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852                                                                        Appendix H: Dissolved Gases



                               Pressure of point A

                                                p gage

                             Atmospheric pressure              Atmospheric pressure
                                                                                  p gage
                                                          P abs
                                                p atm            Pressure of point A        p atm


                                   P abs =P atm +P gage                            P
                                                                     P abs =P atm +P gage  abs

                              (a)      Datum                   (b)       Datum

            FIGURE H.1  Graphical and algebraic pressure relations for conversions between absolute pressure and gage pressure; pressure at level ‘‘A’’
            is the focus. (a) Positive gage pressure and (b) negative gage pressure (vacuum).


                   3. Substitute NTP values.                         101,325  N    V ¼ n(air)   8:314  N   m    293:15 K
                     For  NTP,  substitute,  T ¼ 08C ¼ 273.15  K,           m 2               K mol
                                   5
                                                     3
                     p ¼ 1.013250   10 Pa, and let V ¼ 1.0 m , with            n(air)      mol air
                                        1
                     R ¼ 8.314510 J K  1  mol , to give                             ¼ 41:57   3
                                                                                V           m
                                                                                                 3
                                                                                          3
                          n          1:013 250   10 Pa
                       h      i                 5                   2. Convert density in mol=m to kg=m , i.e.,
                                             1
                        1:00 m 3  ¼  (8:314510 J K )   (273:15 K)
                                                                                 n(air)
                                                                                        molar density(air)
                                                                                   V
                                                                           r(air) ¼
                     then since n=V is molar density, r(molar, NTP),
                     the result is,                                                   mol air 0:028964 kg air
                                                                               ¼ 41:57
                                                                                        m 3      mol air
                                            mol O 2
                       r(molar, NTP) ¼ 0:000446                                       kg air
                                              m 3                              ¼ 1:20
                                                                                       m 3
                     where r(molar, NTP) is the molar density of gas at
                                 3
                     NTP (mol O 2 =m ).
                                                               H.1.3 DALTON’S LAW
                (b) Convert molar density, r(molar, NTP), to mass dens-
                   ity r(mass, NTP).                           In a mixture of gases, the total pressure in that mixture is the
                   1. Apply chain of conversion equivalents,   sum of the partial pressures of each of the species, i, i.e.,
                                          mol O 2 32:00 g O                            X
                    r(mass, NTP) ¼ 0:0004446             2                          p ¼   p i               (H:3)
                                           m 2    mol O 2                               n
                                         gO
                              ¼ 0:014227    2                  where p i is the partial pressure of gas i (Pa).
                                         m 3
                                                                  The partial pressure is also proportional to the mole frac-
                                      gO 2                     tion of gas, i,
                              ¼ 14:23
                                       L
                                                                                        n i
              Comments                                                              p i ¼  n  p             (H:4)
              The above calculations illustrate the methodology to cal-
              culate densities of gases.                       Since the volume, V, is common for all gases in the mixture, it
                                                               follows that the sum of the mole fractions is 1, i.e.,
              Example H.3 Calculate the Density of Air at Sea                   n 1  n 2      X  n i
                                                                                              1
              Level at 208C                                                  1 ¼  n  þ  n  þ     ¼  n       (H:5)
                                                                                               i
                 1. Apply the ideal gas law for the conditions stated,  Understanding how to apply Dalton’s law and its variations is
                   i.e.,
                                                               useful in dealing with mixtures of gases. Example H.4 illus-
                                                               trates how to apply Dalton’s law to determine the partial
                               pV ¼ n(air)RT           (H:2)   pressure of oxygen.
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