## Introduction to Van Der Waals Equation

In 1873, Johannes Diderik Van Der Waals derived the Van Der Waals equation. The equation is considered as the updated version of the ideal gas law that states that there are some point masses present in gases that undergo perfectly elastic collisions. However, the real gas law is incapable of explaining the behaviour of real gases. Due to this reason, the Van Der Waals equation was derived to define the physical state of a real gas.

The full name of Van Der Waals was Johannes Diderik van der Waals. He was a Dutch thermodynamicist and theoretical Physicist, who had won the Nobel prize in physics in the year 1910. And as you may have already understood by now, the Van der Waals equation is named after him, because he was the one who gave it.

The Van der Waals equation is the equation of state of thermodynamics, it is based on the theory that suggests that the particles with non-zero volumes, makes the fluids. Van der Waals derived it in 1873, based on the traditional set of derivations, which are also derived from the Van der Waals. The equation explains the condensation of the gases to the liquid phase. The equation was so impactful that another famous scientist of the time James Maxwell said that the name of Van Der Waals is going to be famous in molecular science.

Also, if you like to learn about the Van der Waals forces, then follow this link.

### What is Van Der Waals Equation?

Van Der Waals equation is an equation that is used to relate the relationship existing between the pressure, volume, temperature, and amount of real gases. For a real gas containing ‘n’ moles, the real gas equation derivation is as follows.

(P + \[\left ( \frac{an^{2}}{V^{2}} \right )\] ) V−nb = nRT

Where,

P is the pressure,

V is the volume,

T is the temperature,

n is the number of moles of gases,

‘a’ and ‘b’ are the constants that are specific to each gas.

The equation can be written as:-

1. Cube power of volume:

V³ - b+(RT/P) V² + ( a / P) V - ab / P = 0

1. Reduced equation (Law of corresponding states) in terms of critical constants:

( 𝜋 + 3 / φ²) ( 3φ- 1 ) = 8τ

Where:

𝜋 = \[\frac{P}{P_{c}}\]

φ = \[\frac{V}{V_{c}}\]

τ = \[\frac{T}{T_{c}}\]

The units of Van Der Waals constants are:

For unit ‘a’ = atm lit² mol⁻²

For unit ‘b’ = litre mol⁻¹

### Derivation of Van Der Waals Equation For Real Gases

It is easy to derive Van Der Waals equation for real gases but only if the right steps are followed. Any mistakes committed in the process to deduce Van Der Waals equation of state can be crucial and affect the whole process.

Let us discuss the process to derive the Van Der Waals gas equation.

In the case of a real gas when students are using Van Der Waals equation, the volume of a real gas is considered as (Vm - b), where b can be considered as the volume occupied by per mole.

Therefore, when the ideal gas law gets substituted with V = Vm - b, it is given as :

P(Vm - b) = nRT

Due to the presence of intermolecular attraction P was modified as follows.

( \[\frac{P+a}{V^{2}}\] ) ( Vm - b) = RT

( \[\frac{P+an^{2}}{V^{2}}\] ) ( V - nb) = nRT

Where,

Vm: molar volume of the gas

R: universal gas constant

T: temperature

P: pressure

V: volume

Thus, it is possible to reduce Van Der Waals equation to ideal gas law as PVm = RT.

### Van Der Waals Derivation For One Mole of Gas

To derive the Van Der Waals equation or to deduce the Van Der Waal equation of state for one mole of gas can be turned into an easy process if the right steps are followed.

The steps for derivation of the real gas equation for one gas are as follows.

p = RT / V = (RT / v) p = RT / Vm - b

C = Na - Vm (proportionality between particle surface and number density)

a'C² = a' ( Na / Vm)² = a / Vm²

p = RT / (Vm - b) - a / Vm² => [ [ p + (a / Vm²)] ] Vm−b = RT

[ [ p + ( n²a / V²) ] ] V−nb = nRT

( substituting nVm = V )

### Derivation of Real Gas Equation When Applied to Compressible Fluids

Van Der Waals equation derivation, when applied to compressible fluids, can be understood if the concepts are clear.

Compressible fluids like polymers have fluctuating specific volumes and this can be expressed as follows.

(p + A)(V - B) = CT

Where,

P: pressure

V: Specific volume

T: Temperature

A, B, C: Parameters

All this above information is used to derive the Van Der Waals equation of state.

### Merits and Demerits of Van Der Waals Equation of State

Merits:

It can predict the behaviour of gas much better and accurately than the ideal gas equation.

It is also applicable to fluids in spite of gases.

The arrangement is made in a manner of cubic equation in volume. The cubic equation can give three volumes which can be used for calculating the volume at and below the critical temperatures.

Demerits:

It can only get accurate answers for real gases which are above the critical temperature.

Below critical temperature results also get accepted.

In the transition phase of gas, the equation is a failure.

### Conclusion

More importantly, the Van Der Waals equation tends to take into consideration the molecular size and the molecular interaction forces which can be attractive or repulsive forces. Sometimes it is also known as Van Der Waals equation of state. In this article, students will learn how to derive the Van Der Waals equation of state.

## FAQs on Van Der Waals Equation Derivation

**1. What Are A and B Constants in the Van Der Waals Equation?**

The Van Der Waals equation is derived to make an approach to the ideal gas law PV = nRT because the value of these constants stands at zero. The constant A is meant for providing a correction for the intermolecular forces exerted. The constant B is meant for correcting the finite molecular size. The value of constant B is equal to the volume of one mole of the atoms or the molecules. Since we can use constant B for calculating molecular volume, thus it can be used to calculate the radius of an atom or a molecule.

**2. What is the Ideal Gas Law?**

The ideal gas law also expressed as a general gas equation can be considered as the equation of the state of a hypothetical ideal gas. It is used to examine the behaviour of gases in different conditions, but it has several limitations. The ideal gas law does not give any explanation on whether a gas heats or cools during the process of expansion or compression. Most of the ideal gases do not cause any change in temperature except for some of them. It was first derived in 1834 by Benoit Paul Emile Clapeyron.

**3. Is understanding Van der Waals equation difficult?**

No, understanding the Van der Waals equation is not difficult at all. First of all, you have to note the fact that, the inclusion of all the topics in the syllabus is done after the careful consideration of many factors, and the intellectual capacity of the students is one of them, and since the topic is included in the syllabus, it implies that it is of the level that you can grasp. Also, there is one more thing, if you learn the topic of the Van der Waals equation carefully, then it automatically becomes easy.

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