Derivation of Vander Waals  equation of State

Derivation of Vander Waals  equation of State

or
Equation of State for Real (Imperfect) Gases

Many attempts have been made in order to get an equation of state viz. applicable to real gases.
A number of equations of state have been suggested to describe the P-V-T relationship in case of real gases.
The Van-der-Waals equation is the earliest and best known of them.

In 1873, Van der Walls proposed an equation of state for real gas (imperfect gas), viz. Known as vanderwaals equation.

He modified the ideal gas equation (PV=nRT) by suggesting the following points:
• Gas molecules were not mass points but behave like rigid spheres.
• There exists intermolecular forces of attraction between them.

In order to introduce these points the following corrections were made in the ideal gas equation :

1. Correction for the volume of gas molecule

In case of ideal gas , we consider ( assume) the molecules of gas as mass points that’s why we neglect the volume of gas molecules in comparison to the volume of container(V).

But in case of real gases, the gas molecules are considered as rigid spheres that’s why we can not neglect the volume of gas molecules.
So in order to get the ideal volume which is compressible, van der Walls suggested that a correction term nb should be substracted from the total volume V.
∴ Corrected Volume= V – nb

Calculation of b

Consider 2 gas molecules as unpenetrable and incompressible spheres, each having radius r.

Calculation of Exclude volume b
Thus the center of these two spheres can not approach each other more closely than distance 2r.
For this pair of molecules, therefore, a sphere of radius 2r , having volume = \( \frac{4}{3} \) \(𝝅(2r)^3\)
This is known as excluded Volume for the pair of molecules.
∴ Excluded Volume per molecule = half of \( \frac{4}{3} \) \(𝝅(2r)^3\)
= \( \frac{1}{2} \) × \( \frac{4}{3} \) \(𝝅(2r)^3\)

As the Actual volume of one gas molecule having radius r = \( \frac{4}{3} \) \(𝝅(r)^3\)

Excluded Volume per molecule= 4 × Actual volume of one gas molecule

Co-volume (b):  The excluded Volume per mole is known as the co-volume. 
Excluded Volume per mole =  NA × Excluded Volume per molecule
= NA × 4 × \( \frac{4}{3} \) \(𝝅(r)^3\)
⇒b=NA × 4 × \( \frac{4}{3} \) \(𝝅(r)^3\)

where NA = Avagadro number

Thus the compressibe volume per mole of gas = total volume – Excluded Volume per mole
= V – b

If the volume V of gas contains n moles then the excluded Volume = nb

Ideal volume which is compressible = V – nb
In this way Van der waal made correction in volume term by replacing the volume V by ideal compressible volume V-nb

2. Correction due to Intermolecular Forces of Attraction

In case of ideal gas equation, it was assumed that there are no intermolecular forces of attraction between the molecules.
Actually it was not so.×××
Because pressure will affected due to these forces of attraction between the molecules.(As Pressure = force/Area)
• When a molecule approaches the walls of the container, it experiences attractive forces from the bulk of molecules behind it.
Thus In case of real gases the molecules strike the wall with a lower velocity and hence exert a lower pressure than they exert in case of ideal gases.
• Thus Pideal > Preal  (always)
So van der Waal take into account the effect of intermolecular forces of attraction in VWL equation.

Thus In order to get the ideal pressure vanderwaal made correction in the pressure term by adding a correction factor p’ to the pressure of the gas.
p’ is also called internal pressure of the gas
Corrected pressure = P + p’

Calculation of correction factor p’

Calculstion of Correction due to Intermolecular Forces of Attraction

The total inward attractive pull on the molecules which accounts for p’ is directly proportional to ρ2

∵ There are two types of molecules A-type and B-type as shown in fig.
A-type is that single molecule which is about to strike the wall and B-type molecules are those which strike the wall at any instant.
Thus force of attraction exerted on both of these depends upon the number of molecules per unit volume in the bulk of the gas.
i.e., depends directly upon the density of the gas

i.e.,  p’ ∝ ρ2
As we know, density ∝  \( \frac{1}{V} \)
if there are n moles of gas occupying volume V then   ρ   \( \frac{n}{V} \)
p’  \( \frac{n^2}{V^2} \)
p’= \( \frac{an^2}{V^2} \)

Thus,Ideal pressure or Corrected pressure
= P + p’
= P+ \( \frac{an^2}{V^2} \)

Hence van der waals equation for n mole of real gas becomes:

\( \fbox{\( \left( {P + \frac{{an^2 }}{{V^2 }}} \right)\left( {V – nb} \right) = nRT \)} \)

where a & b are van der Walls constants.

Units of Vanderwaal constants a and b

As p’= \( \frac{an^2}{V^2} \)
a= p’ \( \frac{V^2}{n^2} \)
Unit of a = atm L²mol-2

As regards b, it is the incompressible volume per mole of a gas
Unit of b = litre/mole

Significance of Vanderwaal constants a and b

Significance of a :

It is the measure of the magnitude of the intermolecular forces of attraction within the gas.

a intermolecular forces of attraction

i.e., greater the value of a, the greater is the intermolecular forces of attraction among the gas molecules.
more internal pressure generate
more is the value of p’

Value of a for H(0.024) and He(0.246) are very small because van der Walls forces are very weak in these gases.
Value of a for NH3(4.17), CO2(3.59), SO2(6.71) are very high. Thus van der Walls forces in these gases are very strong.
As the value of a increases, intermolecular forces of attraction increases; and the liquification tendency also increases, i.e., Greater the value of a, greater is the ease with which a gas can be liquified

 

Significance of b :

b represents the Excluded Volume per mole of the gas. It is also called co-volume
It is the measure of effective size of the gas molecules.

b ∝ size of gas molecules

i.e., more is the size of gas molecule, more will be the value of b, i.e., more volume will have to be excluded

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