Graham’s Law of Diffusion & Gas Eudiometry

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:      
rNH3 / rHCl = (MHCl / MNH3)1/2  

= (36.5 / 17)1/2 = 1.46 or rNH3 = 1.46 rHCl  

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

Reagesnt 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  

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   

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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