Van der Waals equation for real gas

Derivation of van der Waals equation for real gases

A real gas that does not behave as an ideal gas due to intermolecular attraction. Van der Waals modified the Ideal gas and formulate Van der Waals equation for real gases.

By incorporating the size effect and intermolecular attraction effect of the real gas. These above two effects are discussing under the volume correction and pressure correction of the ideal gas law.

P_{i}V_{i}=RT

Volume correction in the ideal gas equation

Real gas molecules are assumed to be a hard rigid sphere, 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 gas molecules,

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

where V is the molar volume of the gas and b is the volume correction factor. Vi = molar volume of the ideal gas where the gas molecules are regarded as point masses.

Questions set for Van der Waals

4

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?

Preal = 4.904 atm, Pideal = 4.92 atm

\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.

Van der Waals equation for n moles gas

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

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

=\frac{N\, m^{-2}\times \left ( m^{3} \right )^{2}}{mol^{2}}=N\, m^{4}\, mol^{-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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How do you find the molar volume of a gas?

Let us take, r is the radius of the molecule and σ = 2r is the diameter using the molecule as a rigid sphere.

How do you find the molar volume of a gas?
Molar volume of a gas

Then it can be shown that the volume correction term or effective volume of one-mole gas molecules,

b=4\times N_{0}\times \frac{4}{3}\pi r^{3}

When two molecules encounter each other the distance between the centers of the two molecules would be σ.

They can not approach beyond this distance. Thus the sphere of radius σ will occupy a space unavailable for a pair of molecules.

Excluded volume for a pair of molecules

=\frac{4}{3}\pi \sigma ^{3}Single-molecule,

=\frac{1}{2}\times \frac{4}{3}\pi \sigma ^{3}

Volume for Avogadro number of molecules,

b=N_{0}\times \frac{2}{3}\pi \sigma ^{3}=N_{0}\times \frac{2}{3}\times \left ( 2r \right )^{3}

{\color{Blue} \therefore b=4N_{0}\times \frac{4}{3}\pi r^{3}}

Measurement of b helps to calculate the radius or diameter of the gas molecule. Thus the gas equation,

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

Pressure correction in the ideal gas equation

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

The pressure exerted by the molecules in real gas will be less if the gas molecules have no intermolecular attraction as in ideal gas pressure.

P_{i}> P

\therefore P_{i}=P+P_{a}

Where Pa is the pressure correction term originating from attractive forces. Higher the intermolecular attraction in a gas, greater is the magnitude of Pa. Pressure depends on both the frequency of molecular collisions with the walls and the impulse exerted by each collision.

The pressure of Van der Waals gas

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

P_{a} \propto\frac{a}{V^{2}}

density \propto\frac{1}{V}

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

Where a is a constant for the gas that measures the attractive force between the molecules.

{\color{Blue} P_{i}=P+\frac{a}{V^{2}}}

Van der Waals equation for n moles of real gas

Using the two corrections Van der Waals equation for a 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}

Why the pressure of real gas is less than ideal gas?

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 by the ideal gas law where 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}

Thus, the intermolecular attraction effect 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

Significance of a and b in Van der Waals equation

‘a’ term originates from the intermolecular attraction and Pa = an2/V2. Thus, ‘a’ is a measure of internal pressure of the gas and so the attractive force between the molecules.

Higher the value of ‘a’ grater is the intermolecular attraction and more easily the gas could be 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 a measure of 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

Define Boyle’s temperature

The mathematical definition of Boyle temperature,

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

Van der Waals equation for 1-mole real gas,

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

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

or,PV=\frac{RTV}{V-b}-\frac{a}{V}

\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}

Z vs P graph for real gas

Amagat curves for real gases
Amagat curves

Van der Waals equation for one-mole real gas

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

or,PV-Pb+\frac{a}{V}+\frac{a}{V^{2}}=RT

Neglecting the small-term (a/V2)

PV=RT+Pb-\frac{a}{V^{2}}

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 )

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.

Hydrogen and helium gas

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.