1. The problem statement, all variables and given/known data A saturated water vapor mix in a 0.05m3 piston-cylinder at 200C with a quality of 0.5 expands adiabatically, producing 500kJ work with a final temperature of 50C. Find: a) final quality b) change in entropy 2. Relevant equations Q-W=ΔU x=(vavg-vf)/(vg-vf) uavg=uf+xufg savg=sf+xsfg 3. The attempt at a solution A) For an adiabatic process Q=0, so -W=ΔU=-500kJ If I can used the starting parameters to find u1 and the total mass, I can find U1 , U2, then u2 and use the equation uavg=uf+xufg to find the final value for x, the quality. relevant values from the steam tables: @ 200C: uf=850.65 kJ/kg ufg=1744.7 kJ/kg vg=0.13736 m3/kg vf=0.001157 m3/kg @ 50C: uf=209.32 kJ/kg ufg=2234.2 kJ/kg x=0.5=(vavg-vf)/(vg-vf) vavg=x(vg-vf)+vf=0.06926 m3/kg V/vavg=m=0.7219 kg uavg=uf+xufg=850.65+0.5(1744.7)=1723 kJ/kg U1=mu1=0.7219*1723=1243.86 kJ U2-U1=ΔU=-500kJ ΔU+U1=U2=-500+1243.86=743.86kJ U2/m=u2=1030.42 kJ/kg uavg=uf+xufg (uavg-uf)/ufg=(1030.42-209.32)/2231.2=0.368=X2 b) On to entropy: savg=sf+xsfg Δs=savg 2+savg 1 ΔS=mΔs from the steam tables: @ 200C sf=2.3309 kJ/kg*K sfg=4.1014 @ 50C sf=0.7038 kJ/kg*K sfg=7.3725 ΔS=mΔs=0.7219((0.7038+0.368*7.3725)-(2.3309+0.5(4.1014)))=-0.69 So this was the first red flag for me. I thought ΔS is always supposed to be greater than 0. Then the second realization hit me, isn't ΔS 0 for an adiabatic process? But then, if ΔS=0, I solved for x sf1-0.5sfg1=sf2-xsfg2 and came out with x=0.499, so essentially no Δs means no Δx. So is final x 0.368 or 0.5? I've checked over and over my calculations in part a. I've quadruple checked to make sure the values were correctly copied from the tables. I've redone the calculations to make sure I copied the values correctly into my calculator. Is some part of the process in part a incorrect, do the equations I used not apply in this situation for some reason?