1. The problem statement, all variables and given/known data A 1 m3 rigid tank has propane at 100 kPa and 300 K. The tank is connected to another 0.5 m3rigid tank which has propane at 250 kPa and 400 K by a ball valve. The valve is opened and both tanks come to a uniform state at 325 K. Calculate the change in internal energy and enthalpy of propane in each tank. 2. Relevant equations PV=ZmR(specific)T U = u*m [internal energy = specific internal energy * mass] H = h*m [same as above, but for enthalpy] ΔU = U2 - U1 ΔH = H2 - H1 U = Q - W [energy = heat transfer - work] W = ∫Pdv 3. The attempt at a solution *note - work is irrelevant here because the tanks are rigid, so no forced volume change So this problem had other sub problems where I found the final pressure of the opened system, the initial Z factors, and the initial separated masses of propane. I labeled this system as 3 parts - Tank A (the m^3 tank), Tank B (the .5 m^3), and System O (the newly connected, opened valve system). Let's just assume for now that my calculations up to this point are correct, I just want to verify if I'm approaching the energy/enthalpy part correctly. Po = 139.87 KPa Ma = 1.785 Kg Mb = 1.69 Kg So I went online and found a steam table for superheated propane gas after double checking that the conditions for both tanks fall well within the gas phase on propane's phase diagram data. I found the following data. Initial specific internal energy of Tank A = ua1 = 478.2 KJ/Kg Initial specific enthalpy of Tank A = ha1 = 534 KJ/Kg Initial specific internal energy of Tank B = ub1 = 644.8 KJ/Kg Initial specific enthalpy of Tank B = hb1 = 717 KJ/Kg *note - the steam tables didn't have the exact conditions I was looking for for a few numbers, but I saw that at constant temperature, even very huge pressure increases didn't have much effect on specific energy/enthalpy properties. So I just used the closest numbers that made sense to use. If this approach is correct then I can just redo it with more exact data via interpolation. So my approach was to find the initial internal energy and enthalpy of a tank, and think of these as properties of the molecules in that tank. Then when the tanks are connected, find the final internal energy and enthalpy of those same molecules (so that same mass of propane), at the final pressure and temperature of the newly connected "System O". I would then find the difference, and this would give me the change in energy/enthalpy that the mass from this tank experienced. Ua1 = ua1 x Ma = 478.2 KJ/Kg x 1.785 Kg = 853.58 KJ Ua2 = 509 KJ/Kg x 1.785 Kg = 908.56 KJ Ha1 = ha1 x Ma = 534 KJ/Kg x 1.785 Kg = 953. 19 KJ Ha2 = 569.7 KJ/Kg x 1.785 Kg = 1016.91 KJ Ub1 = u1b x Mb = 644.8 KJ/Kg x 1.69 Kg = 1089.7 KJ Ub2 = 509 KJ/Kg x1.69 Kg = 860.21 KJ Hb1= hb1 x Mb = 717 KJ/Kg x 1.69 Kg = 1211.73 KJ Hb2 = 569.7 KJ/Kg x 1.69 Kg = 962.79 KJ ΔUa = Ua2 - Ua1 = 54.98 KJ ΔHa = Ha2 - Ha2 = 63.72 KJ (tank A's molecules experienced an increase in temperature and pressure conditions, so this seems reasonable to me) ΔUb = - 226.79 KJ ΔHb = - 248.9 KJ (tank B's molecules experienced a decrease in temperature and pressure conditions, so this seems reasonable to me) I'd like to know if there are any immediate, glaring issues with how I thought of this conceptually. Does this approach look sensible / correct? Are there any issues? If you see any issues with my numbers and just point them out that'd be a nice bonus. Any and all help is greatly appreciated, thanks a lot.