Graham’s Law of Diffusion & Gas Eudiometry

Table of Content

Graham’s Law of Diffusion / Effusion
- Effect of Pressure on State of Diffusion
Gas Eudiometry
Related Resources

Graham’s Law of Diffusion / Effusion

Diffusion in the ability of a gas to spread and occupy the whole available volume irrespective of other gases present in the container.
Effusion is the process by which a gas escapes from one chamber of a vessel through a small opening or an orifice.

$R=\frac{Volume\ Diffused}{Time\ Taken}=\frac{V}{T}$
Thomas Graham in 1831 proposed the law of gaseous diffusion.
The law states that under similar conditions of temperature and pressure, the rates of diffusion of gases are inversely proportional to the square roots of their densities.
Mathematically
r ∝ 1 / √d
where r is the rate of diffusion and d is the density of the gas.
Now, if there are two gases A and B having r₁ and r₂ as their rates of diffusion and d₁ and d₂ their densities respectively. Then

r₁ ∝ $\frac{1}{\sqrt{d_1}}$ ₁
and
r₂ ∝ $\frac{1}{\sqrt{d_2}}$
or
, $\frac{r_1}{r_2}=\sqrt{\frac{d_2}{d_1}} \ (at\ same\ T\ and\ P)$

$\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}$

Here M₁ and M₂ are the molecular masses of the gases having densities d1 and d2 respectively.

Graham’s law of diffusion also holds good for effusion.

Effect of Pressure on State of Diffusion

The rate of diffusion (r) of a gas at constant temperature is directly preoperational to its pressure

r ∝ P at constant temperature

r ∝ 1 / √d at constant temperature

$\frac{r_1}{r_2}=\frac{P_2}{P_1}=\sqrt{\frac{M_2}{M_1}}$

Refer to the following video for Dalton’s Law of Partial pressure & Ghraham’s Law of Diffusion

Example 1:

Question:

Which of the two gases ammonia and hydrogen chloride will diffuse faster and by what factor?

Solution:
r_NH3 / r_HCl = (M_HCl / M_NH3)^1/2

= (36.5 / 17)^1/2 = 1.46 or r_NH3 = 1.46 r_HCl

Thus ammonia will diffuse 1.46 times faster than hydrogen chloride gas

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.

Volume-volume relationship amongst Gases or simple Gaseous reactions.
Composition of Gaseous mixtures.
Molecular formulae of Gases.
Molecular formulae of Gaseous Hydrocarbons.

The various reagents used for absorbing different gases are

Gas	Reagent Used
O₃	Turpentine oil
O₂	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.
$C_xH_y \ + (x+\frac{y}{4})O_2 \rightarrow XCO_2 +\frac{y}{2}H_2O$

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.

Example 2:

Question:
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

C_xH_y + (x + y/4) O₂ → xCO₂ + y/2 H₂O

1 volume (x + y/4) volume x + y/4 = 6 (by equation)

or 4x + y = 24

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

16 + y = 24 or y = 8 formula of hydrocarbon C₄H₈

Question 1: According to Ghraham’s Law of Diffusion

a. $\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}$

b. $\frac{P_2}{P_1}=\sqrt{\frac{M_2}{M_1}}$

c. $\frac{M_2}{P_1}=\sqrt{\frac{P_2}{M_1}}$

d. $\frac{M_1}{M_2}=\sqrt{\frac{r_2}{r_1}}$

Question 2: During Gas analysis, the Eudiometer tube filled with

a. hydrogen

b. oxygen

c. carbon dioxide

d. mercury

Question 3: Graham’s law of diffusion also holds good for

a. viscosity

b. osmosis

c. effusion

d. expansion

Question 4: Which of the following reagents is used for absorbing oxygen gas?

a. Water

b.Turpentine oil

c. FeSO₄ solution

d. Alkaline pyrogallol

Q.1	Q.2	Q.3	Q.4
a	d	c	d

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