4.  Equations must be dimensionally consistent. What does this mean? Let us take the eq p = gh + p 0 Is it correct ? How do we make it correct?  Same for units, each term must have the same unit(s) as all other terms

5.  The mole unit  What is 1 mole of Na ? Units? Are we using grams for mass ? or Kg? or lb? Then we call the mole by the mass we use  What is the mass of lbmole of H 2 SO 4 ? 98 lb H 2 SO 4 = 44.45 Kg  How many gms in 6 gmoles of H 3 PO 4 ? How many moles of H are available? Of P ? of O ?

6.  100 gmoles of O 2 = gms = lbmoles O 2 = 438 g H 3 PO 4 = 18 gmol H = 24 gmol O

7. Density, Specific gravity, sp. Volume, mole fraction, mass fraction, volume fraction  = mass/unit vol sp volume= 1/  = in units ?  it is volume/unit mass ( vol of one unit of mass of material. sp. gravity = density A / density ref Does  change with Temp? YES

8.  We need to specify the temp used for A and for the ref. Example sp. gr = 0.73 20 o What does this mean ? 4 o

9. In the Petroleum Industry sp gr is measured in o API o API = 141.5. sp gr 60 o 60 o EXAMPLE (mass fr., mole fr and Average MWt ) What is the avg MWt of a gas mixture that contains: Mass, Kg H 2 3 CH 4 8 C 2 H 4 56 Total 67 - 131.5

10. = 16.75 kg/kgmol DO YOU HAVE QUESTIONS ?

11.  Concentrations of solutions, mass/unit vol, PPM, molar, molal, etc  Concentration is : Qty of solute per specified amount of solvent or solution How do we specify concentration? 1.Mass of solute per unit volume solution For example 0.10 kgNaCl/L soln.

12. 2.Moles of solute per unit volume solution.For example : 0.10 kgmolNaCl/L soln 3.Molarity (mol solute/L soln) 4.Molality (mol solute/Kg solvent) 5.For dilute soln : we use PPM (parts per million).

13. . Q1: 700 mg salt /m 3 water is PPM 0.7 PPM Note: total mass of solution = 1000 kg+ 700 mg  1000 kg

14. FLOW RATES How do we measure the flow rate of liquid in the pipe ? L/s (Volumetric flow rate) Kg/s (Mass flow rate) How do we change from one to the other? We use the density  How do we calc the velocity ? Example: 40% NaCl soln is flowing at 10L/min in a pipe. If the sp gr of the soln is 1.4,  calculate the concentration of NaCl in Kg/L  calculate the flow rate of soln in kgmol soln/min

15. How do we calc the velocity ? Example: 40% NaCl soln is flowing at 10L/min in a pipe. If the sp gr of the soln is 1.4, then  calculate the concentration of NaCl in Kg/L  calculate the flow rate of soln in kgmol soln/min Q = A. v

16. 0.56

17. Example A gas mixture contains : 40% N 2, 45% O 2 and 15%CH 4. This mixture is hazardous, so to make it safe, it was diluted by adding equal amount of N 2 to it. What is the analysis of the new safe mixture ?

18. + Equal amount of N 2  ? How do we solve? What should we start with? 40 mol% N 2 45 mol% O 2 15 mol% CH 4 ? mol% N 2 ? mol% O 2 ? mol% CH 4

19. BASIS: 1000 gmoles of starting gas mixture BASIS: 100 gmoles of starting gas mixture BASIS: 1 lbmole of starting gas mixture BASIS: 50 kgmoles of starting gas mixture

20. BASIS: 100 gmoles of starting gas mixture 40 gmol N 2 45 gmol O 2 15 gmol CH 4 + 40 gmol N 2  80 gmol N 2 45 gmol O 2 15 gmol CH 4 140gmoles % 80 gmol N 2 57 45 gmol O 2 32 15 gmol CH 4 11 140gmoles 100%

21. SOLUTION How do we solve? What should we start with?

22. SOLUTION Basis 100 kg of the solution, thus kg NaCl 5 KCl 10 HCl 5 H 2 SO 4 soln80 100 H 2 SO 4 solution (water + acid ) How much H 2 SO 4 and how much water in the solution ? Basis: ?

23. Let us change the molality to mass % H 2 SO 4 Molality = mol H 2 SO 4 /Kg water  If we have 1 Kg water, there is 1.5 gmol of H 2 SO 4 Therefore, in 1 Kg water there is 1.5 gmol x 98g gmol = 147 g H 2 SO 4 1.0 kg water 1.5 gmol acid fraction 1000 g water 0.128 147 g acid 0.872 1147 1.000 Basis: ?

24. kg NaCl 5 KCl 10 HCl 5 H 2 SO 4 soln80 100 H 2 SO 4 80x 0.128 =10.24 kg H 2 O 80x 0.872 =69.76 kg 80.00 kg How much H 2 SO 4 and how much water in the solution ?

25. THEREFORE,