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Jan 8, 2019

Soft and Hard Acids and Bases (SHAB) Principle

SHAB Principle and Stability of the Complex:

According to Lewis Concept an Acid-Base reaction involves an interaction of a vacant orbital of an acid (A) and a filled or unshared orbital of a base (B).
A + :B
A : B 
Lewis Acid
(Acceptor)

Lewis Base
(Donor)

Adduct or Complex
The species A is called Lewis Acid or a generalized acid and B is called Lewis Base or a generalized base. A strong acid A and a strong base B, will form the stable complex A : B.
A concept known as Principle of Soft and Hard Acids and Bases(SHAB) Principle is very helpful in making a stability of the complex A : B.
According to this principle the complex A : B is most stable when A and B are either both soft or both hard.
The complex is least stable when one of the reactants (namely A and B) is very hard and the other one is very soft. 
In order to arrive at a comparative estimate of the donor properties of different bases, the preferences of a particular base to bind a proton H+ and methyl mercury (II) ion, [CH3HgB]+ was determined. Both the proton and methyl mercury cation can accommodate only one coordinate bond but the two cation vary widely in their preferences to bases. This preferences was estimated from the experimental determination of equilibrium constants for the exchange reactions:
BH+ + [CH3Hg(H2O)]+
[CH3HgB]+ 
+ H3O+
The results indicate that bases in which the donor atom is N, O or F prefer to coordinate to the proton. Bases in which the donor atoms is P, S, I, Br, Cl or C prefer to coordinate to mercury.

Hard and Soft Bases:

The donor atoms in the first group have high Electronegativity, low Polaris ability and hard to oxidize. Such donors have been named ‘hard bases by Pearson, since they hold on to their electrons strongly. 
The donor atoms of second category are of low electronegativity, high polarisability, and are easy to oxidize. Such donors have been called ‘soft’ bases since they are hold on to their valence electrons rather loosely.
In simple terms hardness is associated with a tightly held electron shell with little tendency to polarise. On the other hand softness is associated with a loosely bound polarisable electron shell.
It will be seen that within a group of the periodic table softness of the Lewis bases increases with the increase in size of the donor atoms. 
Thus, among the halide ions softness increases in the order:
-Cl -Br -I -
Thus F - is the hardest and I - is the Softest base.
Classification of Bases
(R = alkyl or aryl group)
Hard Borderline
Soft
H₂O, OH⁻, F⁻, CH₃COO⁻, PO₄⁻³, SO₄⁻², Cl⁻, CO₃⁻², ClO₄⁻, NO₃⁻, ROH, RO⁻, R₂O, NH₃, RNH₂, N₂H₄
C₆H₅NH₂, C₅H₅N, N₃⁻, Br⁻, NO₂⁻, SO₃⁻²,N₂
R₂S, RSH, RS⁻, I⁻, SCN⁻, S₂O₃⁻², R₃P, R₃As, (RO)₃P, CN⁻, RNC, CO, C₂H₄, C₆H₆, H⁻

Hard and Soft Acids:

After having gone through a classification of bases, a classification of Lewis acids is necessary. The preferences of a given Lewis acid towards ligands of different donor atoms is usually determined from the stability constant values of the respective complexes or from some other useful equilibrium constant measurements. When this is done, metal complexes with different donor atoms can be classified into two sets based on the sequences of their stabilities.
Hard acids have small acceptor atoms, are of high positive charge and do not contain unshared pair of electrons in their valence shell, although all these properties may not appear in one and the same acid. These properties lead to high electronegativity and low polarisability. In keeping with the naming of the bases, such acids are termed as 'Hard'Acids.
Hard Acids:
NPOSFClBrI
Soft acids have large acceptor atoms, are of low positive charge and contain ushered pairs of electrons in their valence shell. These properties lead to high polarisability and low electronegativity. Again in keeping with the naming of the bases, such acids are termed 'Soft' Acids
Soft Acids:
NPOSFClBrI
Classification of Lewis Acids as Hard, Intermediate and Soft Acids.
Hard Borderline
Soft
H⁺, Li⁺, Na⁺, K⁺, Be⁺², Mg⁺², Ca⁺², Sr⁺², Mn⁺², Al⁺³, Ga⁺³, In⁺³, La⁺³, Lu⁺³, Cr⁺³, Co⁺³, Fe⁺³, As⁺³, Si⁺⁴, Ti⁺⁴, Zr⁺⁴, Th⁺⁴, U⁺⁴, Ce⁺³, Sn⁺⁴, VO⁺², UO₂⁺², MoO⁺³, BF₃, AlCl₃, SO₃, Cr⁺⁶
Fe⁺, Co⁺², Ni⁺², Cu⁺², Zn⁺², Pb⁺², Sn⁺², Sb⁺³, Bi⁺³, Rh⁺³, B(CH₃)₃, SO₂, NO⁺, GaH₃
Cu⁺, Ag⁺, Au⁺, Tl⁺, Hg⁺, Pd⁺², Cd⁺², Pt⁺², Hg⁺², CH₃Hg⁺, Tl⁺³, BH₃, GaCl₃, InCl₃, I⁺, Br⁺, I₂, Br₂, Zerovalent Metal atoms

Classify the following as Hard and Soft Acids and Bases. 
(i) H⁻ (ii) Ni⁺⁴ (iii) I⁺ (iv) H⁺

(i) The hydride ion has a negative charge and is far too large in size compared to the hydrogen atom. Its electronegativity is quite low and it will be highly polarisable by virtue of its large size. Hence it is Soft Base.
(ii) Quadrivalent nickel has quite a high positive charge. Compared to bivalent nickel its size will be much smaller. Its electronegativity will be very high and polarisability will be low. Hence it is Hard Acid.
(iii) Mono-positive iodine has a low positive charge and has a large size. It has a low electronegativity and a high polarisability. Hence it is a Soft Acids.
(iv) H⁺ has the smallest size with a high positive charge density. It has no unshared pair of electrons in its valence shell. All these will give a high electronegativity and very low polarisability. Hence H⁺ is a Hard Acid.

Soft and Hard Acids and Bases (SHAB) Principle:

This principle also means that if there is a choice of reaction between an Acid and two Bases and two Acids and a Base,
A Hard Acid will prefer to combine with a Hard Base and a Soft Acid will prefer to combine with Soft Base and thus a more stable product will be obtained.
Hard Acid - Hard Base may interact by strong ionic forces. Hard Acids have small acceptor atoms and positive charge while the Hard Bases have small donor atoms but often with negative charge. Hence a strong ionic interaction will lead to Hard Acid - Base combination.
On other hand a Soft Acid - Soft Base combination mainly a covalent interaction. Soft Acids have large acceptor atoms, are of low positive charge and contain ushered pair of electrons in their valence shell.

Application of Soft and Hard Acids and Bases (SHAB) Principle:

(i) [CoF₆]⁻³ is more stable than [CoI₆]⁻³
It will be seen that Co⁺³ is a hard acid
F⁻ is a hard base and I⁻ is a soft base
Hence [CoF₆]⁻³ (Hard Acid + Hard Base) is more stable 
than [CoI₆]⁻³ (Hard Acid + Soft Base).

AgI₂⁻ is stable, but AgF₂⁻ does not exist. Explain.

We know that Ag⁺ is a soft acid, F⁻ is hard base and I⁻ is soft base
Hence AgI₂⁻ (Soft Acid + Soft Base) is a stable complex and AgF₂⁻ (Soft Acid + Hard Base) does not exist.
(ii) The existence of certain metal ores can also be rationalised by applying SHAB principle. Thus hard acids such as Mg⁺², Ca⁺² and Al⁺³ occur in nature as MgCO₃, CaCO₃ and Al₂O₃ and not as sulphides (MgS, CaS and Al₂S₃), since the anion CO₃⁻² and O⁻² are hard bases and S⁻² is a soft base.
Soft acids such as Cu⁺, Ag⁺ and Hg⁺² on the other hand occurs in nature as sulphides.
The borderline acids such as Ni⁺², Cu⁺² and Pb⁺² occur in nature both as carbonates and sulphides. 
The combination of hard acids and hard bases occurs mainly through ionic bonding as in Mg(OH)₂ and that of soft acids and soft bases occurs mainly by covalent bonding as in HgI₂.

Explains why Hg(OH)₂ dissolved readily in acidic aqueous solution but HgS does not?

In the case of Hg(OH)₂ and HgS, Hg is a soft acid and OH⁻ and S⁻² is hard base and soft base respectively. Evidently HgS (Soft acid + Soft base) will be more stable than Hg(OH)₂ (Soft Acid + Hard base). More stability of HgS than that of Hg(OH)₂ explain why Hg(OH)₂ dissolved readily in acidic aqueous solution but HgS does not. 
If we arrange the donor atoms of most common Lewis bases in an increasing order of their electronegativity values we get,
Soft and Hard Acids and Bases and Hard Soft Acid Base Principle
Donor Atoms of Most Common Lewis Bases





Jan 3, 2019

Acids and Bases Questions and Answers


What is the acidic order of oxoacids of phosphorus?

When the oxidation state rule applied to the oxoacids of phosphorus (H₃PO₂, H₃PO₃ and H₃PO₄) it is predicted that the acidic Character of these acids should be in the order:
H₃PO₂ㄑH₃PO₃ㄑH₃PO₄, but the experimental observation suggested the reverse order:
H₃PO₂ ≥ H₃PO₃〉H₃PO₄.
The experimental order can be explained when we consider the structures of these acids given as,
Acids and Bases Questions and Answers
Oxoacids-of-Phosphorus
H₃PO₂ is a mono-basic acid. The proton attached to an oxygen has a far greater chance of dissociation than any directly bonded hydrogen. the structure of H₃PO₂ involves one protonated oxygen and another unprotonated oxygen. H₃PO₃ is dibasic and hence has two protonated oxygen's and one unprotonated oxygen. H₃PO₄ is tribasic, has three protonated oxygen's and one unprotonated oxygen. 
In this series therefore the number of unprotonated oxygen's, which are the vehicles for the enhancement of acidity, is the same for all the three acids. But dissociable protons increase from one in H₃PO₂ to three in H₃PO₄. Therefore the overall inductive effect of the unprotonated oxygen decreases from H₃PO₂ to H₃PO₄. Hence the acidity slightly falls off in the order:
H₃PO₂ ≥ H₃PO₃H₃PO₄

pH of a solution is 4.5. Calculate the concentration of H⁺ ion.

We have from the definition,
pH = -log[H⁺] =4.5
or, log[H⁺] = -4.5
∴ [H⁺] = 3.16×10⁻⁵

Calculate the [H⁺], [OH⁻] and pH of a solution prepared by diluting 20 ml of 0.1M HCl to one lit. 

[H⁺] = (20×0.1)/1000 
=0.002 
= 2×10⁻³
[OH⁻] = (1×10⁻¹⁴)/[H⁺]
=(1×10⁻¹⁴)/(2×10⁻³) 
= 0.5×10⁻¹¹
pH = -log[H⁺] 
= -log2×10⁻³ 
= 3-log2 
= 2.7

Calculate the [H⁺], [OH⁻] and pH of a solution obtained by dissolving 28 gm KOH to make 200 ml of a solution.

[OH⁻] = (28×1000)/(56×200) 
= 2.5M 
(Molecular Weight of KOH = 56gm)
[H⁺] = (1×10⁻¹⁴)/[OH⁻] 
= (1×10⁻¹⁴)/(2.5) 
= 4×10⁻¹⁵ 
pH = -log[H⁺] 
= - log4 × 10⁻¹⁵ 
= (15 - log4) 

What will be the effect of adding KNH₂ to liquid ammonia with respect to acidity?

2NH₃ NH₄⁺ + NH₂⁻
KNH₂ K⁺ + NH₂⁻
Due to common ion effect the equilibrium shift to left and decrees the acidity of NH₃.

Arrange in order of Lewis acidity :
(i) BF₃ (ii) BCl₃ (iii) BI₃ (iv) BBr₃

These boron halides have pi-interaction between filled p- orbitals of the halogen and empty p-orbital of boron. The effectiveness of the pi-interaction falls off with increasing size of halogens so that it is the strongest in BI₃. When boron halides receives an electron pair the pi-bond between boron and halogen has to be ruptured in order to make room for a coordinate bond.
H₃NBF₃
Thus with BF₃ it will be hardest to rupture the pi-bond. Therefore the order of the Lewis acidity is:
BF₃BCl₃BBr₃BI₃

Why all alkali are bases but all bases are not alkali?

All the bases are dissolved in water to produce alkali whereas all the bases are not dissolved in water but all the alkali are dissolved in water. Thus all alkali are bases but all bases are not alkali.
Na₂O is a base because it dissociate in water to produce NaOH. NaOH, KOH and Ca(OH)₂ dissolved in water to produce OH⁻ ion. Thus all these hydroxide are Alkali.
NaOH Na⁺ + OH⁻
KOH K⁺ + OH⁻
Ca(OH)₂ Ca⁺² + 2OH⁻
Al(OH)₃, Fe(OH)₃, Zn(OH)₂ etc does not dissolved in water but reacts with acids to produce Salt and water. Thus These are bases but but not alkali.





Explain - NH₃ behaves as a base but BF₃ as an Acid.

In NH₃, the central N atom have lone pair of electrons This lone pair coordinate to empty orbital, is termed as a base according to Lewis concept. 
The compounds with less than an octet for the central atom are Lewis acids. In BF₃B in BF₃, contain six electrons in the central atom thus it behaves as an acid.
F₃B + :NH₃ F₃BNH₃
Acid Base Adduct


Explain why tri-covalent phosphorus compounds can serve both as Lewis acids and also as bases?

Tri-covalent phosphorus compounds like PCl₃ have a lone pair of electrons in phosphorus. This lone pair may coordinate to a metal ion thus allowing the compound to serve as a Lewis base. 
Ni + 4PCl₃   [Ni(PCl₃)₄]⁰
Again the quantum shell of phosphorus has provision for d-orbitals which can receive back donated electrons from electron reach low oxidation state of a metal ion. In this latter case tri-covalent phosphorus compound serves as a Lewis acid.

Bi-positive tin can function both as a Lewis acid and a Lewis base. Explain?

The Lewis representation of SnCl₂ shows a lone pair on tin through it is as yet short of an octet. Ligands, particularly donor solvents with lone pairs, may coordinate to tin giving complexes. Here SnCl₂ behaves as a Lewis acid.
Again interaction of platinum group metal compounds with SnCl₂ (SnCl₃⁻) as donor leads to the formation of coordinate complexes. As for examples,
[(Ph₃P)₂PtCl(SnCl₃)], [RuCl₂(SnCl₃)₂]⁻². These are examples of Lewis base behavior of SnCl₂.

Can SiCl₄ and SnCl₄ function as Lewis acids?

Both silicon and tin are members of Group IV of the periodic table and their quantum shells admit of d-orbitals. As a result they can expand their valence shell through SP³d² hybridisation and can give rise to six coordinate complexes. In fact complex like [SiCl₄{N(CH₃)₃}₂], [SiCl₂(Py)₄]Cl₂, [SnCl₄(Bpy)] are known. Thus SiCl₄ and SnCl₄ can function as Lewis acids.

Name the conjugate acids and the conjugate bases of HX⁻ and X⁻².

Conjugate acid of a species is the one that is obtained on the addition of a proton and conjugate base of a species is one that is obtained on the release of a proton.
H₂O + HX⁻   OH⁻ + H₂X
Acid₁ Base₂ Base₁ Acid₂
In the above reaction HX⁻ acts as a base and its conjugated acid is H₂X.
HX⁻ + H₂O   X⁻² + H₃O⁺
Acid₁ Base₂ Base₁ Acid₂
In this reaction HX⁻ acts as an acid thus its conjugated base is X⁻².
In the same way conjugate acid of X⁻² is HX⁻ but X⁻² cannot have any conjugated base because there is no proton that can release.

Arrange these oxides in order of their acidic nature : N₂O₅, As₂O₃, Na₂O, MgO.

Acidic oxides react with water to give oxoacids. The higher the oxidation number and the higher the electronegativity the grater the central element will force to reaction with water to give the oxoacid. Of all these oxoacids nitrogen has the highest oxidation number and electronegativity. Thus the order of acidic nature and oxidation number are-
Acidic nature :
Na₂OMgOAs₂O₃N₂O₅
Oxidation number of Oxide forming element: +1  +2  +3  +5. 

Bisulphate ion can be viewed both as an acid and a base. Explain.

Bisulphate ion is HSO₃⁻. It may lose a proton to give the conjugate base SO₃⁻², thus behaving as an acid. Again it may add on a proton to give the conjugate acid, thus showing its base character.

SO₃ Behaves as an acid and H₂O as a base. Explain.

SO₃ like BF₃ has less than an octet,and will be termed as an acid according to Lewis concept.
But in H₂O oxygen atom contain lone pairs to donate the Lewis acid.
Oxygen and sulphur contain six electrons in their valence shell and therefore regarded as Lewis Acids. The oxidation of SO₃⁻² to SO₄⁻² ion by oxygen and S₂O₃⁻² ion by sulphur are the acid-base reactions.
SO₃⁻²  [O]   ⇆   [OSO₃]⁻²
SO₃⁻² +  [S]       [SSO₃]⁻²

Explains why Hg(OH)₂ dissolved readily in acidic aqueous solution but HgS does not?

In the case of Hg(OH)₂ and HgS, Hg is a soft acid and OH⁻ and S⁻² is hard base and soft base respectively. Evidently HgS (Soft acid + Soft base) will be more stable than Hg(OH)₂ (Soft Acid + Hard base). More stability of HgS than that of Hg(OH)₂ explain why Hg(OH)₂ dissolved readily in acidic aqueous solution but HgS does not. 

Classify the following as Hard and Soft Acids and Bases.
(i) H⁻ (ii) Ni⁺⁴ (iii) I⁺ (iv) H⁺

(i) The hydride ion has a negative charge and is far too large in size compared to the hydrogen atom. Its electronegativity is quite low and it will be highly polarisable by virtue of its large size. Hence it is Soft Base.
(ii) Quadrivalent nickel has quite a high positive charge. Compared to bivalent nickel its size will be much smaller. Its electronegativity will be very high and polarisability will be low. Hence it is Hard Acid.
(iii) Mono-positive iodine has a low positive charge and has a large size. It has a low electronegativity and a high polarisability. Hence it is a Soft Acids.
(iv) H⁺ has the smallest size with a high positive charge density. It has no unshared pair of electrons in its valence shell. All these will give a high electronegativity and very low polarisability. Hence H⁺ is a Hard Acid.

Jan 1, 2019

Van der Waals Equation

Van der Waals Equation :

In 1873, Van der Waals modified the Ideal Gas Equation,
PiVi = RT
By incorporating the size effect and inter molecular attraction effect of the real gases. These above two effects are discussing under the Volume Correction and Pressure Correction of the ideal gas equation.

Volume Correction:

Molecules are assumed to be a hard rigid spheres 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.
How can we derived Ven der Waals Equation?
Representation of Gas Molecules
Then it can be shown that the volume correction term or effective volume of one mole gas molecules,
b = 4 × N0 (4/3) πr3
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 σ3 for a pair of molecules.
Thus the effective volume for a single molecule,
= 1/2 × 4/3 π σ3
The effective volume for Avogadro number of molecules (present in 1 mole gas),
How can we derived Ven der Waals Equation?
The effective volume for Avogadro number of molecules
Thus b measures and helps to calculate the radius ( or diameter) of the gas molecule.
The equation becomes, 
Pi(V - b) = RT

Pressure Correction:

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 pressure exerted by the molecules in real gas will be less than that if there had not been inter molecular attraction as in ideal gas Pi
Thus, Pi 〉P
or, Pi = P + Pa
Where Pa is the pressure correction term originating from attractive forces. Higher the inter molecular 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. 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 Pa is proportional to the square of the density.
Pa ∝ 1/V2  Since, density ∝ 1/V 
Pa = a/V2 
Where a = constant for the gas that measures the attractive force between the molecule. Thus, Pi = P + (a/V2
Using the two corrections we have Van der Waals equation for 1 mole real gas is, 
(P + a/V2 )(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,
what is the Van der Waals equation?
Van der Waals equation for n Moles Real Gas

Calculate the pressure of 2 moles of Nitrogen gas occupying 10 lit volume at 27°C. Given, a = 1.4 atm lit² mol⁻² and b=0.04 lit mol⁻¹ using Van der Waals equation. 
Also calculate the pressure of the gas using ideal gas equation and find the extent of deviation from ideal behavior.

P = 4.904 atm, Pi = 4.92 atm and Deviation=0.325%

Let us consider two hypothetical cases to show the size effect and attraction effect on the pressure of the gas:


(a)For the real gas a=0 (that is no inter molecular attraction ) but b≠0 (size considers):

We have Van der Waals Equation,
(P+ an²/V²)(V - nb) = nRT but a=0
Hence, P = nRT/(V - nb) ) 〉Pi
Since, Pi = nRT/V only.

(b)For the real gas a≠0 (existence of attractive force) but b=0 (no size consider):

It means that the molecular size effect (repulsive interaction) creates higher pressure then that observed by the ideal gas where molecules have no volume. 
We have Van der Waals Equation, 
(P+ an²/V²)(V - nb) = nRT
 but b=0(no size). 
 Hence, P = nRT/V - (an²/V²ㄑPi 
Since, Pi = nRT/V only.
 Thus, inter molecular attraction effect reduces the pressure of the real gases.

Units of a and b:

From the Van der Waals Equation, 
(P+ an²/V²)(V - nb) = nRT 
where, Pa = an²/V²
Where Pa is the pressure correction term, called often the internal pressure of the gas. 
Then, a = Pa × (V²/n²)
Thus the unit of a = atm lit² mol⁻²
Again, nb = Unit of volume (say liter) 
Hence, the unit of b = lit mol⁻¹

Find the Units and Dimensions of  Van dar Walls constant 'a' and 'b'
from Van der walls Equation.

In SI system,unit of 'a' 
= N m⁻² m⁶ mol⁻² 
N m⁴ mol⁻²
In CGS system,unit of 'a' 
= dyne cm⁻² cm⁶ mol⁻² 
= dyne cm⁴ mol⁻²
In SI system,unit of 'b' = m³ mol⁻¹
In CGS system,unit of 'b' = cm³mol⁻¹

Significance of a and b:

a’ term originates from the inter molecular 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 inter molecular attraction and more easily the gas could be liquefied.
Thus, a of CO₂ = 3.95 atm lit² mol⁻² while, a of H₂ = 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.
Thus, b of CO₂ = 0.04 lit mol⁻¹ while, b of H₂ = 0.02 lit mol⁻¹.

Calculation of Boyle Temperature (TB):

Mathematical condition for Boyle’s temperature is,
TB = [d(PV)/dP]T When P0
From the Van der Waals Equation for 1 mole gas,
(P + a/V²)(V - b) = RT
or, P = (RT/V - b) - a/V²
Hence, PV = {RTV/(V - b)} - a/V
Thus, TB = [d(PV)/dP]T 
= [RT/(V-b)-{RTV/(V-b)²}+a/V²][(dV/dP)]T
= [{RT(V - b) - RTV/(V - b)²} + a/V²] [(dV/dP)]T
=[{- RTb/(V - b)²} + a/V²] [(dV/dP)]T
But when T = TB
[d(PV)/dP]T = 0 and [(dV/dP)] ≠ 0
Thus We have, 
RTBb/(V - b)² = a/V²
or, T= (a/Rb) {(V - b)/V}² 
Since P → 0, V is large.
Thus, (V - b)/V ≃ 1 
Hence, TB = a/Rb

Calculate the Boyle temperature of nitrogen gas 
given a=1.4 atm lit² mol⁻² and b=0.04 lit mol⁻¹.

TB = 427 K

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

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 Walls 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.

Explanation of Amagat’s Curves:

We have Amagat Curves,
How can we Explain Amagat Curves By Ven der Walls Equation?
(a) Amagat Curve For Different Gases
Van der Waals equation for 1 mole real gas, 
(P + a/V₂ )(V - b) = RT 
or, PV - Pb + (a/V) + (a/V₂) = RT 
Neglecting the small term (a/V₂), 
we get , PV = RT + Pb - (a/V
Using ideal gas equation for 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 uses to explain Amagat curve qualitatively at low pressure and moderate pressure region.
(a) For CO₂a’ is very high since the gas is easily liquefied. 
What is the Ven der Waals Equation?
(b) Amagat Curve For CO₂
Thus (a/RT) 〉b in the equation and {b - (a/RT)} = - ve . That is slope of Z vs P curve is ( -) for carbon dioxide at moderate pressure region. That is the value of Z decreases with increase of Pressure and it is also found in the curve of carbon dioxide.
For H₂a’ is very small and It is not easily liquefied.
 so a/RT〈 b and slope of Z vs P curve for H₂ becomes (+) ve and the value of Z increases with pressure.
(b) (i) When T〈 Tв that is 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 = Tв = a/Rb or, 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 inter molecular attraction of the gas. Z runs parallel to P axis up to certain range of pressure at low pressure region. 
(iii) When T 〉TB, that is T 〉a/Rb or, b 〉a/RT and {b - (a/RT} = (+)ve. 
That is value of Z increases with increase of P when T 〉TB. The size effect dominates over the effect due to inter molecular attraction. For hydrogen and helium, 0°C is grater then their TB 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/RTP are negligible and Z=1 that is the gas obeys ideal behavior.