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Nov 10, 2018

Ionization Potential

Ionization Potential:

If enough energy is supplied to an electron of a particular atom, it can be promoted to successively higher energy levels. If the supplied energy is large enough it may possible to remove the electron completely from the pull of the positive nucleus.
The amount of energy required to remove the most loosely bound electron, that is outermost electron from an isolated gaseous atom of an element in its lowest energy state, that is ground state to produce cation is known as Ionization Potential or Ionization Energy of that element.
M (isolated gaseous atom) + Ionisation Energy M⁺ ( gaseous cation) + e (electron)
It is generally represented as I or IP and it is measured in electron volt (eV) or Kilo calories (Kcal) per gram atom.
One electron volt (eV) being the energy gained by an electron falling through a potential difference of one volt.
∴ 1 eV = (charge of an electron) × (one volt)
= 1.6 × 10⁻¹⁹ coulomb volt = 1.6 × 10⁻¹⁹ Joule = 1.6 × 10⁻¹² erg
Problem:
Calculate the ionization potential of hydrogen atom in eV.
Answer:
We know that, the energy corresponding to the transition from n = ∞ to n = 1 gives the ionization potential of the hydrogen atom.
Thus the Ionization Potential of the Hydrogen atom = (2π²me⁴/h²) {(1/1²) - 0}
 = 2.179 × 10⁻¹¹ erg = 2.179 × 10⁻¹⁸ Joule 
= (2.179 × 10⁻¹⁸)/(1.6 × 10⁻¹⁹) eV = 13.6 eV
Problem:
Calculate the second ionization potential of helium (given ionization potential of hydrogen = 13.6 eV)
Answer:


The ground state electronic configuration of helium is 1S². The second ionization potential means removal of the second electron from the 1S orbital against the nuclear charge of +2.
So we have: Ionization Potential = (2π²me⁴/){(1/n₁²) - (1/n₂²)} = Z² ×
∴ Second Ionization Potential of Helium = 2² × 13.6 = 54.4 eV

Successive Ionization Potentials:

The electrons are removed in stages one by one from an atom. The energy required to remove the first electron from a gaseous atom is called its first Ionization potential. The energy required to remove the second electron from the cation is called second Ionization potential. Similarly we have third, fourth Ionisation potentials.
Thus, M (g) + IP₁ M⁺ (g) + e
M⁺ (g) + IP₂ M⁺² (g) + e
M⁺² (g) + IP₃ M⁺³ (g) + e
The values show that these increase in order:
IP₁〈 IP₂ 〈 IP₃〈 IP₄
The successive increase in their values is due to the fact that it is relatively difficult to remove an electron from a cation having higher positive charge than from a cation having lower positive charge or from neutral atom.

Factor Affecting the Magnitude of Ionisation Potential and its Periodic Variation: 

The following factors influence the magnitude of the Ionization Potentials:
(1) The distance of the electron from the nucleus.
(2) The charge on the nucleus, that is, atomic number.
(3) The screening effect of lower lying inner electrons.
(4) Overall charge on the ionizing species.

The distance of the electron from the nucleus:

The greater the distance of an element from the positive charge nucleus of an element, the weaker will be the attraction and hence the lower the ionization potential of that element. If an atom is raised to an exited state by promoting one electron to a higher energy level, then that electron is further from the nucleus and it is observed that this excited electron is more easily detached than when the electron is in its ground state.
Thus on moving from top to bottom in a group the ionization potential of the elements decreases with the increase of their nucleus electron separation.
The distance of the electron from the nucleus
First Ionization Potential in a Group

The Charge on the Nucleus that is, Atomic Number:

The higher the charge on the nucleus the more difficult it is to remove an electron and hence the higher the value of ionization potential.
Thus the value of ionisation potential generally increases in moving from left to right in a period, since the nuclear charge of the elements (atomic number) also increases in the same direction.
The Charge on the Nucleus that is, Atomic Number
Variation of Ionization Potential along the 2nd Period
With increasing atomic number electrons are added to orbitals of the same principal quantum number. These electrons add little to the size of the atom, the increasing nuclear charge bring about a contraction in size. In effect therefore ionisation potential steadily increases along a period.

Completely - filled and half - filled orbitals:

According to the Hund's rule atoms having half - filled or completely filled orbitals are comparatively more stable and hence move energy is needed to remove an electron from such atom. The ionization potential of such atoms is therefore relatively higher than expected normally from their position in the periodic table.
A few regulation that are seen in the increasing vale of ionization potential along a period can be explain on the basis of the concept of the half - filled and completely filled orbitals.
Example:
Be and N in the second period and Mg and P in the third period have slightly higher vale of ionization potentials than those normally expected.
This is explained on the basis of extra stability of the completely - filled 2S - orbital in Be (2S²) and 3S - orbital in Mg (3S²) and of half - filled 2P - orbital in N (2S²2P³) and 3P - orbital in P (3S²3P³).

The Screening effect of the lower lying inner electrons:

Electrons provide a screening effect on the nucleus. The outermost electrons are shielded from the nucleus by the inner electrons. The radial distribution functions of the S, P, d orbitals shows that for the same principal quantum number the S - orbital is the most penetrating., next is P - orbital and least penetrating is the d - orbital. Screening efficiency falls off in the order S〉P〉d.
As we move down a group, the number of inner - shells increases and hence the ionisation potential tends to decreases.
Elements of II A Group: BeMgCaSrBa
Problem:
In first transition series electron filling up processes begins in the 3d level below a filled 4S² level. During ionization process will a 4S electron or a 3d electron be lost first ? Explain with reference to Chromium.
Answer:


Electronic Configuration of Chromium : 1S² 2S² 2P⁶ 3S² 3p⁶ 3d⁵ 4S¹.
Screening constant (σ) for 4S electron is : 
σ = (2 × 1.0) + (8 × 1.0) + (8 × 0.85) + (5 × 0.85)= 21.05.
∴Effective nuclear charge = (24 - 21.05) = 2.95.
Screening constant (σ) for 3d electron is : 
σ = (2 × 1.0) + (8 × 1.0) + (8 × 1) + (4 × 0.35)= 19.4.
∴Effective nuclear charge = (24 - 19.4) = 2.95.
Hence a 3d electron is more tightly held than a 4S electron. So during ionization the 4S electron lost first.

Overall Charge on the Ionizing Species:

An increasing in the overall charge on the ionizing species (M⁺, M⁺², M⁺³, etc) will enormously influence the ionization potential since electron withdrawal from a positive charged species is more difficult than from a neutral atom.
The first ionization potentials of the elements very with their positions in the periodic table. In each of the table the noble gas has the highest value and the alkali metals the lowest value for the ionization potential.
Overall Charge on the Ionizing Species
First Ionization Potentials as Functions of Atomic Number
Again the Ionization Potentials of an elements in a group gradually decreases with increasing atomic number.
As for example, Li〉Na〉K〉Rb〉Cs
Problem:
Arrange the following with the increasing order of Ionization Potentials Li, Be, B, C, N, O, F, Ne.
Answer:
LiㄑBㄑBeㄑCㄑOㄑN〈FㄑNe

Related Posts:

Electronic Configuration of Elements
Structure of Atom

Nov 8, 2018

Formulation of Kinetic Gas Equation

Formulation of Kinetic Gas Equation:

After knowing the experimental gas lows, it is of interest to develop a theoretical model based on the structure of gases, which can correlated to the experimental facts. Fortunately such theory has been developed and known as the Kinetic Theory of Gases based upon the certain postulates which are supposed to be applicable to an ideal gas.
The essential Postulates are:
1.The gas is composed very small discrete particles, now called molecules. For a gas the mass and size of the molecules are same and different for different gases.
2.The molecules are moving in all directions with verity of speeds. Some are very fast while other are slow.
3.Due to random motion, the molecules are executing collision with the walls of the container (wall collision) and with themselves (inter molecular collision). These collision are perfectly elastic and so there occurs no loss of kinetic energy or momentum of the molecules by this collision.
4.The gas molecules are assumed to be point masses, that is their size is very small in comparison to the distance they travel.
5.There exist no inter molecular attraction specially at law pressure, that is one molecule can exert pressure independent of the influence of other molecules.
6.The pressure exerted by a gas is due to the uniform wall collisions. Higher the frequency of the wall collision greater will be the pressure of the gas. 
This explains Boyle's Law since when volume is reduced, wall collision becomes more frequent and pressure is increased.
7.Through the molecular velocity are constantly changing due to inter molecular collision, average kinetic energy(є) of the molecules remains fixed at a given temperature. This explain the Charl's law that when T is increased , velocity are increased, wall collision become more frequent and pressure(P) is increased when T kept constant or Volume(V) is increases when P kept constant.
Root Mean Square Speed: 
Root mean square(RMS) speed is defined as the square root of the average of the squares of speeds. 

That is, RMS = (N₁C₁² + N₂C₂² + N₃C₃²......)/N
Derivation of the Kinetic Gas Equation:
Let us take a cube of edge length l containing N molecules of a gas of molecular mass m and RMS speed is CRMS at temperature T and Pressure P. 
Molecular Velocity and its Components
Molecular Velocity and its Components
Let, in a gas molecules, N₁ have the velocity C₁, N₂ have the velocity C₂, N₃ have the velocity C₃, and so on.
Let us concentrate our discussion to a single molecule among N₁ that have resultant velocity C₁ and the component velocities are Cx, Cy and Cz.
So that, C₁² = Cx² + Cy² + Cz²
The Molecule will collide walls A and B with the Component Velocity Cx and other opposite faces by Cy and Cz.
Change of Momentum along X, Y and Z Direction
Change of Momentum along X, Y and Z Direction
Change of momentum along X-direction for a single collision = mCx - (-mCx) = 2 mCx.
Rate of change of momentum of the above type of collision = 2 mCx × (Cx/l) = 2 mCx²/l.
Similarly along Y and Z directions, the rate of change of momentum of the molecule are 
2 mCy²/l and 2 mCz²/l respectively.
Total rate change of momentum for the molecule, 
= (2 mCx²/l) + (2 mCy²/l) + (2 mCz²/l
= 2m/l (Cx² + Cy² + Cz²)
= 2 mC₁²/l
For similar N₁ molecules, it is 2 mN₁C₁²/l.
Taking all the molecules of the gas, the total rate of change of momentum equal to
(2 mN₁C₁²/l) + (2 mN₂C₂²/l) + (2 mN₃C₃²/l) + .......
Thus Rate of Change of Momentum = 2mN/l{(N₁C₁² + N₂C₂² + N₃C₃²)/N} = 2 mNC²/l
Where C = Root Mean Square Velocity of the Gases.

According to the Newton's 2nd Low of Motion, rate of change of momentum due to wall collision is equal to force developed within the gas molecules.
That is, P × 6l² = 2 mNC²/l
or, P × l³ = (1/3)mNC²
or, PV = (1/3)mNC²
The other from of the equation is P = (1/3)(mN/V)C²
P = (1/3)dC²
Where (mN/V) is the density(d) of the gas molecules.
The Kinetic Gas Equation
The Kinetic Gas Equation
These equation are also valid for any shape of the gas container.

Problem:
Calculate the pressure exerted by 10²³ gas particles each of mass 10⁻²² gm in a container of volume 1 dm³. The root mean square speed is 10⁵ cm sec⁻¹.
Solution:

From the given data, we have N = 10²³, m = 10⁻²² gm = 10⁻²⁵ Kg, V = 1 dm³ = 10⁻³ m³ and Root Mean Square(C) = 10⁵ cm sec⁻¹ = 10³ m sec⁻¹.
Therefore, from Kinetic Gas Equation, PV = (1/3)mNC²
or, P = (1/3)(mN/V)C²
Putting the value We have P = (1/3)(10⁻²⁵ Kg × 10²³/10⁻³ m³) × (10³ m sec⁻¹)²
or, P =  0.333 × 10⁷ Pa
Expression of Root Mean Square Velocity:
Let us apply the kinetic equation for 1 mole Ideal Gas. In that case mN = mN₀ = M 
and PV = RT.
Hence from Kinetic Gas Equation PV = (1/3)mNC²
RT = (1/3)MC²
or, C² = 3RT/M
or, C = √(3RT/M)
Thus root mean square velocity depends on the molar mass(M) and temperature(T) of the gas.
Problem:
Calculate the root mean square speed of oxygen gas at 27°C.
Solution:

We know that Crms = √(3RT/M).
Here, M = 32 gm mol⁻¹ and T = 27°C = (273+27)K = 300 K.
Thus Crms = √{(3 × 8.314 × 10⁷ erg mol⁻¹K⁻¹ × 300 K)/(32 gm mol⁻¹)} 
= 48356 cm sec⁻¹
Problem:
Calculate the rms speed of ammonia in N.T.P.
Answer:
Here V = 22.4 dm³ mol⁻¹ = 22.4 × 10⁻³ m³ mol⁻¹, P = 1 atm = 101325 Pa and M = 17 × 10⁻³ Kg mol⁻¹.
Thus Crms = √(3PV/M) = √{(3 × 101325 × 22.4 × 10⁻³)/(17 × 10⁻³)} 
= 632 m sec⁻¹
Expression of Average Kinetic Energy(Ē):
The average kinetic energy(Ē) is defined as Ē = (1/2)mC².
Again from Kinetic Gas Equation PV = (1/3)mNC²
or, PV = (2/3)N{(1/2)mC²}
or, PV = (2/3) N Ē
For 1 mole ideal gas PV = RT and N = N₀
Thus RT = (2/3)N₀ Ē
or, Ē = (3/2)(R/N₀)T = (3/2)kT
Where k = R/N₀ and is known as the Boltzmann Constant. Its value is 1.38 × 10⁻²³ JK⁻¹.
The total kinetic energy for 1 mole of the gas is, Etotal = N₀(Ē) = (3/2)RT.
Thus Average Kinetic Energy is dependent of T only and it is not dependent on the nature of the gas.
Problem:
Calculate the kinetic energy of translation of 8.5 gm NH₃ at 27°C.
Solution:

We know that total kinetic energy for 1 mole of the gas is, Etotal = (3/2)RT 
= (3/2)(2 cal mol⁻¹ K⁻¹ × 300 K) = 900 cal mol⁻¹
Again 8.5 gm NH₃ = (8.5/17) mol = 0.5 mol
Thus Kinetic Energy of 8.5 gm NH₃ at 27°C is (0.5 × 900) cal = 450 cal

Related Post:
Ven der Waals Equation

Nov 6, 2018

Rutherford Experiment

Rutherford Experiment:

Rutherford and his students describes some alpha particle scattering experiments. The alpha particles were already established by the group as He⁺² from their behavior in electric and magnetic fields. A beam of alpha particles obtained from spontaneously disintegrating Polonium, were directed on to a very thin foil of Platinum or Gold.
With the help of fluorescent zinc sulphide screen around the Platinum or Gold foil, any deflection of alpha particle was observed. 
Scattering of Alpha Particles from a Metal Foil
Scattering of Alpha Particles from a Metal Foil
The vast majority of the alpha particles passed straight line through the foil. But a very limited few were found to be deflected back from the foil, some even appearing on the side of incidence.
Rutherford concluded that since most of the alpha particles passed straight through there must be a very large volume of empty space in the atom of the platinum or gold.
A very small part of the platinum or gold atom must be responsible for the large scattering of the few alpha particles, central part was called atomic nucleus
Conclusions of Rutherford experiment:
The following Conclusions and an atomic model emerged from the Rutherford's experiment.
1. All the positively charged and almost entirely mass of the atom was concentrated in very small part of the atom and these central core called atomic nucleus.
2.The large deflection of alpha particle from its original path was due to Coulombic repulsion between the Alpha particle and Positive Nucleus of an atom. Simple impact between the two such massive particles can lead to a Scattering of the order of only .
3. An alpha particle suffers little deflection while passing by an electron
4.The Radius of atomic nucleus is ∼ 10⁻¹³ being the same as that of an electron. Since the radius of an atom is ∼ 10⁻⁸ it is obviously that an atom must have a very empty structure. From the above Conclusions an atomic structure is proposed by Rutherford.
Rutherford's Atomic Model:
According to Rutherford's Model the entire mass of the atom was concentrated in a tiny, positively charged nucleus around which the extra nuclear electrons were moving in a circular orbits.   
Rutherford divided the atom into two part,
(i) Centre of Atom or The Nucleus and (ii) Extra nuclear part of atom or The extra nuclear Electrons.
Rutherford's Atomic Model
Rutherford Atomic Model
(i) Centre of Atom or The Nucleus:
Almost the entire mass of the atom is concentrated in a very small, central core called the atomic Nucleus.

Since the extra nuclear electrons contribute negligible to the total mass of the atom ans since the atom is electrically neutral it follows that the nucleus must carry particles which will account both for the mass and positive charge of the atom. 
(ii) The Extra Nuclear Electrons:
A very small positive nucleus was considered surrounded by electrons. Such a system cannot be stable if the election were in rest. Therefore it is proposed that the electron moving in circular orbits around the nucleus so that the Coulombic attraction between the nucleus and the electron was equal to the centrifugal force of attraction.
Rutherford's Model of the Hydrogen Atom
Rutherford's Model of the Hydrogen Atom
Defects of Rutherford's Model:
(i) The Rutherford model is however, not in conformity with the classical model of electromagnetic radiation. A moving charged particle will emit radiation, will then loss kinetic energy and eventually will hit the nucleus.
(ii) If the electrons lose energy continuously, the observed atomic spectra should be continuous, consisting of broad bands merging one into the other.

Related Post:

Atomic Structure

Atomic Structure:

Constituent Particles of an Atom:
Dalton’s Model to Modern Structure Of Atom: 
1.All the matter is made of atoms are indivisible and indestructible. 
2. All the atoms of a given element are identical mass and properties. 
3. Compounds are formed by combination of two or more same or different kind of atoms. 
4. A chemical reaction is rearrangement of atoms. 
Rutherford has remarked that it is not in the nature of things for any one man to make sudden violet discovery. Science goes step by step and every man depends on the work of his predecessor. 
The journey from Dalton model of the atom to modern structure of atom was long and arduous one. At turn this century many valuable information were being compiled. 
This clearly indicates that Dalton’s atomic model no longer enjoyed the exalted position to grant it. 
Today an atom is considered to made up of a tiny nucleus carrying neutrons and protons. This tiny nucleus has around itself a certain number of negatively charged particle carrying negligible mass, called electrons, arranged in a definite order.

Discussion on the Fundamental Particles of Atom:
Cathode rays - Discovery of Electron:
Gases at low pressures, when subjected to high potential, becomes conducting and various luminous effect were observed. When the pressure is quite low(0.01 mm), the tube remains dark (Crooks dark space) but a streak of rays, named cathode rays, emanate from the cathode. This is confirmed from the fact that a fluorescence is produced on the opposite wall where the rays impinge. The cathode rays have been very carefully studied for their many definite characteristics.
Cathode rays
Deflection of Cathode Rays on Electric Field
Characteristics of Cathode Rays:
(i) Cathode rays excite fluorescence on the glass walls where they impinge.
(ii) The rays have travel in straight lines, confirmed by the shadows of objects placed in their path.
(iii) The rays have penetrating power and can pass through thin metal foils.
(iv) They also possess considerable momentum, small paddle wheels placed in their path rotate from the impact with the rays.
(v) The cathode rays are deflected from their path by the application of magnetic or electrostatic field. From the direction of deflection, the charge accompanying the rays is a negative one.
(vi) When the rays impinge on a metal target, called anti cathode, and placed on the path , a different type of radiation, the X-rays is produced. these new radiation not deflected in an electric or magnetic field. X-rays are really electromagnetic radiation of very short wave length. 
Thus the cathode rays is a stream of negatively charged particles, called electrons and this was first established by J.J. Thomson in 1897
Charge of an Electron:
The electron carrying negatively charged and the charge of an electron(e), 4.8 × 10 ⁻¹⁰ esu = 1.60 × 10⁻¹⁹ coulomb.
Mass of an Electron:
Let the mass of an electron = m and charge = e, then e/m = 1.76 × 10⁸ Coulomb/gram.
Mass of an electron = (1.60 × 10⁻¹⁹)/(1.76 × 10⁸) gram
= 9.11 × 10⁻²⁸ gram.
Coulometric Determination of the Electronic Charge:


Electrode position of silver from an aqueous solution of a silver salt is a suitable experiment for the determination of the electronic charge.
Faraday's Low are readily interpreted by reference to the electrolysis of Silver Nitrate. The change at the cathode requires one electron for every Silver ion reduced. 
That is, Ag⁺ + e Ag 
If the electrons consumed at this electrode is equal to Avogadro Number (6.023 × 10²³ mol⁻¹), 1 mole of Silver metal (107.9 gm) is produced. At the same time, 1 mole of electrons is removed from the anode and 1 mole of nitrate ions is discharged.
Thus, 96500 coulomb of electricity which is necessary to produce 1 equivalent mass of a substance at the electrode will be total charge carried by 1 mole of electrons.
Hence Charge carried by each electron is given by,
e = (96500 C mol⁻¹)/(6.023 × 10²³ mol⁻¹)
= 1.60 × 10⁻¹⁹ C

Positive Ray - Discovery of Protons: 
Since the Electrons contribute negligible to the total mass of the atom and the atom is electrically neutralThat the nucleus must carry particles which will account both for the mass and positive charge of the atom.
We have so far deliberately restricted the discussion on the discharge phenomena at low gas pressure. The production of cathode rays in discharge tubes inspired physicists to look for the oppositely charged ions, namely positive ions.
Goldstein added a new feature to the discharge tubes by using holes in the cathode. With this modification it is observed that on operating such discharge tubes there appeared not only cathode rays travelling from the cathode to anode but also a beam of positively charged ions travelling from around anode to cathode. Some of the positively charged particles passed through the hole in the cathode and produces a spot on the far end of the discharge tube.
Positive Ray - Discovery of Protons
Goldstein Experiment (Discharge Tubes Using Holes In The Cathode)
The nature of these positive rays is extensively investigated by Thomson. It proved much more difficult to analyse the beam of the positive rays than to analyse a beam of electrons.
On deflection by a magnetic and electric field the positive ray beam produced a large defuse spot indicating that the e/m ratio of the constituents of the beam was not same and that the particles moved with different velocities.
Thomson further demonstrated that each different gas placed in the apparatus gave a different assortment of e/m.
Since a H⁺ is produced from a hydrogen atom by the loss of one electron, which has but a negligible mass, it follows that the mass of a H⁺ is same as that of the hydrogen atom(=1).
The particle represented by H⁺ is called a Proton and is considered a fundamental constituent that accounts for the positive charge of the nucleus.
Charge of a Proton:
The Proton carrying Positively charged and the charge of a proton(p), 4.8 × 10 ⁻¹⁰ esu = 1.60 × 10⁻¹⁹ coulomb.
Mass of an Proton:
Let the mass of an Proton = and charge = e, then e/m = 9.3 × 10⁴ Coulomb/gram.
∴ Mass of an Proton = 1.6725 × 10⁻²⁴ gm.
Discovery Of Neutron:
Attempts were now directed towards a correlation of atomic mass number ( = integer nearest to the atomic weight) and nuclear charge (= atomic number). If A stands for the mass number and Z for the nuclear charge of an element, then Z units of nuclear charge means Z number of proton inside the nucleus. But Z protons can contribute only Z mass units. 
The shortfall of the (A - Z) mass units bothered chemists and physicists for the quite same time. Rutherford then suggested this shortfall must be made up by another fundamental particle. This particle has electrically neutral, and mass equal to that of the proton, namely 1. He named the particle in advance as neutron.
The glory of discovering the neutron went to Chadwick, one of Rutherford Students.
An interpretation of the atomic nuclei on the basis of neutrons and protons is now a simple affair. Taking oxygen of mass number 16, for example and recalling that the atomic number of the element is 8, we have an atomic nucleus composed of 8 protons and 8 neutrons.
Since neutrons contribute only to the mass of the element but does nothing towards charge it follows that there may exist species with the same number of protons but varying numbers of neutrons inside the nucleus. 


Such species must belong to the same element and must vary only in their mass numbers. They are called isotopes.
Thus ₁H¹, ₁H², 1H³ are three isotopes of hydrogen, and ₈O¹⁶, ₈O¹⁷ and ₈O¹⁸ are three isotopes of oxygen.

Related Post:

Critical Constants of a Gas

Critical Constants of a Gas:

A gas can be liquefied by lowering temperature and increasing of pressure. But influence of temperature is more important. Most of gases are liquefied at ordinary pressure by suitable lowering of the temperature. But a gas can not be liquefied unless its temperature is below of a certain value depending upon the nature of the gas whatever high the pressure may be applied.
This temperature of the gas is called its Critical Temperature (Tc). A gas can only be liquefied when the temperature is kept below Tc of the gas.

Definitions of Critical Constants:

Critical Temperature (Tc):
Critical Temperature (Tc) is the maximum temperature at which a gas can be liquefied, that is the temperature above which a liquid cannot exist.
Critical Pressure (Pc):
Critical Pressure (Pc) is the maximum pressure required to cause liquefaction at the temperature (Tc).
Critical Volume (Vc):
Critical Volume (Vc) is the volume occupied by one mole of a gas at critical temperate (Tc)  and Critical Pressure (Pc).

Andrews Isotherms:

In 1869, Thomas Andrews carried out an experiment in which P - V relations of carbon dioxide gas were measured at different temperatures
Andrews Isotherms
Andrew's Graphs of P Vs V
Following are observed from this graph:
(1) At high temperature, such as T₄, the isotherms look like those of an ideal gas.
(2)At low temperatures, the curves have altogether different appearances. Consider, for examples, a typical curve abcd. As the pressures increases, the volume of the gas decreases (curve a to b). At point b liquefaction commences and the volume decreases rapidly as the gas converted to a liquid with much higher density. This conversion takes place at constant pressure P. At the point C, liquefaction is complete and thus the cd is evidence of the fact that the liquid cannot be easily compressed. Thus, we note that ab represents the gaseous state, bc, liquid and vapour in equilibrium, and cd shows the liquid state only.
(3) At still higher temperatures we get a similar type of curve as discussed in the above point, except that the horizontal portion is reduced. The Pressure corresponding to this portion being higher then at lower temperatures.
(4) At temperatures Tc the horizontal portion is reduced to a mere point. At temperatures higher then Tc there is no indication of qualification at all.
Thus for every gas, there is a limit of temperature above which it cannot be liquefied, no matter what the pressure is.

Continuity of State:

It appears from the Amagat curve at Tthere is discontinuity or break during the transformation of gas to liquid. But it is not so.
The continuity of the states from the gas to liquid can be explains from the following isotherm pqrs at T
Continuity of State
Schematic Representation of the Continuity of State
The gas at A is heated to B at constant volume (V) along AB. Then the gas is gradually cooled at constant P along BC, the volume is reduced considerably. The gas is again cooled at constant V until the point D is reached. 
No where in the process liquid would appear. At D, the system is highly compressed gas. But from the curve, this point is the representation of the liquid state. Hence there is hardly a distinction between the liquid state and the gaseous state. There are no line of separation between the two phases. This is known as the principle of continuity of state.

Critical Phenomena and Van der Walls Equation:

For one mole of a gas the Van der Waals equation,
(P + a/V²)(V - b) = RT 
This equation can be written as,V³ - (b + RT/P)V² + (a/P)V - ab/P = 0
This equation has three roots in V for given values of a, b, P and T. It found that either all the three roots are real or one is real and the other two are imaginary.
Critical Phenomena and Van der Walls Equation
Van der Waals Isotherms
The main characteristics of the above isotherm are as follows:
(a) At higher temperature and higher volume region, the isotherms looks much like the isotherms for a Ideal gas.
(b) At the temperature lower than Tc the isotherm exhibits a maximum and a minimum. For certain values of pressure, the equation gives three roots of volume as V₁, V₂ and V₃ at pressure P₁. The section AB and ED of the Van der walls curve at T₂ can be realized experimentally. ED represents Supersaturated or Super cooled vapour and AB represents super heated liquid. Both these states are meta-stable. These states are unstable in the seance that slight disturbances are sufficient to cause the system to revert spontaneously into the stable state with the two phases present in equilibrium.
(c) The section BCD of the Van der Walls isotherms cannot be realized experimentally. In this region the slope of the P - V curve is positive. Increasing or decreasing the volume of such a system would increase or decrease of pressure. The line BCD represents the meta- stable state.

Expression of Critical Constants in Terms of Van der Waals Constants:

Again with increase of temperature the minimum and maximum points come close to each other and at the critical point (C) both maximum and minimum coalesce. The slope and curvature both becomes zero at this point
Thus the mathematical condition of critical point is,
(dP/dV) = 0 and (d²P/dV²) = 0
Van der Waals equation for 1 mole gas is
(P + a/V²)(V - b) = RT
or, P = RT/(V - b) - a/V²
Differentiating Van der Waals equation with respect to V at constant T,
We get Slope = (dP/dV)ᴛ = - {RT/(V - b)²} + 2a/V³

And Curvature = (d²P/dV²)ᴛ = {2RT/(V - b)³} - 6a/V⁴

Hence at the critical point,

- {RTc/(Vc - b)²} + 2a/Vc³ = 0

or, RTc/(Vc - b)² = 2a/Vc³
 and {2RTc/(Vc - b)³} - 6a/Vc⁴ = 0

or, 2RTc/(Vc -b)³ = 6a/Vc

Thus, (Vc - b)/2 = Vc/3

or, Vc = 3b

Putting the value of Vc = 3b in RTc/(Vc - b)² = 2a/Vc³.
We have, RTc/4b² = 2a/27b³
or, Tc = 8a/27Rb
Again the Van der Walls equation at the critical state is
Pc = RTc/(Vc - b) - a/Vc²
Putting the value of Vc and Tc,
We have Pc = a/27b²
Problem:
The Critical constants for water are 647 K, 22.09 MPa and 0.0566 dm³mol⁻¹. Calculate the value of Van der Waals constants a and b.
Answer:
a = 212.3 KPa dm⁶ mol⁻² and b = 0.0189 dm³ mol⁻¹
Full Details Click Here
Compressibility Factor at the Critical State (Zc):
The Critical coefficient is defined as RTc/PcVc
Thus the value of Critical Coefficient = {R × (8a/27Rb)}/{(a/27b²) × 3b} = 8/3 = 2.66
Thus the value of Compressibility factor at the critical state (Zc) = PcVc/RTc = 3/8 = 0.375
Van der Waals constants in terms of critical constants:
Van der Waals constants can be determind from critical constants Tc and Pc of the gas. Vc in the expression is avoided due to difficulty in its determination.
We have, b = Vc/3 but PcVc/RTc = 3/8 or, Vc = (3/8) × (RTc/Pc)
hence, b = (1/8)(RTc/Pc)
Again, a = Pc × 27b²  = 3 × Pc × (3b)² = 3 PcVc² = 3Pc × (3RTc/8Pc)²
∴ a = (27/64)(R²Tc²/Pc) 
Problem:
Calculate Van der Waals constants for Ethylene. (Tc = 280.8 K and Pc = 50 atm)
Answer:
a = 0.057 lit mol⁻¹ and b = 4.47 lit² atm mol⁻²
Full Details Click Here

Problem:
Argon has (Tc = - 122°C, Pc = 48 atm). What is the radius of the Argon atom?
Answer:
1.47 × 10⁻⁸ cm
Full Details Click Here
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