*Derivation of van der Waals equation *

*Derivation of van der Waals equation*

In 1873, Van der Waals modified the Ideal Gas Equation and formulate

*of state for real gases.***Van der Waals equation**- P

_{i}V

_{i}= RT

- 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 Equation__.

**Volume C**orrection by Van der Waals

**Volume C**orrection by Van der Waals

- Molecules are assumed to be a hard rigid sphere and in a

*real gas*, the available space for free movement of the molecules becomes less then V. let us take the available space for free movement of 1-mole gas molecules,

Vi = (V-b)

- 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. Let us take, r is the

__radius of the molecule__and σ = 2r is the diameter using the molecule as a rigid sphere.

An encounter of the gas molecules. |

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

- b = 4 × N

_{0}(

^{4}/

_{3}) π 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.

- This excluded volume =

^{4}/

_{3}π σ

^{3}for a pair of molecules. Thus a single molecule, =

^{1}/

_{2}×

^{4}/

_{3}π σ

^{3}The effective volume for Avogadro number of molecules (present in 1-mole gas),

b = N

or, b = N

or, b = 4 N

Thus, b = N

∴ b =

_{0}×^{2}/_{3}× π σ^{3}or, b = N

_{0}×^{2}/_{3}× π (2r)^{3}or, b = 4 N

_{0}×^{4}/_{3}× π (r)^{3}Thus, b = N

_{0}×^{2}/_{3}× π (2r)^{3}∴ b =

^{2}/_{3}× N_{0}× π σ^{3}- Thus b measures and helps to calculate the radius ( or diameter) of the gas molecule. The equation becomes,

- P

_{i}(V-b) = RT

**Pressure Correction by Van der Waals**

**Pressure Correction by Van der Waals**

- The pressure of the gas is developed due to the wall collision of the gas molecules. But due to inter-molecular attraction, the colliding molecules will experience an inward pull and so the pressure exerted by the molecules in

*real gas*will be less than that if there had not been an intermolecular attraction as in

*ideal gas*P

_{i}.

- Thus, P

_{i}〉P or, P

_{i}= P + P

_{a}

- Where P

_{a}is the pressure correction term originating from attractive forces. Higher the intermolecular attraction in a gas, greater is the magnitude of P

_{a}. Pressure depends on both the frequency of molecular collisions with the walls and the impulse exerted by each collision. Both contributions are diminished by the attractive forces and the strength with which these operate is roughly proportional to the density of the molecules. Therefore the average pressure exerted by the molecules is decreased by

**P**is proportional to the square of the density.

_{a}- P

_{a}∝ 1/V

^{2}Since density ∝ 1/V ∴ P

_{a}= a/V

^{2}

- Where

**a**= constant for the gas that measures the attractive force between the molecule.

- Thus, P

_{i}= P + (a/V

^{2})

- Using the two corrections we have

*Van der Waals equation*for 1 mole

*Real Gas*is,

- (P + a/V

^{2})(V-b) = RT

- To convert the equation for n moles volume has to change as 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*we have,

Van der Waals equation for n Moles Real Gas |

*For the real gas a=0 but b≠0*

*For the real gas a=0 but b≠0*

- We have

*Van der Waals Equation*, (P+ an

^{2}/V

^{2})(V - nb) = nRT but a=0 Hence,

**P = nRT/(V - nb) 〉P**Since,

_{i}**P**only.

_{i}= nRT/V*For the real gas a≠0 but b=0*

*For the real gas a≠0 but b=0*

- It means that the molecular size effect (repulsive interaction) creates higher pressure than that observed by the

*ideal gas*where molecules have no volume.

- We have

*Van der Waals Equation*, (P+ an

^{2}/V

^{2})(V - nb) = nRT but b=0(no size). Hence, P = (nRT/V - an

^{2}/V

^{2})ㄑP

_{i}Since, P

_{i}= nRT/V only. Thus, intermolecular attraction effect reduces the pressure of the

*real gases*.

**Units of Van der Waals Constant**

- From the

*Van der Waals Equation*, (P+ an

^{2}/V

^{2})(V - nb) = nRT where P

_{a}= an

^{2}/V

^{2}Where P

_{a}is the pressure correction term, called often the internal pressure of the gas. Then, a = P

_{a}× (V

^{2}/n

^{2})

- Thus the unit of a = atm lit

^{2}mol

^{-2}

- Again, nb = Unit of volume (say liter)

- Hence, the unit of b = lit mol

^{-1}

**Significance of Van der Waals constants**

- ‘a’ term originates from the intermolecular attraction and P

_{a}= an

^{2}/V

^{2}. 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. Thus, a of CO

_{2}= 3.95 atm lit

^{2}mol

^{-2}while, a of H

_{2}= 0.22 atm lit

^{2}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. Thus, b of CO

_{2}= 0.04 lit mol

^{-1}while, b of H

_{2}= 0.02 lit mol

^{-1}.

**Boyle's temperature from Van der Waals equation**

- Mathematical condition for

__Boyle’s temperature__is, T

_{B}= [d(PV)/dP]

_{T}When P→0 From the

*Van der Waals Equation*for 1 mole gas, (P + a/V

^{2})(V - b) = RT or, P = RT/(V - b) - a/V

^{2}Hence, PV = {RTV/(V - b)} - a/V Thus, T

_{B}= [d(PV)/dP]

_{T}= [RT/(V-b)-{RTV/(V-b)

^{2}}+a/V

^{2}][(dV/dP)]

_{T}= [{RT(V - b) - RTV}/(V - b)

^{2}+ a/V

^{2}] [(dV/dP)]

_{T}=[{- RTb/(V - b)

^{2}} + a/V

^{2}] [(dV/dP)]

_{T}But when T = T

_{B}, [d(PV)/dP]

_{T}= 0 and [(dV/dP)] ≠ 0 Thus, We have, RT

_{B}b/(V - b)

^{2}= a/V

^{2}or, T

_{B}= (a/Rb) {(V - b)/V}

^{2}Since P → 0, V is large. Thus, (V - b)/V ≃ 1

Hence,T_{B} = a/Rb |

**Amagat curve from Van der Waals Equation**

Amagat Curves |

*Van der Waals equation*for 1-mole

*real gas*,

(P + a/V

^{2})(V - b) = RT

or, PV - Pb + (a/V) + (a/V

^{2}) = RT

Neglecting the small term (a/V

^{2}), we get,

PV = RT + Pb - (a/V)

Using the

*ideal gas equation*for the small term,

a/V = aP/RT and taking Z = (PV/RT),

We have, Z = 1 + (1/RT){b - (a/RT} P

This Shows that Z = ∫(T, P)

- This equation can be used to

__explain Amagat curve__qualitatively at low pressure and moderate pressure region.

**For CO**

_{2}Van der Waals constant ‘a’ is very high- Thus (a/RT) 〉b in the equation and {b - (a/RT)} = - ve .
That is the slope of Z vs P curve is ( -) for carbon dioxide at moderate pressure region. That is the value of Z decreases with the increase of Pressure and it is also found in the curve of carbon dioxide.

**For H**

_{2}Van der Waals constant ‘a’ is very small- So, a/RT〈 b, slope of Z vs P curve for H

_{2}becomes (+) ve and the value of Z increases with pressure.

(i) When T〈 T

or, T〈 a/Rb

or, b〈 a/RT

and {b−(a/RT)} = (-)ve.

_{B}or, T〈 a/Rb

or, b〈 a/RT

and {b−(a/RT)} = (-)ve.

- That is the value of Z decreases with increases with P at the moderate pressure region of the curve of CO2, Z〈 1 and more compressible.

(ii) When T = T

or, b = a/RT

hence, {b - (a/RT)} = 0

_{B}= a/Rbor, b = a/RT

hence, {b - (a/RT)} = 0

- That is Z = 1 the gas shows

*ideal*behavior. The size effect compensates the effect due to the intermolecular

__attraction of the gas__. Z runs parallel to P axis up to a certain range of pressure at the low-pressure region.

(iii) When T 〉TB

or, T 〉a/Rb

or, b 〉a/RT

hence {b - (a/RT} = (+)ve.

or, T 〉a/Rb

or, b 〉a/RT

hence {b - (a/RT} = (+)ve.

- That is the value of Z increases with the increase of P when T 〉T

_{B}. The size effect dominates over the effect due to the intermolecular attraction. For hydrogen and helium, 0°C is greater then their T

_{B}values and Z vs P slope becomes (+)ve. At very low-pressure P→0 and high temperature, the volume is very large and both the size effect and attraction effect becomes negligible that is Pb and (a/RT) P are negligible and Z=1 that is the gas obeys

*ideal behavior*.

**Questions and Answers of Van der Waals equation **

*Question*

- Calculate the pressure of 2 moles of Nitrogen gas occupying 10 lit volume at

**27**. Given, a = 1.4 atm lit

^{0}C^{2}mol

^{-2}and b=0.04 lit mol

^{-1}using

*Van der Waals equation*. Also, calculate the pressure of the gas using the

*ideal gas equation*and find the extent of deviation from

*ideal behavior*.

*Answer*

- P = 4.904 atm, P

_{i}= 4.92 atm and Deviation=0.325%

*Question*

- Find the Units and Dimensions of

*Van der Waals*constant 'a' and 'b' from

*Van der Waals Equation*.

*Answer*

- In SI system,unit of 'a' = N m

^{4}mol

^{-2}In CGS system,unit of 'a' =dyne cm

^{4}mol

^{-2}In SI system,unit of 'b' = m

^{3}mol

^{-1}In CGS system,unit of 'b' = cm

^{3}mol

^{-1}

*Question*

- Calculate the Boyle temperature of nitrogen gas given a=1.4 atm lit

^{2}mol

^{-2}and b = 0.04 lit mol

^{-1}.

*Answer*

- T

_{B}= 427 K

*Question*

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

*Answer*

- When, a = 0,
*Van der Waals Equation*, P (V - b) = RT or, PV = RT + Pb Differentiating with respect of pressure at constant temperature, [d(PV)/dP]_{T}= b, but b ≠ 0 hence [d(PV)/dP]_{T}≠ 0. The gas has no Boyle temperature since at any temperature since at any temperature, [d(PV)/dP]_{T}≠ 0. - Again when a = b = 0,
*Van der Waals equation*becomes PV = RT; or, [d(PV)/dP]_{T}= 0.

That is at any temperature this becomes zero and so all the temperature are Boyle Temperatures.