1. The problem statement, all variables and given/known data Two pounds of water vapor at 30 psia fill the [itex]4ft^3[/itex] left chamber of a partitioned system. The right chamber has twice the volume of the left and is initially evacuated. Determine the pressure of the water after the partition has been removed and enough heat has been transferred so that the temperature of the water is [itex]40^°F.[/itex] 2. Relevant equations The change in internal energy is heat minus work, [itex]ΔU=Q-W[/itex] where work done by the system is positive. Steam tables [itex]water = 0.0397 lb/mol[/itex] 3. The attempt at a solution I begin with my known properties of state: for state one (before partition removal), I have the following: m = 2 lbs water P = 30 psia V = 4ft^3 T = ? U = ? State two: m = 2 lbs water P = ? V = 12ft^3 (the original space plus eight more cubic feet) T = 40 °F. U = ? I want the initial temperature, so using the steam tables and interpolation, I find 30 psia falling between values as follows: 29.823 psia at 250 °F and 35.422 psia at 260 °F. My calculation results in 250.316 °F, which looks reasonable. My questions begin...... My understanding of the steam table values is lacking clarity I believe. Here is what I think: When I look at these values of pressure and temperature for Saturated water, I see specific volumes between 13 and 11 cubic feet per molar pound, between 29.8 and 35.4 psia. With the problem giving me 2 lbs. of water, it seems to me I need nearly 26 cubic feet of volume to achieve the stated temperature of 250.316 °F. Pushing the vapor into a smaller area, 4 cubic feet, should require work done on the system, adding to the internal energy. This internal energy should be seen as increased temperature, if the pressure is to be held at the 30 psia stated. I need to look at the superheated vapor tables. Assuming this is all correct, I don't think I understand how to read the superheated table. two sections are headed as follows: 20psia(227.96 °F) and 40psia(267.26 °F). The first two columns are Temp (°F) and specific volume V (cubic feet per molar pound). Do I use these section headings and my 30psia to interpolate my initial temperature? This would give me an initial temperature of 247.61 °F. How does the vapor being in 4 cubic feet play into my initial state? State two pressure should be less, with a larger volume, and the cooling to the final temperature, but I am unclear regarding moving from state one to state two. 40 °F has a specific pressure of 0.122psia in the saturated water table....and a much higher specific volume rating, 2445.1 for saturated vapor, suggesting a higher pressure for the temp to be held constant at the 12 cubic feet of volume allowed. I can determine my internal energy, U of state one, find the difference between that and the U at state two, but this comes after the above. I am not looking for a specific answer to the problem here, but help understanding where I am going awry would be appreciated.