--%>

Avogadro's hypothesis Law Principle

Avogadro's hypothesis Law Principle- Berzelius, a chemist tried to correlate Dalton's atomic theory & Gay-Lussac's Law of gaseous volumes. According to his Berzelius hypothesis Equal volumes of all gases under similar conditions of temperature & pressure contain equal number of atoms e.g.

Hydrogen (1 vol) + Chlorine (1 vol)->HCl (2 vol)

Acc to Berzelius hypothesis:

Hydrogen (1/2 atom) + Chlorine (1/2 atom)->HCl (1 compound atom)

But this is indirect conflict of Dalton's atomic theory, so it was rejected.

So a new hypothesis was given by Avogardo.

According to him, An atom is a smallest particle of an element which can take part in a chemical reaction which may or may not be capable of independent existence.

molecule is the smallest particle of an element or of a compound which have an independent existence. So the smallest particle of a gas is a molecule not an atom, so the volume of gas must be related to the number of molecules rather than atoms.

According to Avogrado's Hypothesis-Equal volume of all gases under similar conditions of temperature & pressure contain equal number of molecules. This is able to explain all the gaseous reactions & now known as Avogrado's Law or Avogrado's principle.

For example-

Hydrogen (1 vol) + Chlorine (1 vol)->HCl (2 vol)

By Avogrado's hypothesis:

n molecule+n molecule gives 2n molecule

1/2molecule of both [Hydrogen + Chlorine] ->HCl (1 molecule)

 Applications of this hypothesis-

(1)In the calculation of atomicity of elementary gases-Atomicity is defined as the number of atoms of the element present in one molecule of the substance e.g. atomicity of N2 is two & O3 is three.

(2)To find the relationship between molecular mass & vapour density of gas-(relative density)

Vapour density of gas=Density of gas/density of hydrogen

           =Mass of [certain vol of gas/same volume of H2] at STP

            =Mass of [n molecule of gas/ n molecule of H2] at STP

            =Mass of [1 molecule of gas/ 1 molecule of H2] at STP

 

                          Vapour density=Molecular Mass/2

(3)To find the relationship between mass & volume of gas-As the

Molecular Mass=Vapour density x 2

Or Molecular Mass=Mass of 22.4 L of gas at STP

Thus 22.4 L of any gas at STP weight is equal to the molecular mass of the gas expressed in grams which is called Gram-Molecular Volume Law (G.M.V.).

   Related Questions in Chemistry

  • Q : Donnan Membrane Equilibria The electric

    The electric charge acquired by macromolecules affects the equilibrium set up across a semipermeable membrane.Laboratory studies of macromolecule solutions as in osmotic pressure and dialysis studies confine the macromolecules to one compartment while allo

  • Q : Problem on reversible process a. For a

    a. For a reversible process involving ideal gases in a closed system, Illustrate thatΔS = Cv ln(T2/T1) for a constant volume process ΔS = Cp ln(T2/T1) for a constant pressu

  • Q : Benefits of soapy detergents over the

    What are the benefits of soapy detergents over the soap less detergents? Briefly state the benefits?

  • Q : Soluation of Ideal Gas Law problems

    Explain the method, how do you solve Ideal Gas Law problems?

  • Q : Cations Explain how dissolving the

    Explain how dissolving the Group IV carbonate precipitate with 6M CH3COOH, followed by the addition of extra acetic acid.

  • Q : Formula of diesel Write a short note on

    Write a short note on the formula of diesel, petrol and also CNG?

  • Q : Real vapour pressure Choose the right

    Choose the right answer from following. The pressure under which liquid and vapour can coexist at equilibrium is called the : (a) Limiting vapour pressure (b) Real vapour pressure (c) Normal vapour pressure (d) Saturated vapour pressure

  • Q : Problems related to entropy change A)

    A) Two compartments each of 1 m3 capacity are joined by a valve and insulated from the surroundings and from one another. One compartment has saturated steam at 683.6 kPa and the other contains steam at the same temperature but at a pressure of 101.3 kPa. T

  • Q : Group Cations Explain how dissolving

    Explain how dissolving the Group IV carbonate precipitate with 6M CH3COOH, followed by the addition of extra acetic acid, establishes a buffer with a pH of approximately

  • Q : Mole fraction of hydrogen Give me

    Give me answer of this question. In a mixture of 1 gm H2 and 8 gm O2 , the mole fraction of hydrogen is: (a) 0.667 (b) 0.5 (c) 0.33 (d) None of these