Gas Eudiometry

The relationship amongst gases, when they react with one another, is governed  by two laws, namely Gay-Lussac law and Avogadro’s law.

Gaseous reactions for investigation purposes are studied in a closed graduated tube open at one end and the other closed end of which is provided with platinum terminals for the passage of electricity through the mixture of gases. Such a tube is known as Eudiometer tube and hence the name Eudiometry also used for Gas analysis.

During Gas analysis, the Eudiometer tube filled with mercury is inverted over a trough containing mercury. A known volume of the gas or gaseous mixture to be studied is next introduced, which displaces an equivalent amount of mercury. Next a known excess of oxygen is introduced and the electric spark is passed, whereby the combustible material gets oxidised.

The volumes of carbon dioxide, water vapour or other gaseous products of combustion are next determined by absorbing them in suitable reagents.

For example, the volume of CO2 is determined by absorption in KOH solution and that of excess of oxygen in an alkaline solution of pyrogallol.

Water vapour produced during the reaction can be determined by noting contraction in volume caused due to cooling, as by cooling the steam formed during combustion forms liquid (water) which occupies a negligible volume as compared to the volumes of the gases considered.

The excess of oxygen left after the combustion is also determined by difference if other gases formed during combustion have already been determined.

From the data thus collected a number of useful conclusions regarding reactions amongst gases can be drawn.

(a) Volume-volume relationship amongst Gases or simple Gaseous reactions.

(b) Composition of Gaseous mixtures.

(c) Molecular formulae of Gases.

(d) Molecular formulae of Gaseous Hydrocarbons.

The various reagents  used for absorbing different gases are

O3 turpentine oil

O → alkaline pyrogallol

NO → FeSO4 solution

CO2,SO2 → alkali solution (NaOH, KOH, Ca(OH)2, HOCH2CH2NH2, etc.)

NH3 → acid solution or CuSO4 solution

Cl2  water

Equation for combustion of hydrocarbons.

CxHy + (x+y/4)O2 → xCO2 +(y/2) H2O

General Assumptions: In all problems, it is assumed that the sparking occurs at room temperature. This implies that water formed would be in liquid state and  that nitrogen gas is inert towards oxidation.

Illustration :   A gaseous hydrocarbon requires 6 times its own volume of O2 for complete oxidation and produces 4 times its volume of CO2 . What is its formula ?

Solution: The balanced equation for combustion

Cxy + (x + y/4)O2 → xCO2 + (y/2)H2O

1 volume  (x + y/4) volume

∴ (x + y/4)= 6 (by equation)

or 4x + y = 24             …(1)

Again x = 4 since evolved CO2 is 4 times that of hydrocarbon

∴ 16 + y = 24 or y = 8

∴ formula of hydrocarbon C4H8

Also Read:

The Gas Laws
Combined Gas Law
The Kinetics Theory of Gases
Graham’s Law of Effusion & Diffusion
Imperfect or Real Gases
Real Gases and the van der Waals Equation
Vander Waals equation in different forms
Mean Free Path

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