Lesson 1b: changing the amount of solute


Content

The effect of the total amount of solute (NH4NO3, HNO3) on water uptake is examined, and also the implications for gas/aerosol partitioning of varying amounts of HNO3 in the system.


Part 1

First, we look again at an ammonium nitrate (NH4NO3) particle at 99% relative humidity, but this time for a larger amount of salt. Return to the data input page in the right hand browser window, and fill in the form as follows:

1st Calculation
  1. Enter the values and select the options under the following headings:

    Ambient Conditions
    Relative humidity = 0.99 (i.e., 99%).

    Ionic Composition in Moles
    Ammonium = 2.0, Nitrate = 2.0.

    Solid Phases
    There are no entries under this heading.


  2. Click on the "Run" button at the end of the page to do the calculation.
Note:  the above should be entered on the "simple" calculations page of Model III (http://www.aim.env.uea.ac.uk/aim/model3/model3a.php).


Interpreting the Results

The results are similar to those for the first run in Lesson 1a. Looking in the "Aqueous Phase" section, the number of moles and grams of ammonium and nitrate have doubled, since that is what was specified on input.

The water content has also doubled so that the molalities and mole fractions for this run are exactly the same as for the 1.0 mole case. This is because the relative humidity determines the water activity of the aqueous phase - the two quantities are the same in a system at equilibrium - and hence the concentration of the solutes.

In most aqueous atmospheric aerosols, concentration is independent of the amount of solute for a system at fixed relative humidity and temperature. However, surface tension effects start to become important for particles less than about 0.1 × 10-6 m in size . For these small aerosols the equilibrium vapour pressures of water and other volatile constituents are enhanced. This is the Kelvin effect, which is important in the nucleation and growth of cloud particles, for example.

The result we have just described appears very simple. Apart from the size effect noted above it is true for aqueous particles or bulk solutions of any fixed composition. However, in an aerosol system in which gas/aerosol partitioning is occuring, the amount of material present in the system can have an effect. To show this, we will next re-run the HNO3 example from Lesson 1a, but for a smaller amount of material.


Part 2

Repeat the calculation above, but for 1E-7 moles (0.1 × 10-6 mol) of HNO3, by following the instructions in the box below:

2nd Calculation
  1. Enter the values and select the options under the following headings:

    Ambient Conditions
    Relative humidity = 0.99 (i.e., 99%).

    Ionic Composition in Moles
    Hydrogen = 1.0E-7, Nitrate = 1.0E-7.

    Solid Phases
    There are no entries under this heading.


  2. Click on the "Run" button at the end of the page to do the calculation.
Note:  the above should be entered on the "simple" calculations page of Model III (http://www.aim.env.uea.ac.uk/aim/model3/model3a.php).


Interpreting the Results

The aqueous phase has the same 0.2982 mol kg-1 HNO3 molality as was the case for 1 mole of HNO3, and the 0.18612E-4 moles of aqueous phase water is simply a factor of 107 lower than before.

Under the "Gases" heading the equilibrium partial pressure of HNO3 above the liquid solution is unchanged from the 1.0 mole case at 1.805E-8 atm. However, the equivalent number of moles of gas phase HNO3 that would exist in the 1 m3 system volume (0.7379E-6 moles) is greater than the amount we have specified in the system. What does this imply for gas/aerosol partitioning?

The thermodynamic calculations have shown that a pure aqueous HNO3 solution at 298.15 K and 99% relative humidity has a concentration of 0.2982 mol kg-1, and this solution has an equilibrium HNO3 partial pressure of 1.805E-8 atm (0.7379E-6 mol m-3). If we were to place an amount of HNO3 less than 0.7379E-6 moles in a 1 m3 container (at 298.15 K and 99% relative humidity) then the partial pressure of the HNO3 would be less than that over the equilibrium aqueous phase for that relative humidity and temperature: consequently no HNO3 would condense and it would remain as a vapour.

Looking at it another way, if a volume of 0.2982 mol kg-1 aqueous HNO3 containing less than 0.7379E-6 moles were placed in the container then the HNO3 would start to evaporate to the gas phase phase (and liquid water with it, to maintain the same concentration). However, because the amount of HNO3 placed in the system is insufficient to produce the required partial pressure, the liquid phase would keep on evaporating until nothing was left and all the water and HNO3 exist only in the vapour phase.

This simple example is enough to show that the thermodynamic properties and physical state of a gas/aerosol system are not always as simple as for bulk solutions. A later lesson considers this in more detail.


You should now proceed to Lesson 1c, or return to the main page for this lesson.