Van der waals equation of state

Van der Waals equation derivation

To rectify error for in ideal gas law Van der Waals in 1973 introduces two correction terms in the ideal gas law, one for pressure and other for volume.

Because the ideal gas law has been derived from the postulates of the kinetic theory. It is obvious that some of the postulates in this theory need modification. The primary two postulates which need modification.

  1. The molecule was considered as point masses, practically having no volume.
  2. The existence of the forces of attraction between the molecules was ignored.

Thus for the derivation of Van der Waals equation for real gas, Van der Waals introduced these two correction terms. So the ideal gas equation is written as


Volume correction for real gases

Real gas molecules are assumed to be a hard rigid sphere thus the available space for free movement of the molecules becomes less than the original volume.

Let us take the available space for free movement of one-mole real gas,

V_{i}=\left ( V-b \right )

where V = molar volume of real gases
b = volume correction factor
Vi = molar volume of the ideal gases.

Molar volume of real gases

Let us take a gas molecule with radius r and diameter σ = 2r, assuming the gas molecules as a rigid sphere.

So the volume of the body of each gas molecule

volume\, =\, \frac{4}{3}\pi r^{3}

But when two molecules encounter each other the distance between the centers of the two gas molecules would be σ or the diameter of the molecule.

Thus the space indicated by the dashed circle with diameter 2σ will be unavailable for colliding a pair of molecules.

\therefore excluded\, volume\, for\, a\, pair=\frac{4}{3}\pi \sigma ^{3}For\, a\, single\, molecule=\frac{1}{2}\times \frac{4}{3}\pi \sigma ^{3}Thus for one gm-mol of the gas, such excluded volume is the volume correction factor of the Van der Waals equation for one-mole gas.

b = N0 × volume excluded for a single molecule.

\therefore b=N_{0}\times \left (\frac{1}{2}\times \frac{4}{3}\pi \sigma ^{3} \right )=\frac{2}{3}\, N_{0}\, \pi \sigma ^{3}

Again diameter σ = 2r

b=\frac{2}{3}\, N_{0}\, \pi \sigma ^{3}=4\, N_{0}\, \frac{4}{3}\pi r^{3}

Measurement of b helps to calculate the diameter or radius of the gas molecule. So after the volume correction, the equation becomes

P_{i}\left ( V-b \right )=RT

Pressure correction for real gases

The pressure of the gas is developed due to the wall collision of the gas molecules. But due to intermolecular attraction, the colliding molecules will experience an inward pull.

So the pressures exerted by the molecules in real gases will be less than the ideal gases because the ideal gas molecules have no intermolecular attraction.

\therefore P_{i}> P

or, P_{i}=P+P_{a}

where Pa = pressure correction term originating from attractive forces.

Higher the intermolecular attraction in the gas molecules greater is the magnitude of Pa. Thus the pressure depends on both the frequency of molecular collisions with the walls and the impulse exerted by each collision.

Pressure correction factor formula

Therefore the average pressure exerted by the molecules decreased by Pa, which is proportional to the square of the density.

P_{a} \propto\frac{1}{V^{2}}\, becuse\, density\propto\frac{1}{V}

\therefore P_{a}=\frac{a}{V^{2}}

where a = constant for the gas and it measures the attractive force between the molecules.

Thus, P_{i}=P+\frac{a}{V^{2}}

Van der Waals equation for n moles of real gas

Using these two corrections factor Van der Waals equation for one-mole real gas

{\color{Blue} \left (P+\frac{a}{V^{2}} \right )\left ( V-b) \right )=RT}

For n moles real gases, the volume has to change because it is the only extensive property in the equation.

Let V be the volume of n moles gas hence, V = v/n. Putting this term in the Van der Waals equation.

\left \{ P+\frac{a}{\left ( \frac{V}{n} \right )^{2}} \right \}\left ( \frac{V}{n}-b \right )=RT

{\color{Blue} \therefore \left ( P+\frac{an^{2}}{V^{2}} \right )\left ( V-nb \right )=nRT}

Ideal vs real gas pressure

Van der Waals equation,

\left ( P+\frac{an^{2}}{V^{2}} \right )\left ( V-nb \right )=nRT\therefore P=\frac{nRT}{V-nb}> P_{ideal}{\color{Blue} \because P_{ideal}=\frac{nRT}{V}}

It means that the molecular size effect (repulsive interaction) creates higher pressure than that observed from the ideal gas law, where the gas molecules have no volume.

Van der Waals equation when b=0

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

P=\frac{nRT}{V}-\frac{an^{2}}{V^{2}}< P_{i}

\because P_{i}=\frac{nRT}{V}

So the intermolecular attraction reduces the pressure of real gases.

Units of Van der Waals constant a and b

Van der Waals equation,

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

where Pa is called the internal pressure of a gas.

\therefore a=P_{a}\times \frac{V^{2}}{n^{2}}

Thus the unit of a = atm lit2 mol-2

Unit of n × unit of b = unit of volume.

∴ Unit of b = lit mol-1

The significance of Van der Waals constants

Van der Waals constant a term originates from the intermolecular attraction.

∴ Pa = an2/V2

Thus internal pressure of the gas measured by a and so the attractive force between the molecules.

But higher the value of a grater is the intermolecular attraction and more easily the gas liquefied.

Carbon dioxide gas = 3.95 atm lit2 mol-2
Hydrogen gas = 0.22 atm lit2 mol-2

Another constant b measures the molecular size and also the repulsive forces. The value of b can also be utilized to calculate the molecular diameter(σ). The grater, the value of b larger the size of the gas molecule.

Carbon dioxide gas = 0.04 lit mol-1
Hydrogen gas = 0.02 lit mol-1

Boyle’s temperature formula

The mathematical formula of Boyle temperature,

T_{B}=\left [ \frac{\partial \left ( PV \right )}{\partial P} \right ]_{T}\, when\, P\rightarrow 0

But Van der Waals gas formula

\left ( P+\frac{a}{V^{2}} \right )\left ( V-b \right )=RT

or, P=\frac{RT}{V-b}-\frac{a}{V^{2}}


\therefore \left [ \frac{\partial \left ( PV \right )}{\partial P} \right ]_{T}=\left [\frac{RT}{V-b}-\frac{RTV}{\left ( V-b \right )^{2}}+\frac{a}{V^{2}} \right ]\left ( \frac{\partial V}{\partial P} \right )_{T}

=\left [\frac{RT\left ( V-b \right )-RTV}{\left ( V-b \right )^{2}}+\frac{a}{V^{2}} \right ]\left ( \frac{\partial V}{\partial P} \right )_{T}

=\left [\frac{RTb}{\left (V-b \right )^{2}}+\frac{a}{V^{2}} \right ]\left ( \frac{\partial V}{\partial P} \right )_{T}

when\, T=T_{B},\, T_{B}=\left [ \frac{\partial \left ( PV \right )}{\partial P} \right ]_{T}=0

\therefore \frac{RT_{B}b}{\left ( V-b \right )^{2}}=\frac{a}{V^{2}}

or, T_{B}=\left (\frac{a}{Rb} \right )\left ( \frac{V-b}{V} \right )^{2}

Since P → 0, V is large

\left ( \frac{V-b}{V} \right )\simeq 0

\therefore T_{B}=\frac{a}{Rb}

Amagat curves for real gases

Amagat curves for van der waals gases
Amagat curves

Van der Waals equation for one-mole real gas

\left ( P+\frac{a}{V^{2}} \right )\left ( V-b \right )=RT


Neglecting the small-term (a/V2)


Ideal gas law for the small-term,
a/V = aP/RT and taking Z = PV/RT,

Z=1+\left ( \frac{1}{RT} \right )\left \{ b-\left ( \frac{a}{RT} \right ) \right \}P

\therefore Z=\int \left ( T,P \right )

Thus this equation can be used to explain the Amagat curve qualitatively at low pressure and moderate pressure region.

Compare carbon dioxide and hydrogen gas

When a is very high

\frac{a}{RT}> b

\therefore \left \{ b-\left ( \frac{a}{RT} \right ) \right \}=-ve

The slope of the Z vs P curve for carbon dioxide at a moderate pressure region is negative. That is the value of Z decreases with the increase of pressure and it is also found in the curve of carbon dioxide.

When a is very small

\frac{a}{RT}< b

\therefore \left \{ b-\left ( \frac{a}{RT} \right ) \right \}=+ve

The slope of the Z vs P curve for hydrogen becomes (+) ve and the value of Z increases with pressure.

When T〈 TB
or, T〈 a/Rb

\therefore b< \frac{a}{RT}\, and\, \left (b-\frac{a}{RT} \right )=-ve

That is the value of Z decreases with increases with pressure at the moderate pressure region of the curve for carbon dioxide, Z〈 1 and more compressible.

When T = TB = a/Rb
or, b = a/RT

\therefore \left ( b-\frac{a}{RT} \right )=0

That is Z = 1, the gas is an ideal gas. The size effect compensates for the effect due to the intermolecular attraction of the gas.
Z runs parallel to the pressure axis up to the low-pressure region.

When T 〉TB
or, T 〉a/Rb

\therefore b> \frac{a}{RT}\, and\, \left (b-\frac{a}{RT} \right )=+ve

That is the value of Z increases with the increase of pressure.. The size effect dominates over the effect due to the intermolecular attraction.

Van der Waals equation quiz question

The value of a is extremely small for hydrogen and helium gases as they are difficult to liquefy. Thus the equation of state P(V – b) = RT obtains from van der Waals equation by ignoring small-term a/V2.

Hence Z is always greater than one and it increases with the increase in the pressure of the gas.


1 / 7

Comment on the possibility of defining a Boyle Temperature if a=b=0

T_{B}=\frac{a}{R\times b}

2 / 7

The Boyle temperature of nitrogen gas given a=1.4 atm lit2 mol-2 and b = 0.04 lit mol-1

3 / 7

Comment on the possibility of defining a Boyle temperature if a=0

4 / 7

What are the CGS and SI units of Van der Waals constant b?

\therefore deviation=\frac{4.92-4.904}{4.904}\times 100=0.326

5 / 7

Find the extent of deviation from the ideal behavior of 2 moles of nitrogen gas occupying 10 lit volume at 270°C
Given, a = 1.4 atm liter2 mol-2 and b=0.04 lit mol-1.

unit\, of\, a=\frac{unit\, of\, P\times \left ( unit\, of\, V \right )^{2}}{\left (unit\, of\, n \right )^{2}}

6 / 7

What is the SI unit of Van der Waals constant a?

7 / 7

If a = 1.4 atm liter2 mol-2 and b=0.04 lit mol-1, the pressure of 2 moles of nitrogen gas occupying 10 lit volume at 270°C

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