In the last post we left some questions unanswered. Questions like –

How does one relate the number of particles to volume and mass of the object?

What is Avogadro’s hypothesis?

Let us find answers to these questions in this post.

**Avogadro’s hypothesis**

After studying Gay Lussac’s law of combining volumes of gases, Amedeo Avogadro put forth his own hypothesis , which stated –

“**Equal volumes of all gases under the same conditions of temperature and pressure contain the same number of molecules**.

or* Equal number of molecules of all gases occupy the same volume under the same conditions of temperature and pressure.*“

At STP conditions (* STP = standard temperature and pressure – 0°C and 1 bar pressure*) ,

One mole of any gas occupies 22.4 dm^{3}i.e 22.4 litres volume.∴22.4 litres of any gas at STP has 6.023 × 10^{23}molecules of that gas.

Volume of 1 mole of any gas is called its MOLAR VOLUME.

*e.g.*– If we take 22.4 litres of nitrogen gas in a container, it will contain **6.023 × 10 ^{23} molecules** of N

_{2}gas.

If the amount of gas taken increases then its volume increases too and if the amount of gas decreases, the volume is going to decrease subsequently. Thus, we can say that ,

**V α n** where,

V⇒ Volume of the gas

n ⇒ no.of moles.

Introducing a proportionality constant , we get ,

**V = k n**. (k = constant of proportionality).

∴V/n = k.

Thus, we can state, **(V _{1} / n_{1} ) = (V_{2}/n_{2}).**

Mathematical expression for Avagadro’s law – (V_{1}/ n_{1}) = (V_{2}/n_{2}).

Now isn’t this obvious? Consider a balloon. When you fill in more air , the balloon will grow in size.

How to explain this in a chemical reaction?

In the above reaction, if 1 mole (or 22.4 lit) of hydrogen is used to react with 1 mole of chlorine, then 2 moles(44.8 lit) of hydrogen chloride gas is formed. If the amount of reactants is doubled, the volume of product is also doubled.

The ratio of volumes of reactant and product gases remains constant – 1:1:2.

**Problem – Calculate the number of molecules in 5.6dm ^{3} of CO_{2} at STP.**

* Solution* –

22.4 dm^{3} of CO_{2} at STP contains 1 mole of that gas.

∴ 5.6 dm^{3} of CO_{2} will contain *(5.6 /22.4) = 0.25* moles of the gas.

Also , 1 mole =** 6.023 × 10 ^{23} molecules**

**.**

∴ 0.25 moles will have 0.25 ×6.023 × 10

^{23}

**= 1.506 × 10**

^{23}molecules.Thus, Avogadro’s hypothesis helps us to relate the number of particles ( in moles) and volume of gases. In the next post we shall see how we can relate the mole to another measurable quantity – mass. Till then ,

Be a perpetual student of life and keep learning….

Good day !