1. The problem statement, all variables and given/known data Consider an ideal friction-less cylinder-piston arrangement, in which the piston is free to move between two sets of stops. When the piston rests on the lower stops, the volume available for the contained fluid is 0.5 m3; when the piston reaches the upper stops, the volume is 0.7 m3. The piston has a mass such that an internal pressure of 3.5 bar is required to lift it. Initially (State 1), the system contains a liquid-vapour mixture of water at 1 bar and 14% quality. The system is heated until the water is fully evaporated (i.e., to saturated vapour; State 2). a) Determine the initial temperature in the cylinder. b) Determine the final pressure and temperature in the cylinder. c) Calculate the work done by the fluid. d) What is the heat transferred during the process? If the system in State 1 was cooled, instead of heated, to ambient temperature (25ºC; State 3) what would be the final pressure, and what would be the quality? 2. Relevant equations for closed system, du = q+w w = -PdV 3. The attempt at a solution The initial temperature is 99.6 degrees c according to the steam tables. (because the temperature has to be the temperature of saturation at 1 bar or else there would be no liquid-vapour coexistence?) at quality = 0.14 specific volume of saturated liquid = 0.001043 m3/kg specific volume of saturated vapour = 1.6939 m3/kg using x=(v-vl)/(vg-vl) (v being specific volume of vapour liquid, vl being specific volume of saturated liquid, vg being specific volume of saturated vapour) v = 0.2380 m3/kg taking 1kg water basis, V = 0.2380 m3 but initial volume available in the cylinder for the water is 0.5m3 so actual mass of water present to begin with is 0.5/0.2380 = 2.1kg I'm thinking that at state 2 the specific volume is going to be 0.3333m3/kg as the volume will now be 0.7m3 instead of 0.5m3, so 0.7/2.1 = 0.3333 The pressure is 5.67 bar with a temperature of 156.6 degrees c? If the above is correct, then comes the question of how the process is going to be like (on a pv diagram)....... Please shed some light on this.