|2B-1 :||Condensing Water Vapor by Increasing the Pressure||3 pts|
|A piston-and-cylinder device contains water vapor at 135oC and 1 atm in state 1. The cylinder contains no air and no liquid water. The cylinder is surrounded by a water bath at 135oC that keeps the temperature of the|
|contents of the cylinder constant. The piston is pushed slowly downward causing the pressure to increase steadily. Eventually, a
small droplet of liquid water forms in state 2.
|a.) Describe the state of the contents of the cylinder in state 1 and in state 2.
|b.) Sketch the path of this process on a PV Diagram. Be sure to label all the regions on the diagram and include the two-phase envelope and all relevant isotherms.|
|Read :||The keys to this
1- The process is isothermal because the constant temperature bath keeps the system at 135oC and
2- The final state is saturated vapor because the 1st micro-droplet of lquid water condenses.
|P1||1||atm||P2 > P1|
|Find:||a.)||Describe states 1 & 2|
|b.)||Sketch process path on a PV Diagram|
|Assumptions:||1- The constant temperature bath is prefectly effective in keeping the contents of the cylinder at a constant and uniform temperature of 135oC.|
|Equations / Data / Solve:|
|Part a.)||- In state 1, the cylinder contains all water vapor.
- The pressure must be increased on this vapor in order to cause any liquid water droplets to condense.
- We can conclude that P1 = 1 atm is lower than the vapor pressure (or saturation pressure) of water
at T1 = 135oC.
- Therefore the water vapor in state 1 is a superheated vapor.
|- In state 1, the cylinder contains all water vapor.
- The pressure on this vapor is just exactly high enough to cause a micro-droplet of liquid water to condense.
- We can conclude that P2 = P*(135oC), the vapor pressure (or saturation pressure) of water at T1 = 135oC.
- Therefore the water vapor in state 1 is a saturated vapor.
|Verify:||This assumption cannot be veriefied without experimentation.|
|Answers :||See above.|