Moleculuar Theory of gases 

The behavior of gases when subjected to temperature and pressure changes can be expressed in two simple laws i.e Boyle‘s law and Charles law. These laws apply to ideal gases (i.e. gases with negligible intermolecular forces of attraction and negligible volume)


Boyle’s law
It states that the volume of a given mass of a gas is inversely proportional to its pressure at constant temperature.
Mathematically,

image 160
image 161
image 162
image 163

Gay-Lussac’s law
It‘s the third law describing behavior of gases when involved in chemical reactions
The law states that; when gases react, they do so in volumes which bear a simple ratio to one another and to the volume of the product if gaseous provided the temperature and pressure remains constant.
Gay-Lussac studied chemical reactions between gases and noticed that there is always a very simple ratio between the volumes of gases that react together. For instance

image 164
image 165
image 166
  1. 50 cm3 of oxygen were added to a 40 cm3 mixture of hydrogen and nitrogen. After explosion and cooling to the original temperature,the residual gas occupied 45 cm3. What was the percentage of oxygen in the original mixture.(Answer is 75%)
  2. Calculate the volume of oxygen required for the complete combustion of 200 cm3 of ethane (C2H6). What was the volume of gas product formed.
    Avogadro’s law
    The law states that equal volume of gases at the same temperature and pressure contains the same number of molecules.
    Avogadro‘s law gives an interpretation to Gay-Lussac‘s law in terms of molecules of gases. Consider the following examples
image 167
image 168
image 169
image 170
image 171
image 172
image 173
image 174
image 175
image 176
image 177

b) Calculate the minimum volume of hydrogen at stp needed to reduce the oxide
c) Calculate the volume of carbondioxide at stp produced when the carbonate was completely decomposed.

  1. Calculate the mass of sulphur deposited when 8.4 dm3 of chlorine oxidizes hydrogen sulphide.

Volumetric analysis
In an investigation to determine the nature of matter in a substance, a chemist focuses on two main questions:
a) What are the components of the substance?
b) What amount of each component is present in the substance?
In an attempt to answer question (a) and (b), a chemist carries out qualitative and quantitative analysis respectively. Most of the reactions which a chemist carries out take place in solution.

image 178

If we are to determine the volume of A needed to completely react with a given amount or volume of B, the answer is provided practically through volumetric analysis.
In volumetric analysis, quantities of substances (often acids or alkalis) are estimated by analytical processes involving measurements of volumes of solutions using pipettes, burettes and measuring cylinders (for approximate measurement). Weighing may also be involved. Most of the work in volumetric analysis is based upon molar (M) solutions.


Standard and Molar solutions
A standard solution is a solution of known concentration. Examples of standard solutions are; solution containing 12g of sodium chloride in one litre of a solution; a solution containing 2 moles of solute in 1dm3 e.t.c. The substance that is used to prepare a standard solution is known as a primary standard.

A Molar solution is a solution that contains one mole of a substance in a solution of one litre. In other words, it is a solution containing one mole of solute in one litre.
Other related terms are;


Concentration; this is the amount of solutes in a given volume of solution.

Molarity; this is the number of moles of solute in one litre of a solution. The unit is mol/dm3 or mol/l. The molarity of a solution is commonly denoted by letter M. E.g. 0.2M NaOH which mean 1 litre of a solution containing 0.2 moles of NaOH.
1 litre(1 l)=1cubic decimetre (1dm3) = 1000 cubic centimetre (1000cm3)
Calculations on molarity and masses
Examples

image 179
image 180

When the concentration of a solutr in grams per litre and the RMM are known then the molarity can be calculated from the above expression.
N.B. The use of formula is not so much recommended and workings should be from first principle.

  1. Calculate the mass of the named substance needed to make
    a) 0.1 dm3 of 2M sodium sulphate solution
    b) 1 l of 0.25M sodium hydroxide solution
    c) 25cm3 of 0.1M potassium carbonate solution
    d) 500cm3 of 0.05M sodium carbonate solution
image 181
image 182

Calculating number of moles of ions in standard solutions
Examples

  1. Calculate the number of moles of hydrogen ions in 25cm3 of a 0.2 M sulphuric acid.
  2. Calculate the number of moles of potassium ions in 35cm3 of 0.12 M potassium carbonate solution.
image 183