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. Determine the work done by the gas in the left part during the proce

Step-by-step solution

The vessel is partitioned by a fixed separator, which means the separator cannot move and the volume of each part remains constant throughout the process. Since the volume of the left part does not change, the change in volume is dV = 0 . The work done by a gas is given by the integral of pressure with respect to volume: W = p , dV Substituting dV = 0 into the expression for work done, we get: W = 0 Thus, the work done by the gas in the left part during the process is zero.