas is given by U = 1.5 nRT = 3 2 nRT . The initial total internal energy of the system is: U_i = 3 2 (1) R T_L + 3 2 (2) R T_R U_i = 3 2 R ( pV 2R ) + 3R ( pV 4R ) U_i = 3pV 4 + 3pV 4 = 3pV 2 Let the final common temperature be T_f . The final total internal energy