2019- PG –TRB - UNIT 5- FULL MATERIAL, TESTS & KEY · Web viewPartition functions – Sackurtetrode equation –statistical approach to the third law of Thermodynamics – Maxwell

2019- PG –TRB - UNIT 5- FULL MATERIAL, TESTS & KEY

PG- TRB- CHEMISTRY

UNIT-5COMPLETE MATERIAL,

QUESTION PAPERS&

KEYBY

Dr. C.SEBASTIAN ANTONY SELVAN

ASST. PROFESSOR OF CHEMISTRY

R. V. GOVT.ARTS COLLEGE

CHENGALPATTU

9444040115

2019

C. SEBASTIAN, A/P/CHEM, R.V.GOVT.ARTS COLLEGE, CHENGALPATTU, MOB: 9444040115 Page 1


UNIT V

PART -1- THERMODYNAMICS and GASEOUS STATE

Thermodynamic equations of state – closed and open systems – partial molal quantities –

chemical potential with temperature and pressure – third law of thermodynamics.Fugacity –

methods of determination – activity and activity co-efficient – standard states for gasesliquids –

solids and solutions – mean activity co-efficients of electrolytes.Maxwell’s distribution of

molecular velocities – derivation of expression for average, most probable andrcot mean square

velocities – Microstates Macrostates –

PART -2- STATISTICAL THERMODYNAMICS ,EQUILIBRIUM and PHASE RULE

Partition functions – Sackurtetrode equation –statistical approach to the third law of

Thermodynamics – Maxwell Boltzmann – Bose Einstein and FermiDirace statistics – Heat

capacities of solids – Einstein and Debye Models Low temperature – Negativeabsolute

temperature.Chemical equilibrium – thermodynamic derivation of equilibrium constant –

standard free energy– calculations.Phase equilibrium – thermodynamic derivation of phase rule

application of phase rule – threecomponent systems.

PART -3- ANALYTICAL CHEMISTRY

Chromotography – column, paper, thinlayer, gas-liquid, High pressure liquid

chromatographyHPLC principle and applications.Thermal analysis – different thermal analysis

(DTA) – Principle and applications –thermogravimetric analysis (TGA) Principle and

application.Chemical crystallography – Diffraction methods – X ray Neutron, electron

diffraction methods.Principle and applications.Polarimetry – Circular ichroism – Optical

Rotatory dispersion (ORD) Principle and applications.



PART -13 - THERMODYNAMICS and GASEOUS STATE

1. Thermodynamic Equations Of State

2. Closed And Open Systems

3. Partial Molal Quantities

4. Chemical Potential With Temperature And Pressure

5. Third Law Of Thermodynamics

6. .Fugacity

7. Methods Of Determination

8. Activity And Activity Co-Efficient

9. Standard States For Gases Liquids, Solids And Solutions

10. Mean Activity Co-Efficients Of Electrolytes.

11. Maxwell’s Distribution Of Molecular Velocities

12. Derivation Of Expression For Average,

13. Most Probable And

14. Root Mean Square Velocities

15. Microstates Macrostates –



115THERMODYNAMIC EQUATIONS OF STATE(application of Maxwell’s relations)

Thermodynamic equation relates state variables which describe the state of matter under a

given set of physical conditions, such as pressure, volume, temperature (PVT), or internal

energy.The two thermodynamic equations of state are

The variation of U with volume at constant T for the system

1. ¿)T = T ¿)V - P

The variation of H with pressure at constant T for the system

2. ¿)T = V - T ¿)P

215CLOSED and OPEN SYSTEM:

A system which can exchange matter as well as energy with its surroundings called open

system

dT≠ 0 , dQ≠ 0, dE≠ 0, dm≠ 0

A system which can exchange energy but not matter is called closed system.

dQ≠ 0, dT≠ 0 , dE≠ 0, dm = 0



315PARTIAL MOLAR QUANTITIES [Thermodynamics of open system ]

In the case of open system containing two or more components, there can be change in

the number of moles of various components. In that case the extensive properties ( which

depends on number of moles) like U,H,S,A, and G are not only the function of temperature

and pressure, but also of the number of moles of various components present in the system.

The partial molar propertyX is defined as the rate of variation in the property X , with

a change in number of moles at constant temperature and pressure.

i.e X = f ( T,P, n1,n2,n3...), where X denotes U,H,S,A, and G .

The quantity (∂ X∂ n 1

) T, P, n2,n3... is called partial molar property. Thus for any component i

( ∂ U∂ n 1

) T, P, n2,n3. – partial molar internal energy = U

( ∂ H∂ n 1

) T, P, n2,n3... partial molar enthalpy =H

( ∂ S∂ n 1

) T, P, n2,n3... partial molar entropy = S

( ∂V∂ n 1

) T, P, n2,n3... partial molar volume = V



415CHEMICAL POTENTIAL( Partial Molar Free Energy)

The chemical potential of a given substance is the change in the free energy of the system

that results on the addition of one mole of that particular substance at constant temperature and

pressure . It is expressed as

μi = (∂ G∂ n1

) T, P, n2,n3...

VARIATION OF CHEMICAL POTENTIAL WITH TEMPERATURE

The variation of chemical potential with temperature is given by

( ∂ μ∂ T )P , N= -¿ T, P

= - partial molar entropy of component i.

= - S

dμ = - S dT

This shows that when temperature increases, chemical potential decreases.

The graph of chemical potential versus temperature for a substance in solid, liquid and

gaseous state shows that at the melting point the chemical potential of solid and liquid are

same.

Similarly at the boiling point, the chemical potential of liquid and vapour are same.



VARIATION OF CHEMICAL POTENTIAL WITH PRESSURE

The variation of chemical potential with pressure is given by

( ∂ μ∂ P )T,n1,n2.. = ¿T, P

= partial molar volume of component i.

GIBBS – DUHEM EQUATION:

Consider a system at temperature T and pressure P. The Gibbs free energy of a system

containing n1 moles of constituent 1, n2 moles of constituent 2... is given by

G = f ( T, P, n1,n2,n3....).This can be reduced in to

( n1 dμ1+ n2 dμ2 + n3 dμ3..) = 0



∑ ¿dμi= 0

This is known as Gibbs –Duhem equation.For system involving two constituents the above

equation becomes

n1 dμ1+ n2 dμ2 = 0

n1 dμ1 = - n2 dμ2

This shows that , the change in chemical potential with respect to any constituent is not

independent, but it depends the change in chemical potential of the constituents

515 THIRD LAW OF THERMODYNAMICS

Nernst heat theorem:

When the temperature is lowered towards the absolute zero, the value of ∂ ∆ G∂T

approaches zero gradually limT → 0

[ ∂ ∆ G∂ T

]P = 0

When the temperature is lowered towards the absolute zero, the value of ∂ ∆ H

∂T

approaches zero gradually limT → 0

[ ∂ ∆ H∂ T

]P = 0

thereforelimT → 0

[∆ S ] = 0 and limT → 0[CP] = 0



III Law Statement:

At the absolute zero of temperature, the entropy of every substance is zero

.limT → 0[S] = 0

Entropies calculated by Third law of thermodynamics is called thermal entropies.

Third Law of Thermodynamics is concerned with the limiting behavior of systems as the

temperature approaches absolute zero.

The Third Law states, “The entropy of a perfect crystal is zero when the temperature of the

crystal is equal to absolute zero (0 K).”

Exception to III law:

Entropies of H2,D2, CO, NO, N2O, H2O are not zero at zero K.. These entropies are

called Residual entropy and is due to alternate arrangements of molecules in the solid.



BOLTZMANN LAW

S = k ln Wwhere k – Boltzmann constant, W – probability

Entropies calculated by Boltzmann law ( statistical) is called statistical entropies.

RESIDUAL ENTROPY

Residual entropy = Statistical entropies - Thermal entropies

CO, NO and N2O have two types of arrangements

CO and OC , NO and ON, NNO and ONN

Therefore W = 2N where N is Avagadro number

S = k ln W

= k ln 2N

= N k ln 2

= R ln 2

= ( 8.314) (0.692)

= 5.76 J/K/mol

615 FUGACITY [THERMODYNAMICS OF REAL GASES]

It is the pressure term for real gas. It is represented by the letter f



Relation between fugacity and free energy If f1 and f2 are fugacities of a gas at two different pressures P1and P2 and at constant

temperature T , then

∆ G = nRT ln f 2

f 1

where ‘n’ is the number of moles. R – gas constant

Problem 1.Calculate the free energy change accompanying the compression of one mole of a

gas at 100 K, from 20 to 200 atm. The fugacitities at 20 and 200 atmpressure are 50 and 100

atm respectively.

Solution:

Number of moles = 1

Temperature T = 100K

Initial fugacity f1 = 50 atm

Final fugacity f2 = 100 atm

free energy ∆G = ?

The relation between fugacity and free energy ∆G = nRT ln f 2

f 1

G = nRT ln f 2

f 1

= (1) (8.314) ( 100) ln 10050

= 831.4 ln 2

Fugacity at low pressure can be calculated byf = P2VRT

Problem 2.A gas at 0.1 atmpressure occupies a volume 83.14 lit . Calculateits fugacity at 0.1 K Solution: Fugacity f = ?Pressure P = 0.1 atmVolume V = 83.14 litTemperatureT = 0.1 K

Fugacity f = P2VRT



= (0.1)2× 83.148.314 ×0.1

= 1 atm

Variation of fugacity with temperature:

Variation of fugacity with temperature can be calculatedusing the following expression,

if the fugacity is known at one temperature

lnf 2f 1 = -

H1−H 2

R [

T1−T 2

T 1 T2 ] .

Where ‘H’ represents molar heat content.

Variation of fugacity with pressure:Variation of fugacity with pressure can be calculatedusing the following expressionif the

fugacity is known at one pressure

lnf 2f 1 =

VRT [ P2-P1 ]

715 METHODS OF DETERMINATION

Fugacity is determined by graphical method. In this method, for different pressure

value𝛂 ( departure from ideal behavior) is determined using the formula α = RTP – V

and a graph is plotted with pressure along x- axis and 𝛂 along y – axis .The area under the

curve gives the value of ∫0

P

αdP using this value, fugacity can be calculated by the formula

lnfp = - 1

RT ∫0

P

αdP

= - 1

RT × area under the curve



815 ACTIVITY & ACTIVITY COEFFICIENT:

Activity of a substance in any given state is defined as the ratio of the fugacity of the substance in the mixture to the fugacity of the same substance in the pure state.

Activity = fugacity∈mixture

fugacity∈ pure state

a = f 1

f 0

Problem 7: The fugacity of a pure gas is 20 atm and the same gas in a mixture is 40 atm . Calculate the activity of the gas.Solution:



a = 4020

= 2Activity coefficient:

For real gas the activity is proportional to pressure a ∝ P

a = 𝛄 P𝛄 is called activity coefficient.

Activity coefficient can be defined as the ratio between activity and pressure of the real gas.



activity coefficient 𝛄 = activitypressure

Problem 8: The activity of a gas at 20 atm pressure is 5 . What is its activity coefficient?

Solution:

Activity coefficient 𝛄 = activitypressure

= 202

= 2

Problem 9.The fugacity of a pure gas is 5 atm and the same gas in a mixture is 20 atm Calculate

the activity coefficient of the gas at 40 atm pressure

Solution:

Activity coefficient = activitypressure



a= 205

= 4

γ = 4

40

= 0.1

Variation of activity with temperature: ln a = H−H 0

R¿] .

Variation of activity with pressure: ln a = V

RT [ P2 – P1]

Determination of activity and activity coefficients of non electrolytes.

1. By Nernst distribution law 2. From vapour pressure:

3. From freezing point



4. From emf measurements

915CHOICE OF STANDARD STATES

1. For Gases :

1. Since the fugacities of real gases can be measured, the standard sate can be fixed in

terms of fugacity.

2. The fugacity corresponding to 1 atm, pressure is taken as standard state .

2. For Solution:

The standard state for solvent is based on Raoult’s law and that for solute is based on Henry’s

law

A. Solvent:

According to Raoult’s law f1 = fo x1

B. Solute:

1. If the solute and solvent are completely miscible in all proportions, the standard state

of solute is taken in the same manner as for the solvent.

2. When the solute has a limited solubility , two systems are used.

1. Rational System:

1. If the composition is expressed in mole fraction , the system is known as rational system.

2. Here the standard state is the fugacity , at which the mole fraction of the solute is unity.

3. At this stage it behaves ideally and obeys Henry’s law.

4. According to Henry’s law f2 = K x2 where K is Henry’s constant

2 Practical System:



1. If the composition is expressed in molarity or molality , the system is practical system.

2. Under these conditions, Henrys law becomes f2 = K m2

2. if the activity is plotted against mole fraction of the solute

standard state is the state at which m2 = 1

1015MEAN ACTIVITY CO-EFFICIENTS OF ELECTROLYTES.

activity and mean activity coefficient are related as

a = (a±)2



1115MAXWELL’S DISTRIBUTION OF MOLECULAR VELOCITIES :

The fraction of molecules with velocity between c and c+dc is

ρ(c) dc = 4π( M2 πRT

)32 c2 exp ( −M c2

2 RT ) dc

Important features:

1. The fraction of molecules having velocity greater than zero increases with increase in

velocity, reaches a maximum and then falls off towards zero at higher velocities.

2. The fraction of molecules with too low or too high velocities is very small



3. There is certain velocity for which the fraction of molecules is maximum. This is called

most probable velocity.

4. Most probable velocity increases with rise in temperature.

5. exp ( −M c2

2 RT ) is called Boltzmann factor.

6. Boltzmann factor. increases with increase in temperature.



1215 AVERAGE VELOCITY (Mean Velocity ( <c> :

It is given by the arithmetic mean of different velocities possessed by the molecules at

the given temperature.

If c1,c2,c3 ... cn are the velocities of the gas molecules and n is the total number then

the average velocity is given by

<c> = c1+c 2+c3+ ....+cn

n

Average velocity CAV

= √ 8 RTπM

Derivation:

Average velocity = 4π ( m2 πRT

)32 ∫

0

∞

ce−mc2

2 RT c2 dc

= 4π ( m2πRT

)32 ∫

0

∞

e−m c2

2 RT c3 dc

= 4π ( m2πRT

)32 [

1

2( m2 RT

)2 ∫

0

∞

x2 n+1 e−a x2

dx =

n !2 an+1

= 4π m

2πRT × ( m2 πRT

)12 ×

2(RT )2

m2

= 4 ( m2 πRT

)12 ×

RTm

= √2√2√2 π

× 2 ( mRT

)12 ×

RTm

= √ 2π

× 2 ( RTm

)12

= √ 8 RTπm

Problem: The average velocity of a gas at 100 K is 20 m/s. When the temperature is doubled,

What is the average velocity of the gas?

. Solution:



C1 = √ 8 RTπM

20 = √ 8 R (100)πM

C2 = √ 8 R (200)πM

Dividing equation 2 by 1 we get

C 220 = √ 200

100

¿ 0.414

C2 = 8.28 m/s

1315 MOST PROBABLE VELOCITY ( CP):

It is defined as the velocity possessed by maximum number of molecules of a gas

at a given temperature.

Most probable velocity CMPV

= √ 2 RTM

Derivation:

At the most probable state, the first derivative is equal to zero

c = 4π( m2 πRT

)32 e

−mc2

2 RT c2

differentiating with respect to ‘c’

dc = 4π( m2 πRT

)32 [e

−mc2

2 RT (2c) + c2 ¿ ) ]

equating to zero we get

4π( m2 πRT

)32 [e

−mc2

2RT (2c) + c2 ¿ ) ] = 0

Dividing by 4π( m2 πRT

)32

¿ (2c) + c2 ¿ ) ] = 0

Dividing by e−mc2

2 RT (2c)



1 - m c2

2 RT = 0

m c2

2RT = 1

c2 = 2RT

m

c =√ 2 RTm

Problem: The Most probable velocity of a gas at 225 K whose mass is twice that of gas constant

is

Solution:

Cp = √ 2 RTM

= √ 2 R(225)2 R

= 15 m/s

Problem: At what temperature the Most probable velocity of a oxygen will be doubled if the

Most probable velocity of at 100 K is 20 m/s

Solution:

C1 = √ 2 R T1

M

C2 = √2 R (T ¿¿2)M

¿

Dividing equation 2 by 1 we get

C 2C 1 = √ T 2

T 1

2 = √ T 2T 1

Squaring on both sides

4 = T 2T 1

T2 = 4 T1



= 4 9100)

= 400 K

Problem: The Most probable velocity of a gas at 100 K is 20 m/s. What is the mass of the gas?

Solution:

C1 = √ 2 R T1

M

Squaring on both sides

(C 1)2 = 2 R T 1

M

400 = 2 R(100)

M

M = R2

= 8.314

2

= 4.15

1415 ROOT MEAN SQUARE VELOCITY( RMS VELOCITY):

It is defined as the square root of the mean of the squares of different velocities

possessed by the molecules of the gas at the given temperature.

If c1,c2,c3 ... cn are the velocities of the gas molecules and n is the total number then

(< c2 > ) 1/2 = (c12+c 22+c 3 2+ .....

n) ½

CRMS

= √ 3 RTM

Derivation:

CRMS

= √∫0

∞

c2 p(c)dc

∫0

∞

c2 p(c ) dc = ∫0

∞

c2 4 π ( m2πRT

)32 e

−mc2

2 RT c2 dc

= 4 π ( m2 πRT

)32 ∫

0

∞

c4 e−mc2

2 RT dc



= 4 π ( m2 πRT

)32 ×

1.322+1 √ π

( m2 RT )

4+1 ∫0

∞

x2 n e−a x2

dx = 1.3 ….. 2n−1

2n+1

√ πa2 n+1

= 4 π ( m2 πRT

)32 ×

38 ( 2RT

m)

52 √π

=4π m

2πRT × ( m2 πRT

)12 ×

38 ( 2 RT

m)

52 √π

= m

2 RT × ( m2 πRT

)12 ×

32 ( 2 RT

m)

52 √π

= m

2 RT × ( m2 RT

)12 ×

32 ( 2RT

m)

52

= 32 ×

2 RTm

= 3 RT

m

CRMS

= √ 3 RTm

Problem: The RMS velocity of of a gas at 144 K whose mass is thrice that of gas constant is

Solution:

C RMS = √ 3 RTM

= √ 3 R 1443 R

= 12 m/sRelation between average velocity, RMS velocity and most probable velocity

Average velocity CAV

= √ 8 RTπM

CRMS

= √ 3 RTM

C AV

CRMS = √ 8 RT

πM× √ M

3 RT



= √ 83 π

= 0.9213Average velocity = 0.9213 × RMS velocity

Most probable velocity CMPV

= √ 2 RTM

CRMS

= √ 3 RTM

CMPV

CRMS = √ 2 RT

M× √ M

3 RT

= √ 23

= 0.8165Most probable velocity = 0.8165 × RMS velocity

1515 MACRO STATE AND M1CRO STATE:

A macro state is defined as the specification of number of space points in each cell of

phase space. Micro state is defined as the individual position of the phase point for each

system.

Let there be cell 1, cell 2 and cell 3 in phase space. Suppose there are 4 phase points

ABCD in cell 1 three phase points EFG in cell 2 and two phase points HI in cell 3

Then the number of macro states = 4



Problem: .Find the number of macrostates, and micro states for four particles having a total

energy of 6E , the energy levels are equally spaced. Find also the Thermodynamic probability

when the four particles are distributed as one in each cell

Solution:

ROW 0 E 2E 3E 4E 5E 6E Total



1 3 1 6E2 2 1 1 6E3 2 1 1 6E4 1 2 1 6E5 2 2 6E6 2 2 6E7 1 1 1 1 6E8 1 3 6E9 3 1 6E

Total number of macro states = 9

Number of microstates in macrostate with equal probability = n!

r ! (n−r )!

Row 1: Number of microstates in macrostate ( 3,1) = 4 !

3!1 ! = 4

Row 5: Number of microstates in macrostate ( 2,2) = 4 !2!2 ! = 6


3!1 ! = 4


3!1 ! = 4


2!2 ! = 6

Total number of micro states = 4 + 6 + 4 +4 + 6 +12 + 12 + 12 24

= 84

Problem:2. Three distinguished particles have a total energy of 9 units. But the particles are

restricted to energy levels to 0 to 4. Calculate the number of macro states and micro states.

Solution:

There will be 5 different energy levels

Thus total number of macrostates = 3 ,

. Row 1: Number of microstates in macrostate ( 1,2) = 3 !

2!1 ! = 3


Total 0 E 2E 3E 4E9E 1 29E 39E 1 1 1


Row 2: Number of microstates in macrostate = 3C3 = 1Row 3: Number of microstates in macrostate = 3C3 = 1

Total number of microstates = 3 + 1 + 1

= 5

Problem:3. In a random distribution of 10 particles between two boxes with equal probability

find the a.The total number of microstates and macrostates b. Number of microstates in

macrostate (3,7)

Solution:

Total number of microstates = 2 n = 2 10

Total number of macrostates= number of terms in the binomial expansion

= ( a+ b ) 10

= 11[ a10 +10a9 b + …… b10 ]

The number of microstates in macrostate( 3,7 ) = 10 !

3!7 !

= 120

Problem. A system has 5 different macrostates under which there 6,20,42,12 and 2 microstates.

Calculate the relative probability for each of the micro states .

Solution:

Total number of microstate is = 6 + 20 + 42 + 12+ 2

= 82

Relative probability of each microstate is

682 ,

2082 ,

4282 ,

1282 ,

282

Problem: Three particles are distributed in two boxes. Find the macro state and microstates for

each Solution:



Problem: ten particles are distributed in two boxes. Find the macro state and microstates for

each

ORBox-1 0 1 2 3 4 5 6 7 8 9 10Box -2 10 9 8 7 6 5 4 3 2 1 0

Number of macro states = 11Number of micro states for macro states 1 = Number of micro states for macro states 2 = Number of micro states for macro states 3 = Number of micro states for macro states 4 = Number of micro states for macro states 5 = Number of micro states for macro states 6 = Number of micro states for macro states 7 = Number of micro states for macro states 8 = Number of micro states for macro states 9 = Number of micro states for macro states 10 =Number of micro states for macro states 11 = Problem: Two dices are thrown. The sum of numbers turned up is the even. The number of macrostates is Solution: Sum of numbers turned up will be even if

S = 2,4,6,8,10,12 The number of macrostates is 6

Problem:. The thermodynamic probability when four particles are distributed as one in each cell of four boxes is Solution:



Thermodynamic probability = N !

n1 !n1! …

= 4 !

1!1 !1!1 ! = 24Problem:. Consider a system of three cells such that the cells are filled up by two, three and one particle respectively. The thermodynamic probability for such arrangement is Solution:

Thermodynamic probability = N !

n1 !n1! …

= 6 !

2!3 !1 !

= 6 ×5× 4×3 !

2×3 !

= 6 ×5× 4

2 = 60

UNIT -5- PART -23 - STATISTICAL THERMODYNAMICS , EQUILIBRIUM and

PHASE RULE

1. Partition Functions

2. Sackur Tetrode Equation

3. Statistical Approach to Third Law of Thermodynamics

4. Maxwell Boltzmann Statistics



5. Bose Einstein Statistics

6. Fermi Dirac Statistics

7. Heat Capacities Of Solids

8. Einstein Model

9. Debye Model

10. Low Temperature

11.Negative Absolute Temperature

12.. Chemical Equilibrium

13.Thermodynamic Derivation of Equilibrium Constant

14.Standard Free Energy Calculations

15..Phase Equilibrium

16. Thermodynamic derivation of Phase Rule

17.Application of Phase Rule

18. Three Component Systems.

1

18PARTITION FUNCTIONS

It is the sum of all the energy levels. .It is given by q = ∑ g ie−∈ iKT

1.Translational partition function:

1. It is given by q tra = (2 π m KT )32

h3 × V

k- Bolltzmann constant, T- Temperature, h- Planck constant, V – volume of the container

2. It has no unit

3. The order of magnitude of translational partition function of diatomic gas is1025 - 10 30



4. The quantity h

(2π mKT )1 /2 is called Thermal de-broglie wavelength ( Ʌ)

5. The relation between thermal de-broglie wavelength and translational partition function is

Ʌ = ( Vqtr ) 1/3

Problem The translational partition function of H2 gas confined in a box of 81 m3 is 3× 10 30

at 3000 K. The thermal de Broglie wavelength is

Solution:

Ʌ = ( Vqtr ) 1/3

= ( 81

3× 1030 ) 1/3

= 3× 10−10

= 3A

Problem The translational partition function of a gas confined in a box of 1 m3 with thermal

de Broglie wavelength 10 × 10 -11 m is

Solution:

Ʌ 3 = ( Vqtr )

qtr = ( VɅ3 )

= 1

(10 ×10−11)3

= 1

(10−10)3

= 1030

2.Rotational partition function

1. It is given by qrot = 8 π 2 IKT

h 2

= T

θ rot where θ rot = h2

8 π 2 IK known as Characteristic rotational

temperature



2. . In general, q rot = 8 π 2 IKT

σ h2 . where σ is called symmetry number

3. σ=2for homo nuclear molecule and σ = 1 for hetero nuclear molecule

4. The order of magnitude of rotational partition function of diatomic gas is 10-100

Problem If the value of 8 π 2 Kh2 is 2.0× 10 38 .The rotational partition function of H 2 gas at

100 K ,with moment of inertia 1.5 × 10 -38 Kg /m2 is

Solution:

qrot = 8π π2 Kh2 ( IT)

= 2 × 10 38 (1.5 × 10 -38 × 10 2 )

= 3 × 10 2

Problem The rotational partition function of a gas at 100 K ,with moment of inertia 1.5 × 10 -

38Kg/m2 is 3 × 10 2 and if the value of 8 π 2 Kh2 is 2.0 × 10 38 the gas is

Solution:

qrot = 8π π2 Kσ h2 ( IT)

3 × 10 2 = 2.0× 1038

σ (1.5 × 10 -38 × 100 )

σ = 1

Gas is Mono atomic

Problem . The rotational partition function of a diatomic gas at 100 K is 150 × 10 -2 and if

the value of 8 π 2 Kh2 is 3 × 10 38 , the moment of inertia of the gas is

Solution:

qrot = 8 π π2 Kσ h2 ( IT)

150 × 10 -2 = 3× 1038

2 (100 I)

I = 10−40



Problem At what temperature the rotational partition function of a diatomic gas will be 150 ×

10 -2 The value of 8 π 2 Kh2 is 3× 10 38 . The moment of inertia of the gas is 2 × 10 -40 Kg /m2

Solution:

qrot = 8 π π2 Kσ h2 ( IT)

150 × 10 -2 = 3× 1038

2 (2 × 10 -40 T)

I = 50 K

Problem : If the 8 π 2 Kh2 is 1.5× 10 38 Calculate the rotational partition function of H 2 gas at

100 K , the moment of inertia is 2.2 × 10 -40gm/cm2

Solution:

q rot = 8 π 2 Kσ .h2 ( IT)

= 8 π 2 Kh2 (

I Tσ )

= 1.5 × 10 38 × ½ × 2.2 × 10 -40 × 100

= 165 × 10 -2

Problem : The rotational partition function of a gas at 100 K ,with moment of inertia 2.2 × 10 -

40gm/cm2 is 165 × 10 -2 and if the value of 8 π 2 Kh2 is 1.5× 10 38 . find the atomicity of the

gas

Solution:

q rot = 8π 2 Kh2 (

I Tσ )

165 × 10 -2 = 1.5 × 10 38 × 1σ × 2.2 × 10 -40 × 100

σ = 2

The gas is homo nuclear diatomic

Problem : The rotational partition function of a diatomic gas at 100 K is 150 × 10 -2 and if

the value of 8 π 2 Kh2 is 1.5× 10 38 find the moment of inertia of the gas



Solution: q rot = 8 π 2 Kh2 (

I Tσ )

150 × 10 -2 = 1.5 × 10 38 × 12 × I × 100

I = 2× 150× 10−2

150× 1038

= 2 × 10 -40

Problem . At what temperature the rotational partition function of a diatomic gas will be 150

× 10 -2 The value of 8π 2 Kh2 is 1.5× 10 38 . The moment of inertia of the gas is 2 × 10 -40

gm/cm2

Solution: q rot = 8 π 2 I Kσ h2 × T

150 × 10 -2 = 1.5 × 10 38 × 12 × 2 × 10 -40 × T

T = 100 K

Problem . : If the rotational partition function of a diatomic gas is 600 × 10 2 . If the value of

hcKT is 0.5× 10 -4 find the rotational constant.

Solution: q rot = 8 π 2 IKTσ . h2

= 8 π 2 I ch

× KT

σ hc

= B × KT

σ hc [ B = h

8 π 2 I c]

B = qσ hc

KT

= q × σ h cKT

= 600 × 10 2 × 2 ×0.5× 10 -4

= 6



Problem . If the rotational partition function of a diatomic gas is 600 × 10 2 . If the value of

hcKT is 0.5× 10 -4 what will be the spacing between the lines of microwave spectrum of the gas.

Solution:

q rot = 8π 2 IKTσ .h2

= 8 π 2 I ch

× KT

σ hc

= B × KT

σ hc [ B = h

8 π 2 I c]

B = qσ hc

KT

= q × σ h cKT

= 600 × 10 2 × 2 ×0.5× 10 -4

= 600 × 10 -2

= 6

spacing between the lines of microwave spectrum of the gas. = 2B

= 12 cm-1

Problem The characteristic rotational temperature of H2 gas at 3000 K is 150 K. Its rotational

partition function is

Solution:

qrot = T

θ rot

= 3000150

= 20

Problem . What is the characteristic rotational temperature of a gas whose rotational partition

function at 3000 K is 20.

Solution:

qrot = T

θ rot

θ rot = 3000

20



= 150 K

3.Vibrational partition function

. The vibrational partition function is given by qvib = 1

1−e– θvib/T

where θvib =h ϑK , called characteristic vibrational temperature

The order of magnitude of vibrationalpartition function of diatomic gas is 1-10

Problem. Calculate the vibrational partition function of a gas at 1000 K whose vibrational

frequency is such that e−h ϑKT = 0.5

Solution :

e−h ϑKT = 0.5

q Vib = 1

1−e – h ϑ / KT

= 1

1−0.5

= 1

0.5

= 2

Problem. .Calculate the vibrational partition function of a gas at 1000 K whose vibrational

frequency is such that hϑKT = 2.3 given e−2.3 = 0.100

Solution :

hϑKT = 2.3

q Vib = 1

1−e – h ϑ / KT

= 1

1−e−2.3

= 1

1−0.10

= 1

0.9



= 1.11

4. Nuclear partition function:

The nuclear partition function of any molecule or atom is the degeneracy of the ground state

nuclear energy level.

The degeneracy of ortho state = (I + 1)( 2I + 1)

For para state = I ( I+ 1) where I is the spin quantum number of the nucleus.

5.Electronic partition function:

The electronic partition function is the degeneracy of the ground electronic state which is

given by qel = 2J+ 1

Problem. Find the electronic partition function of 1. Hydrogen 2. Helium

Solution:

Hydrogen atom 1s1

J = L+S

= ½

q0 = 2J+ 1

g0 = 2 ( ½) + 1

= 2

∴ qel = 2

Helium atom 1s2

S = ½ - ½

= 0,

J= 0+0

= 0

qel = 2J+ 1

qel = 1

Problem .Calculate the nuclear partition function of ortho H2 and ortho D2 molecules.

Solution:



The nuclear partition function of any molecule or atom is the degenarcy of the ground state

nuclear energy level.

The degeneracy of ortho state = (I + 1)( 2I + 1) where I is the spin quantum number of the

nucleus. For H-atom I= ½

q nu = ( ½ + 1)( 1+1) = 3

For Deutrium atom I =1

qnu =( 2)(3) = 6

218SACKUR –TETRODE EQUATION ( TRANSLATIONAL ENTROPY )

It is used to calculate the Translational entropy for mono atomic gas ( He, Ne, Ar)

S = 18.605 R + 32R ln m +

32 R ln T + Rln V

In terms of pressure:

S = 2.303 R {32 log m +

52 log T - logP - 0.5055 }

Problem Calculate the entropy change of one mole of He when it is heated from 100 K to

1000 K at constant pressure. [ R = 1.98 cal/deg/mol

Solution:

∆ S=?

S100 = 2.303 R {32 log m +

52 log (100) - logP - 0.5055 }

S1000 = 2.303 R {32 log m +

52 log (1000) - logP - 0.5055 }

∆ S=¿ S1000 - S100

= 2.303 R {52 log (1000) -{

52 log (100) }

= 2.303 × 1.98 × 2.5 × log 10

= 2.303 × 1.98 × 2.5 × 1

= 11.399 cal/deg/mol

Problem Calculate the entropy change of one mole of Ne when it is expanded from a

pressure of 1000 atm to 100 atm at constant temperature. [ R = 1.98 cal/deg/mol



Solution:

∆ S=?

S1000 = 2.303 R {32 log m +

52 log T - log(1000) - 0.5055 }

S100 = 2.303 R {32 log m +

52 log T - log (100) - 0.5055 }

∆ S=¿ S100 - S1000

= 2.303 R {−¿ log (100) + log (1000) }

= 2.303 × 1.98 × log 10

= 2.303 × 1.98 × 1

= 4.56 cal/deg/mol

Problem Calculate the difference in entropy of one mole of Ne (Mol wt = 20.0) and one mole

of He( Mol wt = 4.0) when they are heated at constant temperature and pressure [ R = 1.98

cal/deg/mol , log 5 = 0.6989 ]

Solution:

SNe = 2.303 R {32 log 20 +

52 log T – log P - 0.5055 }

SHe = 2.303 R {32 log 4 +

52 log T – log P - 0.5055 }

SNe - SHe

= 2.303 R { log (20) - log (4) }

= 2.303 × 1.98 × log 5

= 2.303 × 1.98 × 0.6989

= 3.19 cal/deg/mol



318STATISTICAL APPROACH TO THE THIRD LAW OF THERMODYNAMICS

III Law Statement:

At the absolute zero of temperature, the entropy of every substance is zero

.limT → 0[S] = 0

Entropies calculated by Third law of thermodynamics is called thermal entropies.

Exception to III law:

Entropies of H2,D2, CO, NO, N2O, H2O are not zero at zero K.. These entropies are

called Residual entropy and is due to alternate arrangements of molecules in the solid.

BOLTZMANN LAW

S = k ln W where k – Boltzmann constant, W – probability

Entropies calculated by Boltzmann law ( statistical) is called statistical entropies.

RESIDUAL ENTROPY

Residual entropy = Statistical entropies - Thermal entropies

CO, NO and N2O have two types of arrangements

CO and OC , NO and ON, NNO and ONN

Therefore W = 2N where N is Avagadro number

S = k ln W

= k ln 2N



= N k ln 2

= R ln 2

= ( 8.314) (0.692)

= 5.76 J/K/mol

418 MAXWELL BOLTZMANN STATISTICS

STATISTICS

Classical Statistics Quantum Statistics

(Maxwell Boltzmann Statistics) 1.Bose Einstein Statistics

2. Fermi-Dirac Statistics

Maxwell Boltzmann Statistics [ Classical Statistics]:

1. This is based on classical theory where particle has only particle character. This is applicable

to distinguishable particles.

2. Total probability W of the distribution is W = N! ∏1

i (gi)¿

¿!

where N – total number of particles, ∏1

i

❑ represents product , n- number of particles

g- number of states

3. Expression for most probable distribution of ‘ni’ particles in ‘gi’ energy levels is

ni = gieα+β ∈ i



Problem: Find the number of ways of arranging 2 distinguishable particles in two states Solution:

Let the particles be A,B, The possibilities are

Number of ways = 4Aliter:

Number of ways = 2! × 22

2


AB AB A B B A


= 2 × 22

2 [2! = 2]

= 4Problem: Find the number of ways of arranging 2 distinguishable particles in three states Solution

Let the particles be A,BThe possibilities are

Number of ways = 9

Aliter:

Number of ways W = 2× 32

2

= 9

Problem: Find the number of ways of arranging 3 distinguishable particles in two states

Solution

Let the particles be A,B,C,

The possibilities are

Number of ways = 8

Aliter:

Number of ways W = 3!× 23

3!

= 8

Problem: .Calculate the number of ways of distributing distinguishable particles a,b,c among

three energy levels so as to obtain the following set of occupation number n1 = n2 = n3 =1

Solution


AB AB AB

A B A B A B

B A B A B A

ABC ABC

BC AAC BAB C

A BCB ACC AB


g1 g2 g3n1 = 1 n2 = 1 n3 = 1 N = n1 + n2 +n3 = 1 + 1 + 1

= 3

The probability W = N !

n1× n2 × n3

= 3 !

1!1 !1!

= 6

Problem: .Calculate the number of ways of distributing 4 distinguishable particles among four

energy levels so as there are 2 molecules in the level g1 , 1 in g2 1 in g3 and 0 in g4.

Solution

g1 g2 g3 g4n1 = 2 n2 = 1 n3 = 1 n4 = 0

The probability W = = N !

n 1× n2× n3× n 4

= 4 !

2!1 !1! 0!

= 12

Problem: Calculate the number of ways of distributing two particles among four energy levels when the particles are MaxwellonsSolution

Number of ways for Maxwellons = N ! gi¿

¿ !

= 2 !. 42

2!

= 16 Let the particles be a,b 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16a b a b a b ab

b a a b a b ab

b a b a a b ab



b a b a b a ab

Problem: Calculate the number of ways of distributing three particles among four energy

levels when the particles are Maxwellons

Solution


¿ !

= 3 !. 43

3!

= 64

518BOSE EINSTEIN STATISTICS

This statistics deals with

1. Particles which are indistinguishable.

2. Any number of particles may occupy a given energy level.

This statistics is obeyed by particles having integral spin such as hydrogen ( H2),

deuterium (D2), nitrogen ( N2), helium-4 ( 4 He). Particles obeying BE statistics are called

bosons.

The total number of ways in which n particles are to be distributed among g levels

W = ∏1

g [¿+gi−1]!(gi−1) !×∋!



Expression for most probable distribution of ‘ni’ particles in ‘gi’ energy levels is

ni = gi

eα+β ∈ i−1

Problem: Calculate the number of ways of distributing two particles among four energy levels

when the particles are Bosons

Solution

Number of ways for Bosons = [¿+gi−1]!(gi−1)!∗¿ !

= [2+4−1]![ 4−1 ] !∗2 !

= 10.


levels when the particles are Bosons

Solution


1 2 3 4 5 6 7 8 9 10a a a aaa a a aa

a a a aaa a a aa



= [3+4−1]![ 4−1 ] !∗3 !

= 6 !

3!3 !

= 6 ×5 × 4×3 !3 × 2×1 ×3 !

= 20.

Let the particles be a,a,a

1 2 3 4 5 6 7 8 9 10 11 1

2

13 14 15 16 17 1

8

19 20

aa

a

aa aa Aa a a a a a a

aaa a aa aa aa a a a a a

aa

a

a a aa aa aa a a a a

aaa A a a aa aa aa a a a

Problem . Calculate the number of ways of distributing two particles among four energy levels

when the particles are 1. Bosons 2. Maxwellons

Solution

ni = 2

gi = 4


= [2+4−1] ![ 4−1 ] !∗2 !

= 10.


1 2 3 4 5 6 7 8 9 10A a a aaA a a aa

a a a aaa a a aa


Number of ways = 10


¿ !

= 2!. 42

2!

= 16

Let the particles be a,b

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

a b a b a b ab

b a a b a b ab

b a b a a b ab

b a b a b a ab

Number of ways = 16

618 FERMI DIRACE STATISTICS

This statistics is obeyed by

1. Indistinguishable particles of half integral spin

2. Obey pauli’s exclusion principle.

This statistics is obeyed by particles having half integral spin such as electron, proton .

and the particles obeying FD statistics are called Fermions

The number of ways in which gi quantum states will each be occupied by one particle is



W = ∏1

g gi!( gi−¿ )!×∋!

Expression for most probable distribution of ‘ni’ particles in ‘gi’ energy levels is

ni = gi

eα+β ∈ i+1

Problem: Calculate the number of ways of distributing two particles among four energy levels

when the particles are Fermions

Solution

Number of ways for Fermions = [ gi ]!

(gi−¿)!∗¿!

= [4 ]!

(4−2) !∗2!

= 6.Let the particles be a,a

1 2 3 4 5 6a a aa a a

a a aa a a


levels when the particles are Fermions

Solution


(gi−¿)!∗¿!

= [4 ]!

(4−3) !∗3 !

= 4× 3!1×3 !

= 4

Let the particles be a,a,a

1 2 3 4a a a



a a aa a a

a a a

Problem Calculate the number of ways of distributing three particles among four energy levels when the particles are 1. Fermions 2. Bosons 3.. Maxwellons Solution

ni = 3gi = 4


(gi−¿)!∗¿!

= [4 ]!

(4−3) !∗3 !

= 4 × 3!1 ×3 !

= 4 Let the particles be a,a,a


= [3+4−1]![ 4−1 ] !∗3 !

= 6 !

3!3 !

= 6×5× 4×3 !3× 2×1 ×3 !

= 20. Let the particles be a,a,a 1 2 3 4 5 6 7 8 9 1

011

12

13

14

15

16

17

18

19

20

aaa

aa

aa

Aa

a a a a a a

aaa

a aa

aa

aa a a a a a

aa a a aa aa aa a a a a


1 2 3 4a a aa a aa a a

a a a


aaaa

A a a aa aa aa a a a

Number of ways for Maxwellons = N! gi¿

¿ !

= 3 !. 43

3!

= 64





COMPARISON OF MB , BE and FD STATIISTICS

S.NO

MAXWELL –BOLTZMANN BOSE EINSTEIN FERMI-DIRAC

1 Classical statistics ( particle aspect of particle)

Quantum statistics( wave aspect of particle)

Quantum statistics ( wave aspect of particle)

2Particles are distinguishable

Particles are indistinguishable Particles are indistinguishable

3 No restriction for the number of particles in a given state

No restriction for the number of particles in a given state

Only two particles are permitted in a given state.

4 Number of distinguishable

ways is N! (gi¿ )¿!

Number of distinguishable

ways is [¿+gi−1]!

(g i−1)!×∋!

Number of distinguishable

ways is gi!

(gi−¿ ) !×∋!

5Applicable to ideal gas Applicable to Photons, bosons Applicable to Electrons,

fermions

6 most probable distribution is

gi

eα +β∈ i

most probable distribution is gi

eα +β∈ i−1

most probable distribution is gi

eα +β∈ i+1

8. No question of spin of particles arises.

Particle posses zero or integral spin

Particle posses half integral spin



718HEAT CAPACITIES OF SOLIDS

SPECIFIC HEAT CAPACITY:

It is defined as the energy required to raise the temperature of one gram of solid through

one degree kelvin.

DULONG- PETIT’S LAW:

According to Dulong and pettit, the specific heat capacity of sold remains constant at all

temperature.

Cv = 3R (constant) at all temperature

Limitation:

It was not found out that Cv is constant at all temperature. It varies with temperature.

818 EINSTEIN’ S HEAT CAPACITY OF SOLIDS

According to Einstein

1.All solids contain many atoms which are rest at zero kelvin

2.As temperature increases , the atoms vibrate simple harmonically.

3.A crystal can be considered as a system of N non – interacting particles.

4.Each atom vibrates independently and has three independent vibrational degrees of freedom.(X,Y,Z) Therefore the crystal may be treated as a system of 3N independent harmonic oscillators.

Heat capacity Cv = 3R( θE

T¿¿2

eθE

T

(e¿¿θE

T –1)2

¿

This is the expression for heat capacity of solid.Case 1: At high temperature:



Cv = 3R ( θE

T¿¿2

eθE

T

(e¿¿θE

T – 1)2

¿

= 3R (θE

T¿ 2

eθE

T

(1+θE

T +…..−1)2 [ e x = 1 + x +...]

= 3R (θE

T¿ 2

eθE

T

(θE

T )2

= 3R eθE

T

= 3R ( 1 + ……….. )

= 3R Thus at higher temperature it resembles Dulong’s- Petit’s law

Case 2: At low temperature:

Cv = 3R ( θE

T¿¿2

eθE

T

(e¿¿θE

T –1)2

¿

= 3R ( θE

T¿¿2

eθ E

T

(e¿¿θE

T )2

¿ [e

θE

T – 1 = eθE

T ]

= 3R ( θE

T¿¿2

1

eθ E

T

Case 3: At 0 K temperature:

= 3R ( θE

T¿¿2 1

¿¿ [ ex = 1 + x + x2

2 + ….. ]

= 3R 1¿¿

1¿¿

= 3R 1¿¿

= 3R ( 1∞

¿

= 0

Case 1: At low temperature: Cv = 0



Case 2: At high temperature: Cv = 3R

Thus at higher temperature it resembles Dulong’s- Petit’s law

CHARACTERISTIC TEMPERATURE:

The quantity hγK is called Einstein’s characteristic temperature. It is denoted by θE

θE = hγK Where ϑ is vibrational frequency , k- Boltzmann constant .h- Plancs constant

918 DEBYE MODELS OF HEAT CAPACITY OF SOLIDS

Debye suggested that

1.The crystal behaves as a system of coupled harmonic oscillator.

2. The atoms in the crystal is capable of vibrating in many different modes,

3. The crystal containing N atoms, has 3N normal modes of vibrations,

4. Each mode of vibration has a unique frequency

5. The vibrations of atoms , are not independent. ie the vibration of one atom , affects the

vibration of the neighbouring atoms.

6. Crystal properties depends upon , frequencies of the vibrational modes.

The heat capacity of solid is given by

Cv = 3R [4 D ( θD

T) - 3 ¿¿ ]

At High Temperature: T → ∞

Cv = 9 R d

dT ( TxD

3 ∫0

xD x3

ex−1 dx )



= 9 R d

dT ( TxD

3 ∫0

xD x3

1+x+….−1 dx ) [ ex = 1 +x + …. ]

= 9 R d

dT ( TxD

3 ∫0

xD x3

x dx )

= 9 R d

dT ( TxD

3 ∫0

xD

x2 dx )

= 9 R d

dT ( TxD

3

xD3

3 ) [ ∫0

xD3

x2 dx = xD3

3 ]

= 9 R d

dT ( T3 )

= 9 R ( 13 ) [

ddT (

T3 ) = (

13 ) ]

= 3 R

At Low Temperature: T → 0, xD→ ∞

Cv = 9 R d

dT ( TxD

3 ∫0

∞ x3

ex−1 dx )

= 9 R d

dT ( TxD

3 × π4

15 ] [∫

0

∞ x3

ex−1 dx = π4

15 ]

= 9 R × π4

15×

ddT (

T

( h γD

KT )3 ] [ xD

3 = hγ D

KT

]

= 3R π4

5×

ddT (

T

( θD

T )3 ] [

hγ D

K = θD ]

= 3R π4

5

ddT (

T 4

(θD )3 ]

= 3 R π4

5 ( θD )3

ddT ( T 4 ]

= 3 R π4

5 ( θD )3 ×( 4 T 3 ) [

ddT (T 4 ) = 4 T 3 ]



= 12 R π4

5 (θD )3 × T 3

= Constant × T 3 [12 R π4

5 (θD )3 = Constant ]

at high temperature:∴ Cv = 3R

at low temperature: Cv ∝ T 3

DEBYE’S T 3 LAW:

It states that, when temperature tends to zero the heat capacity of solid is directly

proportional to third power ( T 3) of its temperature. i.e Cv ∝ T 3

Problem : The heat capacity of copper at 10 K is 16 KJ/K/mol. Find the heat capacity at 5 K using Debye’s theory at low temperature limitSolution:

According to Debye’s theory , the heat capacity of solid at low temperature is directly proportional to third power ( T 3) of its temperature. i.e Cv ∝ T 3

(C v )5(C v )60

= (5 )3

(10 )3

(C v )5 = (C v )60× (5 )3

(10 )3



= 16 × 5× 5× 5

10× 10× 10

= 2

Problem 2: Calculate the specific heat of Aluminium at 796 K θD for Aluminium = 398 K

The constant of Debye’s T3 law is 450

Solution:

Cv = 12 Nk π 4

5 (θD )3 × T 3

= 450 × ( TθD

)3

= 450 × ( 796398

)3

= 450 × 8

= 3600 J/K/mol

EINSTEIN MODEL DEBYE MODEL1. Einstein solid is composed of single-

frequency quantum harmonic oscillatorslong wavelength modes have lower frequencies than short wavelength modes.

2.

Cv = 3R( θE

T¿¿2

eθE

T

(e¿¿θE

T – 1)2

¿

Cv = d

dT ( 9 RTxD

3 ∫0

xD x3

ex−1 dx )

3. At low temperature the heat capacity is equal to zero

At low temperature the heat capacity, is proportional to T3.

4. graph graph

Debye versus Einstein

1018 LOW TEMPERATURE

Low-temperature physics is also known as cryogenics, from the Greek meaning

"producing cold."

Low temperatures are achieved by removing energy from a substance.


https://en.wikipedia.org/wiki/Quantum_harmonic_oscillator

https://en.wikipedia.org/wiki/Einstein_solid


The simplest way to cool a substance is to bring it into contact with another substance

that is already at a low temperature.

1118NEGATIVE ABSOLUTE TEMPERATURE

A system with a negative temperature on the Kelvin scale is hotter than any system with

a positive temperature.

If a negative-temperature system and a positive-temperature system come in contact,

heat will flow from the negative- to the positive-temperature system.

A standard example of such a system is population inversion in laser physics.

1218 CHEMICAL EQUILIBRIUM

When the rate of forward reaction equals to that of reverse reaction, the concentration of

reactant and product remains unchanged. This state is known as chemical equilibrium

1318THERMODYNAMIC DERIVATION OF EQUILIBRIUM CONSTANT

Thermodynamically the expression for equilibrium constant can be derived using

chemical potential

Derivation:

∆ G=¿ ∑products

G - ∑reactants

G


https://en.wikipedia.org/wiki/Laser_physics

https://en.wikipedia.org/wiki/Population_inversion


Consider a reaction in which ‘a’ moles of A and ‘b’ moles of B combines to form ‘c’ moles of C

and ‘d’ moles of D

Greactants = aμA + bμB

Gproducts = cμC + dμD

∆ G=¿ ∑products

G - ∑reactants

G

= [cμC + dμD] – [aμA + bμB ]

Chemical potential μ = μo + RT ln P

μA = μAo + RT ln PA

Substituting this value we get

∆ G=−RT ln Kp Where k p = PC PD

PA PB

This is the expression for equilibrium constant

This is known as Vant Hoff isotherm

Problem Calculate Kp of a reaction at 100K whose ∆G at 25 0C is - 83.14 KJ/ mol Solution:

∆G = - RT ln Kp

= - 2.303 RT log Kp

log Kp = −∆ G

2.303 RT

= 83.14

2.303× 8.314 ×100

= 1

2.3

Kp = 102.3

Problem The value of ln Kp for a reaction is 2.0 ×10 -2 at 100K . Calculate ∆G for this

reaction.

Solution:

∆G = = - 2.303 RT log Kp

= - (2.303 )( 8.314) ( 100) (2.0 ×10 -2 )



1418STANDARD FREE ENERGY CALCULATIONS

∆ Go=¿ ∑products

G o - ∑

reactantsGo

Problem The standard free energy of products is 20 KJ . The ∆ G0 for this reaction is – 10 J

Find the standard free energy of reactants

Solution

- 10 J = 20000 – G0

G0 = 19990 J

1518 .PHASE EQUILIBRIUM

STATEMENT:

For a heterogeneous equilibrium system, the number of components (C), number of

phases( P) and number of degrees of freedom( F) are related by the following equation

F = C – P + 2 This is known as Gibbs phase rule.

PHASE( P):

It is defined as any homogeneous and physically distinct part of a system which is

bounded by a surface and is mechanically separable from other parts of the system.

Example:

1. A gas mixture constitutes a single phase.

2. Completely immiscible liquid like water and kerosene, forms two different phases.

3. Completely miscible liquids like water and milk forms single phase.

4. Two solids form two phases.

COMPONENT(C):

The number of components of a system at equilibrium is defined as, the minimum

number of constituents , by means of which , the composition of each phase can be expressed

in terms of chemical equations.

Example:

1. Water exists in three phases (ice, liquid and vapour).but the composition of each

phase can be expressed in terms of H2O. Hence it is a one component system.



2. Consider the decomposition of CaCO3

CaCO3 ↔ CaO + CO2

The composition of various phases of the system can be represented as follows

CaCO3 = CaO + CO2

CaO = CaCO3 - CO2

CO2 = CaCO3 - CaO

Thus composition of each phase can be expressed in terms of any two constituents.

Hence the number of component is two.

DEGREE OF FREEDOM (F):

The degree of freedom of a system is defined as the minimum number of independent

variables like temperature, pressure and composition, which must be specified , in order to

define the system completely.

If F = 0 , the system is called in variant or zero variant or non variant. If F = 1 , the

system is called mono variant or uni variant . If F = 2 , the system is called bivariant

Example: Consider the water system

H2O (s) ↔ H2O (l) ↔ H2O (g)

This condition exists at fixed pressure and fixed temperature, therefore no variables are

required to define the system, so F = 0. Hence , at this condition, the system is called in variant

or zero variant or non variant .

Problem. 1. Find the number of phases, components and degree of freedom for the following

equilibrium. CaCO3 (s) ↔ CaO (s) + CO2 ( g)

Solution:

Number of phases (P) = 3 (CaCO3 , CaO, CO2 )

Number of components (C) = 2

Number of degrees of freedom F = C-P +2 ( by phase rule)

= 2-3 +2

= 1



Problem. 2. Find the number of phases, components and degree of freedom for the following

equilibrium. H2O (s) ↔ H2O (l) ↔ H2O (g)

Solution:

Number of phases (P) = 3

Number of components (C) = 1

Number of degrees of freedom F = C-P +2 ( by phase rule)

= 1-3 +2

= 0

1618 THERMODYNAMIC DERIVATION OF PHASE RULE

Using chemical potential concept the phase rule can be derived

Total number of variables = component + Temperature + pressure

= CP + 1+ 1

= CP +2 --------------------------1

Separate equation for each component = P-1

There are C components Therefore Separate equation for ‘C’ component = C ( P-1)

Total number of conditions = P+CP –C

Degree of freedom = number of variables – number of restrictions

F = CP +2 – (P + CP – C)

F = C- P+2

This is known as Gibbs phase rule

1718APPLICATION OF PHASE RULE



1818 THREE COMPONENT SYSTEMS.

For the three component system, the phase rule becomes

F = C-P + 2

= 3 –P + 2

= 5 – P

For a system having only one phase F = 5-1 = 4. Therefore four variables are required to

describe the phase diagram which is difficult. Therefore P and T are fixed for a given diagram

and triangular graph is used to describe the system.

In this diagram the three components are fixed at the corners of the triangle. The side

of the triangle opposite to the corner implies the absence of that substance. The relative

amounts of the three components are represented as percentage by mass. For example consider

the phase diagram of three components A,B and C.

The distance from each apex to the centre of the opposite side of the equilateral triangle

is divided into 100 parts corresponding to percentage composition and the composition

corresponding to a given point is obtained by measuring the perpendicular distance to the three

sides.

The horizontal lines across the triangle show the increasing percentage of A from zero at

the base to 100 % at the apex.

1. Phase diagram of three liquid system

2. System involving two solids and a liquid:

Examples Three liquid systems

1. Vinyl acetate - water – acetic acid

2. Chloroform – water – acetic acid



Two solids and a liquid system

1. Lead nitrate – sodium nitrate – water.

2.Potassium nitrate - sodium nitrate – water.











UNIT -5- PART - 33 ANALYTICAL CHEMISTRY

CRYSTALOGRAPHY/POLARIMETRY

1. Chromotography

2. Column Chromotography

3. Paper Chromotography

4. Thinlayer Chromotography

5. Gas-Liquid Chromotography

6. High Pressure Liquid Chromatography HPLC- Principle And

Applications.

7. Thermal Analysis

8. Different Thermal Analysis (DTA) – Principle And Applications

9. Thermogravimetric Analysis (TGA) Principle And Application.

10.Chemical Crystallography

11. Diffraction Methods

12. X Ray Diffraction

13.Neutron Diffraction

14. Electron Diffraction Methods. Principle And Applications.

15. Polarimetry

16. Circular Dichroism

17. Optical Rotatory Dispersion (ORD) Principle And Applications.



117CHROMOTOGRAPHY

s.no Technique Stationary phase Mobile phase

1 Column Chromatography Solid Liquid

2 Paper Chromatography Liquid Liquid

3 Thin –layer Chromatography Solid or Liquid Liquid

4 Gas – liquid Chromatography Liquid gas

5 HPL Chromatography Solid Liquid

217COLUMN CHROMATOGRAPHY

The rate of adsorption varies with a given adsorbent for different materials. This principle of

selective adsorption is used in column chromatography.

The component which has greater adsorbing power is adsorbed in the upper part of the

column

Examples for adsorbents.:

aluminium oxide, silica gel, activated charcoal, magnesium oxide, magnesium

carbonate, calcium carbonate

Chromatogram:

The banded column of adsorbent is called Chromatogram and the operation is termed

development of Chromatogram



Zone:

The portion of a column which is occupied by a particular substance is called zone.

Eluent:

The solvents used to remove the required content of each zone of the column are

known as eluents.

Eluotropic series.

A series of solvents with an increasing degree of polarity, used to explain solvent

strength in liquid-solid or adsorption chromatography.

... Thus, when developing a method or running a gradient, an eluotropic series is

useful for selecting solvents.



Successive elution: Here single solvent is used

Gradient elution: Here two solvents that are completely miscible in each other but with

different dielectric constant are used.

Development and Developers

The process of development by which the zones of chromatogram are separated to their fullest

extent is called development and the solvent used are called developers

317PAPER CHROMATOGRAPHY:



Migration parameter:

The position of migrated spots on the chromatograms are indicated by Migration parameter

RF, Rx, RM

RF = distance travelled by the solute

distance travelled by the solvent

Rx = distance travelled by the substanedistance travelled by the standard

RM = log ( 1

RF -1)

Rƒ value, solutes, and solvents

The retention factor (Rƒ) may be defined as the ratio of the distance traveled by the solute to the distance traveled by the solvent

. It is used in chromatography to quantify the amount of retardation of a sample in a stationary phase relative to a mobile phase[

If Rƒ value of a solution is zero, the solute remains in the stationary phase and thus it is immobile.

If Rƒ value = 1 then the solute has no affinity for the stationary phase and travels with the solvent front.

For example, if a compound travels 9.9 cm and the solvent front travels 12.7 cm, the

Rƒ value = (9.9/12.7)

= 0.779 .

Rƒ value depends on temperature and the solvent used in experiment, so several solvents offer several Rƒ values for the same mixture of compound.

Visualising agents:

Various spots in the chromatogram are visualized by suitable reagents known as

Visualising agents:

Types:

1. Descending chromatography

2. Ascending chromatography

3. Ascending- Descending chromatography

4. Radial chromatography(circular chromatography)


https://en.wikipedia.org/wiki/Retention_factor


417 THINLAYER CHROMATOGRAPHY (TLC)

Choice of best developing solvent is can be obtained from Stahl’s triangle.

Stahl’s triangle relates 1. adsorbant activity 2. nature of solute 3. nature of solvent



517 GAS-LIQUID CHROMATOGRAPHY



1. It consists of a mobile gas phase and a stationary liquid phase that is coated on the wall of the

capillary tube.

2. Sample mixture in gaseous form is run through the column with a carrier gas.

3. Hydrogen, Helium, Nitrogen and air are the carrier gas used.

Hydrogen may react with unsaturated compound and it creates fire and explosive hazard.

Helium is used because of its excellent thermal conductivity, inertness and low density

4. Separation can be achieved by the differences in the distribution ratios of the components of

the sample between the mobile ( gas) and stationary(liquid) phases.



617 HIGH PRESSURE LIQUID CHROMATOGRAPHY HPLC- PRINCIPLE AND

APPLICATIONS.

1. The efficiency of column in HPLC is expressed as Height Equivalent of a Theoretical

Plate( HETP)

2. Van Dempter equation is

HETP = A +Bμ + Cμ where A – eddy diffusion term ,

Bμ - longitudinal term, Cμ

-non-equilibrium in the mass transfer term.

μ – gas velocity A is constant that involves packing effect in the column B- is a constant that includes the effect of diffusion in the gas phase, C is the constant that reflects the resistance to mass transfer between the gas and the liquid. 3. Van Dempter graph is the plot of HETP versus velocity (ml/min)

3. HETP is inversely related to efficiency of column for separation. The smaller the value of

HETP, the more efficient the column for separation

Isocratic and gradient elution

A separation in which the mobile phase composition remains constant throughout the procedure is termed isocratic (meaning constant composition).

(The example of these the percentage of methanol throughout the procedure will remain constant i.e 10%)

The mobile phase composition does not have to remain constant.

A separation in which the mobile phase composition is changed during the separation process is described as a gradient elution.

One example is a gradient starting at 10% methanol and ending at 90% methanol after 20 minutes.

In reversed-phase chromatography, solvent A is often water or an aqueous buffer, while B is an organic solvent miscible with water, such as acetonitrile, methanol, THF, or isopropanol.

In isocratic elution, peak width increases with retention time linearly according to the equation for N, the number of theoretical plates

A schematic of gradient elution. Increasing mobile phase strength sequentially elutes analytes having varying interaction strength with the stationary phase.


https://en.wikipedia.org/wiki/Isopropanol

https://en.wikipedia.org/wiki/Tetrahydrofuran

https://en.wikipedia.org/wiki/Acetonitrile

https://en.wikipedia.org/wiki/Reversed-phase_chromatography

https://en.wikipedia.org/wiki/Methanol

https://en.wikipedia.org/wiki/Mobile_phase


Gradient elution decreases the retention of the later-eluting components so that they elute faster, giving narrower (and taller) peaks for most components.

7

17THERMAL ANALYSIS

In thermal analysis the physical parameter of the system is recorded as a function of

temperature.

s.no Name of the technique abbrevation Instrument employed Parameters

measured

graph

1 Thermogravimetry TG Thermo balance Mass Mass vs temp

or time



2 Derivative

thermogravimetry

DTG Thermo balance dmdt

dmdt vs temp

3 Differential thermal

analysis

DTA DTA Apparatus ∆T ∆T vs temp

4 Differential scanning

calorimetry

DSC Calorimeter dHdt

dHdt vs temp

5 Thermometric titrimetry Calorimeter Temperature Temp vs

titre volume

6 Dynamic reflection

spectroscopy

DRS spectrophotometer Reflectance Reflectance

vs temp

THERMOGRAVIMETRY(TG)

It is a technique in which weight of a substance in a controlled heated environment is

recorded as a function of temperature or time.

Here weight is plotted against temperature. The plot of weight against temperature

produced in a thermogravimeter is known as thermogram

The horizontal portion in the TG curve indicates the region where there is no mass change

The factors affecting thermogram

1. heating rate 2. furnace atmosphere 3. particle size

Applications:

Thermogravimetry is used

1. for evaluation of suitable standard 2. testing of purity of sample 3. curie point determination

4. Determination of metals present in alloys





‘

817DIFFERENT THERMAL ANALYSIS (DTA) – PRINCIPLE AND APPLICATIONS



It involves the technique of recording the difference in temperature between a substance

and a reference material against temperature. when the specimens are to controlled heating.

The plot of temperature difference between sample and reference against temperature

produced in DTA apparatus is known as differential thermogram Thus differential

thermogram is a plot of ∆T vs temperature. In this curve physical changes give rise to

endothermic curves whereas chemical reactions give rise to exothermic peaks.



Applications of DTA

To determine heat of reaction, specific heat, thermal diffusivity melting points



917THERMOGRAVIMETRIC ANALYSIS (TGA) PRINCIPLE AND APPLICATION

.



1017CHEMICAL CRYSTALLOGRAPHY ( Symmetry In Crystal Systems)

Unit cell :

A unit cell is the smallest repeating unit , in the array of points, which when repeated

over and over again results in a crystal of the given substance.

Elements of symmetry

There are three types of symmetry

1. Plane of symmetry:

It is an imaginary plane, which divides the crystal into two parts, such that one is the

exact mirror image of the other.

2.Axis of symmetry:

It is an imaginary line, about which the crystal may be rotated through an angle of 360n

such that it presents similar appearance.

If the similar appearance is repeated after an angle of 180 o , the axis is called 2- fold axis

of symmetry. If it appears after 120 ,90, 60 o, it is called 3- fold axis of symmetry, 4-fold and 6-

fold axis of symmetry respectively.

In general if a rotation, through an angle of 360

n brings the molecule to similar

appearance, then the crystal is said to have n – fold axis of symmetry

3. Centre of symmetry

Centre of symmetry of a crystal is such a point that any line drawn through it

intersects the surface of the crystal at equal distances in both directions.

The total number of planes, axes and centre of symmetries possessed by a

crystal is termed as elements of symmetry.

Elements of symmetry in a cube:

Rectangular planes of symmetry = 3

Diagonal planes of symmetry = 6

2- fold axis of symmetry = 6



Centre of symmetry = 1



--------------------------------------------

Total = 23

Crystallographic axis:

The straight line drawn through the selected lattice point is known as crystallographic axis.

WEISS INDICES

The coefficients of intercepts of the plane on the co-ordinate axes are called Weiss indices.

Consider a plane LMN. This plane has intercepts OL, OM, ON along x. y

and z- axes at a distances a,2b and 3c respectively. The coefficients of the intercepts ( 1,2 and

3) are called Weiss indices

MILLER INDICES

The reciprocal of coefficients of intercepts of the plane on the co-ordinate axes in

simple integral whole number are called Miller indices.

steps:

1. Write down the intercepts

2. Write their coefficients.

3. Find their reciprocals

4. Multiply each by a commom multiplier

5. Write the values in paranthesis without space and comma.

For example in the above plane

1. a,2b and 3c are the intercepts.

2. 1 , 2 , and 3 are the coefficients of intercepts.

3. 1 ½ 1/3 are reciprocals of the coefficients of intercepts.



4. Multiplying through out by 6 (a smallest common multiple ) to make as integrals

5. Their Miller indices are ( 632)

Find the Miller indices of the plane which makes the intercepts 2a,4b and 5c on crystallographic

axes.

Solution:

1. The intercepts are 2a, 4b and 5c

2. Their coefficients are 2,4 and 5

3. Their reciprocal are ½, ¼ and 1/5

4. Multiplying by 20 we get 10, 5 and 4

5. Mioller indices are ( 1054)

Crystallographic axis:

The straight line drawn through the selected lattice point is known as crystallographic axis.

Bravais lattice

Bravais has shown that there are 14 different ways in which similar points can be

arranged .thus the total number of space lattices belonging to all the 7 crystal systems put

together is 14.

Simple cube:

There is only one lattice point at each of the eight corners of the unit cell and no

lattice point inside the unit cell.

Body centered lattice:

There is only one lattice point at each of the eight corners of the unit cell and one

lattice point at the centre of the unit cell.

Face centered lattice:

There is only one lattice point at each of the eight corners of the unit cell and one

lattice point at the centres of each of the six faces of the unit cell.

s.no Crystal system Bravais lattice Cell dimensions interfacial

angles

1 Tri clinic Primitive -1 a≠ b ≠c α≠β≠γ≠ 902. Monoclinic Primitive, end – centred = 2 a≠ b ≠c α=β= 90 ≠ γ3 Orthorhombic Primitive, end – centred ,Face a≠ b ≠c α=β=γ= 90C. SEBASTIAN, A/P/CHEM, R.V.GOVT.ARTS COLLEGE, CHENGALPATTU, MOB: 9444040115 Page 89


centred, body centred = 4

4. Hexagonal Primitive =1 a= b ≠c α=β=9 0γ=1205. Rhombohedral

( trigonal)

Primitive = 1 a= b =c α= β=γ ≠ 906 Tetragonal Primitive, body centred = 2 a= b ≠c α=β=γ= 907. Cubic Primitive, ,Face centred, body

centred =3

a= b=c α=β=γ= 90Lattice energy :

Lattice energy of an ionic crystal is defined as the energy released when a mole of gaseous

cations and a mole of gaseous anions, separated from each other by an infinite distance are

brought to their equilibrium distance to form one mole of ionic crystal.

Calculation of lattice energy ( Born – Lande equation)

Lattice energy ( U) = - z+¿ z−¿e2 A N

r0¿

¿ ( 1- 1n ) per mole

Where A- Mandelung constant, N- Avagadro number

1117 DIFFRACTION METHODS (Determination Of Crystal Structure)

The following are the diffraction methods used to determine the Crystal Structure)

1. Lane photographic method

2. Rotating crystal method

3. Powder method( Debye- Scherrer method)









1217 X RAY DIFFRACTION Rotating crystal method

This method is used to determine the structure of crystals using diffraction of X- rays

The technique makes use of Bragg’s X-ray spectrometer, where crystal is used as reflecting

grating .

X- rays generated in the tube T are passed through a slit so as to obtain a narrow beam.

This narrow beam is allowed to strike the crystal C mounted on the turn table. The reflected

rays are sent to ionisation chamber where the intensities are recorded.

The crystal is rotated gradually by means of the turn table , so as to increase the incident

angle at the exposed face of the crystal. The process is carried out for each plane of the crystal.



The lowest angle at which , maximum reflection occurs is , called first order reflection which

corresponds to n= 1. The next higher angle , at which maximum reflection occurs again is called

second order reflection.

The lattice constant d is found out using different planes of the crystal as reflecting

surface for the same known wavelength of X – rays.

Applying Bragg’s equation

2dsinθ = n λ

For first order spectrum n= 1, hence the above equation becomes

2dsinθ = λ

1d =

2 sinθλ

If the ratio 1

d 1 :1

d 2 : 1

d 3 = 1 : √2 : √3 the crystal is simple cubic.

If it is 1 : 1

√ 2 : √3 then the crystal is body centred cubic

If it is 1 : √2 : √ 32 , the crystal is face centred cubic (FCC)

Problem: Find the crystal structure of the crystal whose θ values for the first order reflection

from the three faces are 5.22 o, 7.30 o and 9.05 o [ the values of sin 5.22 , sin 7.30, sin 9.05 are

0.0910 , 0.1272 and 0.157 respectively

Solution:

1d 1 :

1d 2 :

1d 3 = sin 5.22 : sin 7.30 : sin 9.05

= 0.0910 : 0.1272 : 0.1570

= 1 : 1.4 : 1.73

= 1 : √2 : √3

: It has simple cubic structure.

Powder method( Debye- Scherrer method)

The substance to be examined is finely powdered and is kept in the form of cylinder

inside a thin glass tube. This is placed at the centre of Debye Scherer camera which consists of a

cylindrical cassette,



X- rays are generated and allowed to fall on the powder specimen. The X- ray beam

enters through a small hole, passes through the sample and the unused part of the beam leaves

through the hole at the opposite end. The powder consists of many small crystals which are

oriented in all possible directions. So the reflected radiation is not like a beam ; instead, it lies

on the surface of a cone whose apex is at the point of contact of the incident radiation with the

specimen.

For each combination of d and θ, one cone of reflection must result. Therefore, many

cones of reflection are emitted by the powder specimen. The recorded lines from any cone are,

a pair of arcs. The first arc on either side of the exit point corresponds to the smallest angle of

reflection.

The distance between any two corresponding arcs on the film ( S) is related to the radius

of the powder camera R

S = 4Rθ where θ is the Bragg angle in radians( 1 rad = 57.3 o ) . ---------1

Combining d(hkl) = a

√h2+k2+l 2 with Bragg equation, we get

nλ = 2 a

√h2+k2+l 2 sinθ

∴ sin 2 θ = λ 2

4 a2 ( h2 + k2 + l2 ) [ for first order reflection n = 1]

Θ values can be obtained from the powder pattern using equation 1 The values of sin2θ are

compared with the below mentioned extinction rules.

1:2:3:4:5:6:8 SC [ 7 cannot be written in the form h2 + k2 + l2 ]

2:4:6:8 BCC [ odd integer for h + k+l are absent]

3:4:8:11:12 FCC [ h,k,l are either all odd or all even 111, 200, 220,311,222]

3:8:11:16 DC

1317NEUTRON DIFFRACTION METHOD

Neutrons have high penetrating power and are useful for structural studies of solids.

Nuclear pile is required for neutron diffraction. This consists of large size and high cost neutron

spectrometer. Therefore it is a less popular method in crystallography.



X – rays are scattered by orbital electrons whereas neutrons are scattered by atomic

nuclei. X – ray scattering power increases with atomic number but there is no regular trend for

neutron scattering.

In neutron diffraction two scattering takes place

1. nuclear scattering due to interaction of neutrons with the atomic nuclei

2. magnetic scattering due to the interaction of magnetic moments of neutrons with

permanent magnetic moments of atoms or ions.

Advantage:

The light element such as H or D which can not be located by X – ray diffraction , can

be located by neutron diffraction.





1417 ELECTRON DIFFRACTION METHODS. PRINCIPLE AND APPLICATIONS.

Electron diffraction studies utilize electrons with energies 40 keV . Since electrons are

charged they are scattered by their interaction with electron and nuclei of atoms of the sample.

Hence they can not be used for studying the interiors of solid samples. But they are used for

studying molecules in the gaseous state held on surfaces and in the films.

Diffraction of X – ray by crystal depends upon the spacing between the layers but the

diffraction of electrons by gaseous molecules depends upon the distances between the atoms..

The total intensity of scattering is given by Wierl equation which is

I ∝∑i , j

f i f j

sin s R ij

s Rijf i and f i are the scattering factors of the i th and j th atoms.

Applications:

The electron diffraction studies are used to evaluate the bond length and bond angle of

gas molecules.





1517 POLARIMETRY

In polarimetry we measure light’s polarization whereas in spectrophotometry we

measure light’s intensity.

In ordinary light ,vibration takes place in all planes whereas when the light is passed

through Nicol prism , the emerging ray has vibration only in one plane. This light is called plane

polarized light.

When organic liquids or quartz crystal or sugar solution are placed in the path of plane

polarized light, the plane of polarization is rotated. This is called optical activity.

The magnitude of rotation depends on

1. nature of the substance

2. concentration of solution

3. nature of solvent

4. wavelength of light used.

Specific rotation is defined as the number of degrees of rotation of the plane polarized light,

produced by one decimeter in length filled with a solution having one gram of substance per ml.

α Dl =

100× θl × c

θ – observed angle of rotation, l- length in decimeter c- gram of substance in 100 ml of

solution.









1617 CIRCULAR ICHROISM

The electric field vector is made up of two components EL and ER which corresponds to

left and right circularly polarized light. In ordinary medium , both EL and ER rotate at same

speed. In optically active medium EL and ER rotate at different speeds because the refractive

index is different for the left and right circularly polarized light (nL ≠ nR ) This medium is said

to be circularly birefringent.

If magnitude of EL ≠ magnitude of ER , E will no longer oscillate along a straight line

and hence it traces out an ellipse rather than circle. This medium is said to exhibit circular

dichroism. Thus in medium exhibiting circular dichroism (nL ≠ nR and magnitude of EL ≠ magnitude of ER

The combination of circularly birefringent and circular dichroism is called Cotton effect.

There are two types of Cotton effects. 1. positive cotton effect 2. negative cotton effect

The sign and magnitude of Cotton effects are determined by Octant rule. This rule is applicable

to substituted cyclohexanones.

The rate of change of specific rotation is known as Optical Rotatory Dispersion

(ORD). The plot of specific rotation vs wavelength is known as ORD curves.

If an optically active medium is kept in a magnetic field, the emerging light will get

rotated by an angle φ = VBl

where V - Verdet constat, B- magnetic induction, ‘l’ path length. This is known as

Farraday effect or magneto optical rotatory dispersion(MORD)

If an optically active medium is kept in Electric field E, the substance become doubly

refracting . that is the refractive index parallel to the direction of the field is not equal to that

perpendicular to the field.

The difference of the two quantities

npl – npr = KE2 (lamda)

This is known as Kerr electric optic effect and K is called Kerr constant



1717 OPTICAL ROTATORY DISPERSION (ORD) PRINCIPLE AND APPLICATIONS.

Optical rotatory dispersion is the variation in the optical rotation of a substance with a change

in the wavelength of light.

Optical rotatory dispersion can be used to find the absolute configuration of metal complexes.

For example, when plane-polarized white light from an overhead projector is passed through a

cylinder of sucrose solution, a spiral rainbow is observed perpendicular to the cylinder


https://en.wikipedia.org/wiki/Sucrose

https://en.wikipedia.org/wiki/Overhead_projector

https://en.wikipedia.org/wiki/Coordination_complex

https://en.wikipedia.org/wiki/Absolute_configuration

https://en.wikipedia.org/wiki/Light

https://en.wikipedia.org/wiki/Wavelength

https://en.wikipedia.org/wiki/Optical_rotation


Principle

When white light passes through a polarizer, the extent of rotation of light depends on

its wavelength.

Short wavelengths are rotated more than longer wavelengths, per unit of distance.

Because the wavelength of light determines its color, the variation of color with distance

through the tube is observed

This dependence of specific rotation on wavelength is called optical rotatory dispersion.

In all materials the rotation varies with wavelength.

The variation is caused by two quite different phenomena.

The first accounts in most cases for the majority of the variation in rotation and should not

strictly be termed rotatory dispersion.

It depends on the fact that optical activity is actually circular birefringence.

In other words, a substance which is optically active transmits right circularly polarized light

with a different velocity from left circularly polarized light.

In addition to this pseudodispersion which depends on the material thickness, there is a true

rotatory dispersion which depends on the variation with wavelength of the indices of refraction

for right and left circularly polarized light.

For wavelengths that are absorbed by the optically active sample, the two circularly polarized

components will be absorbed to differing extents.

This unequal absorption is known as circular dichroism.

Circular dichroism causes incident linearly polarized light to become elliptically polarized.

The two phenomena are closely related, just as are ordinary absorption and dispersion.

If the entire optical rotatory dispersion spectrum is known, the circular dichroism spectrum can

be calculated, and vice versa.


https://en.wikipedia.org/wiki/Elliptically_polarized

https://en.wikipedia.org/wiki/Circular_dichroism

https://en.wikipedia.org/wiki/Circularly_polarized

https://en.wikipedia.org/wiki/Birefringence

https://en.wikipedia.org/wiki/Specific_rotation

https://en.wikipedia.org/wiki/Wavelength

https://en.wikipedia.org/wiki/Polarizer

https://en.wikipedia.org/wiki/Electromagnetic_spectrum#Visible_radiation_.28light.29


APPLICATIONS.

Circular dichroism (CD) is dichroism involving circularly polarized light, i.e., the

differential absorption of left- and right-handed light.

Left-hand circular (LHC) and right-hand circular (RHC) polarized light represent two

possible spin angular momentumstates for a photon, and so circular dichroism is also referred to

as dichroism for spin angular momentum.

CD spectroscopy has a wide range of applications in many different fields.

Most notably, UV CD is used to investigate the secondary structure of proteins.

UV/Vis CD is used to investigate charge-transfer transitions.

Near-infrared CD is used to investigate geometric and electronic structure by

probing metal d→dtransitions.

Vibrational circular dichroism, which uses light from the infrared energy region, is used for

structural studies of small organic molecules, and most recently proteins and DNA.


https://en.wikipedia.org/wiki/Infrared

https://en.wikipedia.org/wiki/Vibrational_circular_dichroism

https://en.wikipedia.org/wiki/D_orbital

https://en.wikipedia.org/wiki/D_orbital

https://en.wikipedia.org/wiki/Transition_metal

https://en.wikipedia.org/wiki/Electronic_structure

https://en.wikipedia.org/wiki/Molecular_structure

https://en.wikipedia.org/wiki/Near-infrared

https://en.wikipedia.org/wiki/Charge-transfer_complex

https://en.wikipedia.org/wiki/Secondary_structure

https://en.wikipedia.org/wiki/Ultraviolet

https://en.wikipedia.org/wiki/Spectroscopy

https://en.wikipedia.org/wiki/Spin_angular_momentum_of_light

https://en.wikipedia.org/wiki/Light

https://en.wikipedia.org/wiki/Absorption_(electromagnetic_radiation)

https://en.wikipedia.org/wiki/Circular_polarization

https://en.wikipedia.org/wiki/Dichroism


TEST

UNIT -5-PART -13- THERMODYNAMICS and GASEOUS STATE

1.Thermodynamic equations of state are

I. ¿)T = T ¿)V - P II. ¿)T = V - T ¿)P

III. ¿)T = T ¿)V + P IV. ¿)T = V + T ¿)P

a. . I and III only b. II and IV only c. III only d. I and II only

2. Which represents closed system?

a. dE≠ 0, dm≠ 0 b. dE≠ 0, dm = 0

c. dT = 0 d. dQ = 0

3.In an open system------- takes place with its surroundings

a. exchange of matter alone b. exchange of energy alone

c. both matter as well as energy d. volume alone

4. The term partial molar property is applicable to

a. open system b. closed system

c. isolated system d. reversible system

5. Extensive property is

a. U b. H c. S d. all

6. Extensive property depends on

a. temperature only b. pressure only

c. number of moles only d all

7. Variation in property with change in number of moles at constant temperature and pressure

is called

a. ideal property b. real property

c. partial molar property d. entropy

8. Chemical potential is

a. partial molar internal energy b. partial molar enthalpy

c. Partial Molar Free Energy d. partial molar volume



9. Which represents chemical potential ?

a. (∂ U∂ n 1

) T, P, n2,n3 b. (∂ H∂ n 1

) T, P, n2,n3

c. ∂ S

∂ n1¿ T, P, n2,n3 d. (

∂ G∂ n 1

) T, P, n2,n3...

10. Variation of chemical potential with temperature is equal to

a. partial molar internal energy b. partial molar entropy


11. When temperature increases, chemical potential

a. increases b. decreases.

c. remains same d. not predicable

12. Chemical potential of a solid at its melting point is X. Then thechemical potential of its

liquid is

a. greater than X b. lesser than X

c. equal to X d. twice of X

13. Chemical potential of a liquid at its boiling point is Y. Then the chemical potential of its

vapour is

a. greater than Y b. lesser than Y

c. equal to Y d. twice of Y

14. Variation of chemical potential with pressure is equal to



15. Gibbs –Duhem equation is

a. ∑ ¿dGi= 0 b ∑ μi dGi= 0

c. ∑ ¿dμi= 0 d. ∑ μi dni= 0

16. For system involving two constituents Gibbs –Duhem equation is

a. n1 dμ2+ n2 dμ1= 0 b. n1 dμ1+ n2 dμ2 = 0

c. n1 dn1+ μ2 dμ2 = 0 d. n1 n2 + dμ1dμ2 = 0

17.According to Nernst heat theorem

a. limT → 0

[ ∂ ∆ G∂ T

]P = 0 b. lim

T → 0[ ∂ ∆ H

∂ T]P = 0



c. limT → 0[CP] = 0 d. all the above

18. According to Third law of thermodynamics

a. limT → ∞

[S ] = 0 b.limT → 0[S] = 0

c.limV → 0[S] = 0 d. limP→ 0

[S ] = 0

19. Third law of thermodynamics is based on

a. Gibbs Duhem equation b. Nernst heat theorem

c. Gay –Lussac law d. none of the above

20. Exception to Third law of thermodynamics is

a. CO b.NO

c. N2O d. all

21. Entropy of solid NO and N2O is not zero at absolute zero kelvin because

a. They belong to oxides of V group element

b. They are gases at ordinary temperature.

c. They have high boiling point

d. They have alternate arrangements of molecules in the solid.

22. The residual entropy of CO is

a. R ln 3 b. R ln 2

c. R ln 4 d. R ln 23

23. If the residual entropy of CO is 5.76 J/K/mol then that of N2O in J/K/mol is

a. 15.76 b. 11. 52

c. 2.88 d. 0.576

24. Entropy of solid H2 and D2 is not zero at absolute zero kelvin because

a. They belong to I group element b. They are gases at ordinary temperature.

c. They have high boiling point d. There exist ortho and para form

25. The Relation between fugacity and free energychange is given by

a. ∆ G = nT ln f 2

f 1 b. ∆ G = nRT ln

f 2

f 1

c. ∆ G = nR ln f 2

f 1 d. ∆ G = nRT ln f1f2



26. The fugacitities at 20 and 200 atm pressure of one mole of gas are 50 and 100 atm

respectively. The free energy changeaccompanying the compression at 100 K is

a. 8.314 ln 2 b. 83.14 ln 2

c. 831.4 ln 2 d. 8314 ln 2

27. A gas at 0.1 atmpressure occupies a volume 83.14 lit . Its fugacity at 0.1 K isa. 0.2 atm b. 1 atm

c. 8 atm d. 8.3 atm

28. Variation of fugacity with temperature can be calculatedusing

a. ln f 2f 1= -

H 1−H 2

R[

T1−T 2

T 1 T2 ] b. ln

f 2f 1=

VRT [ P2-P1 ]

c. ln f = P VRT d. none of the above

29. Variation of fugacity with pressure can be calculatedusing

a. ln f 2f 1= -

H 1−H 2

R[

T1−T 2

T 1 T2] b.ln

f 2f 1=

VRT [ P2-P1 ]

c. ln f = PVRT d. none of the above

30. The relation between activity and fugacity is

a. Activity = fugacity∈ pure state

fugacity∈mixture

b. Activity = fugacity∈mixture


c. Activity = 1


d. Activity =fugacity∈mixture× fugacity∈pure state

31. The fugacity of a pure gas is 20 atm and the same gas in a mixture is 40 atmTthe activity of

the gas is

a. 2 b. 4

c.8 d.10

32. The relation between activity and activity coefficient is

a. activity coefficient = pressuractivity

b. activity coefficient = 1

pressure



c. activity coefficient = activitypressure

d. activity coefficient = activity×pressure

33. The activity of a gas at 5atm pressure is 20 . What is its activity coefficient?

a. 2 b. 4

c.8 d.10

34. Average velocity of a gas molecule is given by

a. √ 8RTπM

b. √ 2 RTM

c. √ 3 RTM

d. √ 3 RTπM

35. Most probable velocity of a gas molecule is given by

a. √ 8RTπM

b. √ 2 RTM

c. √ 3 RTM

d. √ 3 RTπM

36. Root Mean Square velocity of a gas molecule is given by

a. √ 8RTπM

b. √ 2 RTM

c. √ 3 RTM

d. √ 3 RTπM

37. Most probable velocity is

a. velocity greater than zero

b. velocity possessed by maximum number of molecules

c. velocitylesser than zero

d. velocity possessed by minimum number of molecules

38. Four distinguished particles have a total energy of 6 units. But the particles are restricted to

energy levels to 0 to 6. The number of macro states is

a. 3 b. 6

c. 9 d. 12



39. Three distinguished particles have a total energy of 9 units. But the particles are restricted

to energy levels to 0 to 4. The number of macro states is

a. 3 b. 6

c. 9 d. 1240. Number of microstates in macrostate with equal probability a.

n!(n−r )! b.

n !r !

c. n!

r ! (n−r )! d. n

r! (n−r )!


energy levels to 0 to 6. The number of micro states corresponding to the macro state (3,1) is

.a. 3 b. 4

c. 9 d. 12


energy levels to 0 to 6. The number of micro states corresponding to the macro state (2,2) is

.a. 3 b. 4

c. 6 d. 12


to energy levels to 0 to 4. The number of m1cro states corresponding to the macro state (1,2) is

a. 3 b. 6

c. 9 d. 12

44. In a random distribution of 10 particles between two boxes with equal probability the

number of microstates in macrostate (3,7) is

a. 3 b. 6

c. 9 d. 120

45. In a random distribution of 10 particles between two boxes with equal probability The

total number of microstates is

a. 212 b. 310

c. 2 10 d. 2 14

46. A system has 5 different macrostates under which there 6,20,42,12 and 2 microstates. The

relative probability for the micro states 6 is .



a. 6

82 b. 2082

c. 2

82 d.1

82

47. A system has 5 different macrostates under which there 6,20,42,12 and 2 microstates. The

relative probability for the micro states 2 is .

a. 6

82 b. 2082

c. 2

82 d.1

82

48. Ten particles are distributed in two boxes. The number of macro states is

a. 10 b.11

c.21 d. 30

49.Four coins are tossed . The number of macro states is

a. 10 b.11

c.16 d. 30

50. . Which represents variation of chemical potential with pressure?

a. (∂ U∂ n 1

) T, P, n2,n3 b. (∂V∂ n 1

) T, P, n2,n3

c. −( ∂ S∂ n1

) T, P, n2,n3 d. (∂ G∂ n 1

) T, P, n2,n3...

KEY – UNIT-5 – PART-1 - THERMO

1 2 3 4 5 6 7 8 9 10

B B C A D D C C D B

11 12 13 14 15 16 17 18 19 20

B C C D C B A B B D



21 22 23 24 25 26 27 28 29 30

D B D D B C B A B B

31 32 33 34 35 36 37 38 39 40

A C B A B C B C A C

41 42 43 44 45 46 47 48 49 50

B C A D C A C B C B

TEST -PART -2- STATISTICAL THERMODYNAMICS ,EQUILIBRIUM

and PHASE RULE

1. The translational partition function is

a. (2 π mKT )32 V

h3 b. . (2 π KT )

32 V

h3

c. . (2π mT )32 V

h3d. 8 π 2 IKT

σ h

2. Thermal de-broglie wavelength is

a. h

(2 π mKT )32

b. h

(2 π mKT )12

c. h

(2 π mKT )32 d. 8 π 2 IKT

σ h

3.The order of magnitude of translational partition function of diatomic gas is

a. 1 -10 b. 10 25- 1030

c .10 - 1000 d. 0 -1

4. Translational partition function depends upon

a. Volume only b. temperature only

c. both temperature and volume d. none of the above



5. The rotational partition function is given by

a. 8 π IKT

σ b. 8π 2 IKTσ h

c. 8π 2 IKTσ h2 d. (2 π mKT )

32 V

h3

6. Characteristic rotational temperature is

a. 8 π IK

h 2 b. h2

8π 2 IK

c. h

8 π IK d. (2 π mKT )32 V

h3

7. The symmetry number of N2 is

a. 1 b.2

c. 3 d. 0

8..The vibrational partition function is

a. 1

1−e−θT

b. 8π 2 IKTσ h

c. 1

eθT −1

d. (2 π mKT )32 V

h3

9. The order of magnitude of vibrational partition function of diatomic gas is

a. 1 -10 b. 10-100

c . 10 25- 1030 d. 0 -1

10. The degeneracy of the ground state nuclear energy level is given by--- partition function

a. Translational b. Rotational

c. Nuclear. d. vibrational

11. The nuclear partition function is

a. (I + 1)( 2I + 1) b. 8 π 2 IKTσ h2

c. (I + 1)( 3I + 1) d. (2π mKT )32 V

h3



12. The translational partition function of H2 gas confined in a box of 81 m3 is 3× 10 30 at 3000

K. The thermal de Broglie wavelength is

a. 3 A b. 3 nm

c. 27 nm d. 59 nm

13. The translational partition function of a gas confined in a box of 1 m3 with thermal de

Broglie wavelength 10 × 10 -11 m is

a. 3× 10 30 b. 1× 10 30

c. 3× 10 33 d.3× 10 3

14. If the value of 8 π 2 Kh2 is 2× 10 38 .The rotational partition function of H 2 gas at 100 K ,with

moment of inertia 1.5 × 10 -38gKg/m2 is

a. 165 × 10 -2 b. 1.65 × 10 -2

c... 165 × 10 -8 d. 3 × 10 2

15. The rotational partition function of a gas at 100 K ,with moment of inertia 1.5 × 10 -38Kg/m2

is 3 × 10 2 and if the value of 8 π 2 Kh2 is 2× 10 38 the gas is

a. Mono atomic b. homo nuclear diatomic

c. hetero nuclear diatomic d. can not be predicted

16. The rotational partition function of a diatomic gas at 100 K is 150 × 10 -2 and if the value

of 8 π 2 Kh2 is 3 × 10 38 , the moment of inertia of the gas is

a. 165 × 10 -2 b. 2 × 10 -2

c...1 × 10 -40 d. 165 × 10 2

17. At what temperature the rotational partition function of a diatomic gas will be 150 × 10 -2

The value of 8 π 2 Kh2 is 3× 10 38 . The moment of inertia of the gas is 2 × 10 -40 Kg/m2

a. 100 K b. 2 × 10 2 K

c... 3 × 10 3 K d. 50 K

18: If the value of 8 π 2 Kh2 is 1.5× 10 38 Calculate the rotational partition function of H 2 gas at


a. 1 × 10 2 b. 2 × 10 2



c. 3 × 10 3 d. 165 × 10 -2

19: The rotational partition function of a gas at 100 K ,with moment of inertia 2.2 × 10 -

40gm/cm2 is 165 × 10 -2 and if the value of 8 π 2 Kh2 is 1.5× 10 38 .The atomicity of the gas

a. 2 b. 1

c. 3 d 4

20: The rotational partition function of a diatomic gas at 100 K is 150 × 10 -2 and if the value

of 8 π 2 Kh2 is 1.5× 10 38 Find the moment of inertia of the gas

a. 1 × 10 2 b. 2 × 10 2

c.. 3 × 10 3 d. 2 × 10 -40


The value of 8 π 2 Kh2 is 1.5× 10 38 . The moment of inertia of the gas is 2 × 10 -40 gm/cm2

a. 200K b. 100K

c. 300K d 400K

22. If the rotational partition function of a diatomic gas is 600 × 10 2 . If the value of hcKT is

0.5× 10 -4 Find the rotational constant.

a. 1 × 10 2 b. 2 × 10 2

c.. 3 × 10 3 d. 600 × 10 -2


0.5× 10 -4 what will be the spacing between the lines of microwave spectrum of the gas.

a. 2 cm-1 b. 6 cm-1

c. 12 cm-1 d 24 cm-1

24. The characteristic rotational temperature of H2 gas at 3000 K is 150 K. Its rotational partition function is a. 2 b. 1

c. 3 d 2025. What is the characteristic rotational temperature of a gas whose rotational partition function at 3000 K is 20. a. 200K b. 100K c. 300K d.150 K



26.. Calculate the vibrational partition function of a gas at 1000 K whose vibrational frequency

is such that e−h ϑKT = 0.5

a. 2 b. 1c. 3 d 20

27. .Calculate the vibrational partition function of a gas at 1000 K whose vibrational frequency

is such that hϑKT = 2.3 given e−2.3 = 0.100

a. 2 b. 1.11c. 3 d 20

28. The electronic partition function of Hydrogen and Helium isa. 2,2 b. 1,2c. 2,1 d 20,3

29.The nuclear partition function of ortho H2 and ortho D2 molecules.isa. 2,2 b. 1,2c. 3,6 d 20,3

30 The statistics applicable for distinguishable particles is

a. Bose- Einstein b. Maxwell

c. Fermi- Dirac d. none of the above

31. Most probable distribution of Maxwellons is

a. gi

eα+β ∈ i+1 b.

gi

eα+β ∈ i−1

c. g i

eα+β ∈ i d.gi

eα+β ∈ i+2

32. The number of ways of arranging 2 distinguishable particles in two states is

a. 2 b. 4c. 3 d 20

33. The number of ways of arranging 2 distinguishable particles in three states isa. 2 b. 4c. 9 d 20

34.The number of ways of arranging 3 distinguishable particles in two states is a. 2 b. 4c. 9 d 8

35 The number of ways of distributing two Maxwellons among four energy levels is

a. 10 b.64

c. 16 d. 15

36. The number of ways of arranging ‘ni ‘ Maxwellons in ‘gi’ states is



a. ∏ [¿+gi−1]!(gi−1) !×∋! b. N!∏

1

i (gi)¿

¿!

c.∏ [ g]!(gi−1) !×∋! d. ∏ [g ]!

¿¿ ¿

37. Most probable distribution of proton is

a. gi

eα+β ∈ i+1 b.

gi

eα+β ∈ i−1

c.1

eα+β ∈ i+1d.

g i

eα+β ∈ i

38. The statistics applicable for photon gas is

a. Maxwell b. Fermi- Dirac

c. Bose- Einstein d. none of the above

39. The most probable distribution by photons is

a. gi

eα+β ∈ i+1 b.

gi

eα+β ∈ i−1

c. g i

eα+β ∈ i d.gi

eα+β ∈ i+1

40. The number of ways of arranging ni Bosons in gi states is

a. ∏ [¿+gi−1]!(gi−1) !×∋! b. N!∏

1

i (gi)¿

¿!

c.∏ [ g]!(gi−1) !×∋! d. ∏ [g ]!

¿¿ ¿

41. Bose- Einstein statistics is applicable to molecules which have

a. integral spin b. half integral spin

c. negative spin d. zero spin

42. Bosons are particles having

a. half integral spin b. integral spin


43. The number of ways of distributing two Bosons among four energy levels is

a. 10 b.20

c. 30 d. 50

44. Planc’s energy distribution formula can be derived from





45. The statistics applicable for protons is



46. Pauli’s principle is followed in



47. Most probable distribution of electron is

a. gi

eα+β ∈ i+1 b.

gi

eα+β ∈ i−1

c. g i

eα+β ∈ i d.gi

eα+β ∈ i−1

48. The number of ways of distributing two Fermions among four energy levels is

a. 2 b. 4

c. 6 d. 5

49 .The statistics applicable for H2 is



50. The most probable distribution by helium ion is

a. gi

eα+β ∈ i+1 b.

gi

eα+β ∈ i−1

c. g i

eα+β ∈ i d. gi

eα+β ∈ i−151. Fermions are particles having

a. zero spin b. half integral spin c. integral spin d. negative spin52. The number of ways of arranging ni Fermions in gi states is

a. ∏ [¿+gi]!(gi−1) !×∋! b. N!∏

1

i (gi)¿

¿! c.∏ [g ]!(gi−¿)!×∋! d. ∏ [g ]!

¿¿ ¿

53.Thermionic emission is satisfactorily explained by





54.The statistics applicable for electron gas is



55. .The specific heat capacity of sold remains constant at all temperature. This is stated by

a. Einstein b. Debye

c. Dulong and Pettit d. Boltzmann

56. Which is true according to Einstein

I. As temperature increases , the atoms vibrate simple harmonically.

II. A crystal can be considered as a system of non – interacting particles.

III .Each atom vibrates independently .

a. I ,II only b. I,II, III only c. II, III, only d. all

57. The expression for heat capacity according to Einstein is

a. 3R θE

Te

θE

T

(e¿¿θE

T – 1)2

¿ b. 3R

eθE

T

(e¿¿θE

T – 1)2

¿

c. 3R( θE

T¿¿2 e

θE

T

(e¿¿θE

T – 1)2

¿ d.3R(

θE

T¿¿2

eθ E

T

(e¿¿θE

T )2

¿

58. According to Einstein’s theory, the heat capacity of solid at low temperature is

a. zero b. Proportional to T3

c.. infinity d. 3R

59. According to Debye’s theory, the heat capacity of solid at low temperature is

a. independent b. Proportional to T3

c. zero d. 3R

60. ASSERTION(A) : vibrational partition function has very low value .

REASON(R): It depends on temperature

a. Both A and R are true and R is the correct explanation of A

b. Both A and R are true but R is not correct explanation of A

c. A is true but R is false

d. Both A and B are false



61. Match the following

Partition function Expression1 Translational partition function A

8 π 2 IKTσ h2

2 Rotational partition function B(2 π mKT )

32 V

h3

3 Vibrational partition function C gground + g excited

4 Electronic partition function D 1

1−e−h ϑKT

a. 1-A,2-B,3-C,4-D b. 1-B,2-C,3-D,4-Ab. 1-B,2-A,3-D,4-C d. 1-A,2-C,3-B,4-D

62. If ∈t,∈r ,∈v … represents contributions to the energy corresponding to translational ,

rotational ,vibrational and q t , qr , qv represents contributions to the partition function

corresponding to translational , rotational ,vibrational then which is true

a. E = ∈t+∈r+∈v q = q t+qr+qv b. E = ∈t ×∈r ×∈v q = q t+qr+qv

c. E = ∈t+∈r+∈v q = q t ×qr ×qv d. E = ∈t ×∈r ×∈v q = q t ×qr ×qv

63. Sackur- Tetrode equation involves

a. volume dependent factor b. temperature dependent factor

c. both volume and temperature d. none of the above

64. Gibb’s phase rule is applicable to a. Homogeneous equilibrium b. Heterogeneous equilibrium c. Both d. none

65. Gibb’s phase rule is a. F +P = C +2 b. F= P-C+2 c. F+C = P+2 d. F= P +C+2

66. The number of phases present in 2 immiscible liquid system isa. 1 b. 2 c. 3 d. 4

67. The number of phases present in 2 miscible liquid system isa. 1 b. 2 c. 3 d. 4

68. The number of phases present in mixture of 3 non reacting gas system isa. 1 b. 2 c.3 d. 4

69. The number of phases present in CaCO3↔ CaO + CO2 system is

a. 1 b. 2 c. 3 d. 4

70 . SR↔SM . The number of phases present in this system is



a. 1 b. 2 c. 3 d. 4

KEY – UNIT-5 – PART-2 – STAT

1 2 3 4 5 6 7 8 9 10

A B B C C B B A D C

11 12 13 14 15 16 17 18 19 20

A A B D A C D D A D

21 22 23 24 25 26 27 28 29 30

B D C D D A B C C B

31 32 33 34 35 36 37 38 39 40

C B C D C B B C B A

41 42 43 44 45 46 47 48 49 50

A B A C C B A C B A

51 52 53 54 55 56 57 58 59 60

B C B B C D C A B B

61 62 63 64 65 66 67 68 69 70

B C C B A B A C C B



UNIT -5 - PART -3- ANALYTICAL CHEMISTRY

1. The stationary phase and mobile phase of Paper chromatography: is

a. solid- liquid b. liquid- solid

c. liquid – liquid d. liquid – gas

2. Principle is used in column chromatography is

a. diffusion b. adsorption

c. selective adsorption d. osmosis

3. Which is not adsorbent in column chromatography

a. Magnesium oxide b. Magnesium carbonate

c. Calcium carbonate d. Xylene

4. Eluotropic series is the series of solvents according to their increasing

a. eluting power b. adsorbing power

c. dissolving power d. polarity

5. The component which has greater adsorbing power is adsorbed in the -----of the column

a. lower part b. upper part

c. middle part d. lowest part

6. The position of migrated spots on the paper chromatograms are indicated by

a. RF b. Rx

c. RM d. all the above

7.. The Migration parameter RF and RM are related as

a. RM = log ( 1

RF) b. RM = log (

1RF

+1)

c. RM = log ( 1

RF -1) d. RM = log ( RF+1)

8. .Stahl’s triangle relates

1. adsorbant activity 2. nature of solute 3. nature of solvent

a. I and II only b. II and III only c. I and III only d. I,II and III

9.. The carrier gas not used in Gas-Liquid Chromatography is

a. Hydrogen b. Helium



c. Nitrogen d. Oxygen

10. Hydrogen is not used as carrier gas because

I. It may react with unsaturated compound

II. It creates fire and explosive hazard

III .It has low thermal conductivity

a. I only b. II only c. III only d. both I and II

11. The efficiency of column in HPLC is expressed as

a Rf value b. HETP

c. Rx d. none

.12. Van Dempter equation is

a. HETP = A +Bμ - Cμ b. HETP = A -

Bμ

c. HETP = A +Bμ + Cμ d. HETP =

Bμ + Cμ

13. Van Dempter graph is the plot of HETP versus

a. velocity of gas b. density of solid

c. resistivity d. permeability

14.. The horizontal portion in the TG curve indicate the region where there is

a. no pressure change b. no temperature change

c. no volume change d. no mass change

15. Differential thermogram is a plot of

a. Mass vs Temperature b. dmdt vs Temperature

c. ∆T vs Temperature d. Temperature vs Volume

16. The magnitude of rotation depends on

a. nature of the substance b. concentration of solution.

c nature of solvent d. all the above

17. Specific rotation is given by

a. α Dl =

100× θl × c b. α D

l = θ

l× c c. α Dl =

100l× c d. α D

l = 100× θ

c

18. The medium is said to be circularlybirefringent if



a. refractive index is same b. refractive index is different c. very dilute d. none

19. In optically active medium EL and ER rotate at

a. different speeds b. same speed c. perpendicular d. parallel

20. For circularly birefringent medium

a. nL = nR b.. nL≠nR c. nL = 0 d. nR = 0

21. In medium exhibiting circular dichroism

a. nL≠nR and magnitude of EL = magnitude of ER

b. nL≠nR and magnitude of EL≠ magnitude of ER

c. nL = nR and magnitude of EL = magnitude of ER

d. nL≠nR and magnitude of EL = magnitude of ER

22. ORD curves areplot of specific rotation vs

a, concentration b. intensity c. wavelength d. amplitude

23. The rate of change of specific rotation is known as

a. CD b. ORD c. optical activity d. polarization

24. The influence of electric field on optical activity is called

a. Faraday effect b. Kerr electric optic effect c. assymetric effect d. nepheluxity effect

25. The influence of magnetic field on optical activity is called

a. Faraday effect b. Kerr electric optic effect c. assymetric effect d. nepheluxity effect

26. MORD is the influence of

a. electric field on optical activity b. magnetic field on optical activity. solvent

d. temperature

27. Kerr electric optic effect is

a. npl – npr = KE2 (lamda) b. φ = VBl c. α Dl =

100× θl × c d. α D

l = θ

l× c

28. Verdet equation is

a. npl – npr = KE2 ¿ b. φ = VBl c. α Dl =

100× θl × c d. α D

l = θ

l× c

29. Neutron diffraction is less popular method in crystallography because I large size II high

cost neutron spectrometer.

a. I only b. II only c. both I and II d. none

30. X – ray Diffraction differs from neutron diffraction in the sense that X – rays are scattered



a. by orbital electrons whereas neutrons are scattered by atomic nuclei.

b. power increases with atomic number but there is no regular trend for neutron scattering

c. only one scattering but in neutron diffraction two scattering takes place

d. all

317. Electron diffraction studies utilize electrons with energies

a. 40keV . b. 40 eV c. 40V d. 40 MeV

32. Electron diffraction studies are used for studying molecules in

a. solid state b. liquid state c. gaseous state d. all

33. Diffraction of X – ray differs from diffraction of electron in the sense that 1. Diffraction of

X – ray is applicable to crystal whereas diffraction of electron is applicable to gases. 2.

Diffraction of X – ray depends upon the spacing between the layers but the diffraction of

electrons depends upon the distances between the atoms.

a. I only b. II only c. both I and II d. none

34. The total intensity of scattering in Electron diffraction is given by

a. Bragg equation b. Born equation c. Wierl equation d. Duhem equation

35. Wierlequation is

a. U = - z+¿ z−¿e2 A N

r0¿

¿( 1- 1n ) b. I ∝∑

i , jf i f j

sin s R ij

s Rij c. S = 4Rθ d. S = 6 fθ



KEY – UNIT-5 – PART-3 – analytical

1 2 3 4 5 6 7 8 9 10

C C D A A D C D D D

11 12 13 14 15 16 17 18 19 20

B C A D C

21 22 23 24 25 26 27 28 29 30



UNIT – 5 – THERMODYNAMICS / STATISTICAL / ANALYTICAL

1.Thermodynamic equations of state are

I. ¿)T = T ¿)V - P II. ¿)T = V - T ¿)P

III. ¿)T = T ¿)V + P IV. ¿)T = V + T ¿)P

a. . I and III only b. II and IV only c. III only d. I and II only

2. Which represents closed system?

a. dE≠ 0, dm≠ 0 b. dE≠ 0, dm = 0

c. dT = 0 d. dQ = 0

3.In an open system------- takes place with its surroundings

a. exchange of matter alone b. exchange of energy alone

c. both matter as well as energy d. volume alone

4. The term partial molar property is applicable to

a. open system b. closed system

c. isolated system d. reversible system

5. Extensive property is

a. U b. H

c. S d. all the above

6. Extensive property depends on

a. temperature only b. pressure only

c. number of moles only d all the above

7. Variation in property with change in number of moles at constant temperature and pressure

is called

a. ideal property b. real property

c. partial molar property d. entropy



8. Chemical potential is

a. partial molar internal energy b. partial molar enthalpy


9. Which represents chemical potential ?

a. (∂ U∂ n 1

) T, P, n2,n3 b. (∂ H∂ n 1

) T, P, n2,n3

c. ∂ S

∂ n1¿ T, P, n2,n3 d. (

∂ G∂ n 1

) T, P, n2,n3...

10. Variation of chemical potential with temperature is equal to



11. When temperature increases, chemical potential

a. increases b. decreases.

c. remains same d. not predicable

12. Chemical potential of a solid at its melting point is X. Then thechemical potential of its

liquid is

a. greater than X b. lesser than X

c. equal to X d. twice of X

13. Chemical potential of a liquid at its boiling point is Y. Then the chemical potential of its

vapour is

a. greater than Y b. lesser than Y

c. equal to Y d. twice of Y

14. Variation of chemical potential with pressure is equal to



15. Gibbs –Duhem equation is

a. ∑ ¿dGi= 0 b ∑ μi dGi= 0

c. ∑ ¿dμi= 0 d. ∑ μi dni= 0

16. For system involving two constituents Gibbs –Duhem equation is

a. n1 dμ2+ n2 dμ1= 0 b. n1 dμ1+ n2 dμ2 = 0



c. n1 dn1+ μ2 dμ2 = 0 d. n1 n2 + dμ1dμ2 = 0

17.According to Nernst heat theorem

a. limT → 0

[ ∂ ∆ G∂ T

]P = 0 b. lim

T → 0[ ∂ ∆ H

∂ T]P = 0

c. limT → 0[CP] = 0 d. all the above

18. According to Third law of thermodynamics

a. limT → ∞

[S ] = 0 b.limT → 0[S] = 0

c.limV → 0[S] = 0 d. limP→ 0

[S ] = 0

19. Third law of thermodynamics is based on

a. Gibbs Duhem equation b. Nernst heat theorem

c. Gay –Lussac law d. none of the above

20. Exception to Third law of thermodynamics is

a. CO b.NO

c. N2O d. all

21. Entropy of solid NO and N2O is not zero at absolute zero kelvin because

a. They belong to oxides of V group element

b. They are gases at ordinary temperature.

c. They have high boiling point

d. They have alternate arrangements of molecules in the solid.

22. The residual entropy of CO is

a. R ln 3 b. R ln 2

c. R ln 4 d. R ln 23

23. If the residual entropy of CO is 5.76 J/K/mol then that of N2O in J/K/mol is

a. 15.76 b. 11. 52

c. 2.88 d. 0.576

24. Entropy of solid H2 and D2 is not zero at absolute zero kelvin because

a. They belong to I group element b. They are gases at ordinary temperature.

c. They have high boiling point d. There exist ortho and para form

25. The Relation between fugacity and free energychange is given by



a. ∆ G = nT ln f 2

f 1 b. ∆ G = nRT ln

f 2

f 1

c. ∆ G = nR ln f 2

f 1 d. ∆ G = nRT ln f1f2

26. The fugacitities at 20 and 200 atm pressure of one mole of gas are 50 and 100 atm

respectively. The free energy changeaccompanying the compression at 100 K is

a. 8.314 ln 2 b. 83.14 ln 2

c. 831.4 ln 2 d. 8314 ln 2

27. A gas at 0.1 atmpressure occupies a volume 83.14 lit . Its fugacity at 0.1 K isa. 0.2 atm b. 1 atm

c. 8 atm d. 8.3 atm

28. Variation of fugacity with temperature can be calculatedusing

a. ln f 2f 1= -

H 1−H 2

R[

T1−T 2

T 1 T2 ] b. ln

f 2f 1=

VRT [ P2-P1 ]

c. ln f = P VRT d. none of the above

29. Variation of fugacity with pressure can be calculatedusing

a. ln f 2f 1= -

H 1−H 2

R[

T1−T 2

T 1 T2] b.ln

f 2f 1=

VRT [ P2-P1 ]

c. ln f = PVRT d. none of the above

30. The relation between activity and fugacity is

a. Activity = fugacity∈ pure state

fugacity∈mixture

b. Activity = fugacity∈mixture


c. Activity = 1


d. Activity =fugacity∈mixture× fugacity∈pure state

31. The fugacity of a pure gas is 20 atm and the same gas in a mixture is 40 atmTthe activity of

the gas is

a. 2 b. 4

c.8 d.10



32. The relation between activity and activity coefficient is

a. activity coefficient = pressuractivity

b. activity coefficient = 1

pressure

c. activity coefficient = activitypressure

d. activity coefficient = activity×pressure

33. The activity of a gas at 5atm pressure is 20 . What is its activity coefficient?

a. 2 b. 4

c.8 d.10

34. Average velocity of a gas molecule is given by

a. √ 8RTπM

b. √ 2 RTM

c. √ 3 RTM

d. √ 3 RTπM

35. Most probable velocity of a gas molecule is given by

a. √ 8RTπM

b. √ 2 RTM

c. √ 3 RTM

d. √ 3 RTπM

36. Root Mean Square velocity of a gas molecule is given by

a. √ 8RTπM

b. √ 2 RTM

c. √ 3 RTM

d. √ 3 RTπM

37. Most probable velocity is

a. velocity greater than zero

b. velocity possessed by maximum number of molecules

c. velocitylesser than zero

d. velocity possessed by minimum number of molecules




energy levels to 0 to 6. The number of macro states is

a. 3 b. 6

c. 9 d. 12


to energy levels to 0 to 4. The number of macro states is

a. 3 b. 6

c. 9 d. 1240. Number of microstates in macrostate with equal probability a.

n!(n−r )! b.

n !r !

c. n!

r ! (n−r )! d. n

r! (n−r )!

41. The translational partition function is

a. (2 π mKT )32 V

h3 b. . (2 π KT )

32 V

h3

c. . (2 π mT )32 V

h3d. 8 π 2 IKT

σ h

42. Thermal de-broglie wavelength is

a. h

(2 π mKT )32

b. h

(2 π mKT )12

c. h

(2 π mKT )32 d. 8 π 2 IKT

σ h

43.The order of magnitude of translational partition function of diatomic gas is

a. 1 -10 b. 10 25- 1030

c .10 - 1000 d. 0 -1

44. Translational partition function depends upon

a. Volume only b. temperature only

c. both temperature and volume d. none of the above



45. The rotational partition function is given by

a. 8 π IKT

σ b. 8π 2 IKTσ h

c. 8π 2 IKTσ h2 d. (2 π mKT )

32 V

h3

46. Characteristic rotational temperature is

a. 8 π IK

h 2 b. h2

8π 2 IK

c. h

8 π IK d. (2 π mKT )32 V

h3

47. The symmetry number of N2 is

a. 1 b.2

c. 3 d. 0

48..The vibrational partition function is

a. 1

1−e−θT

b. 8π 2 IKTσ h

c. 1

eθT −1

d. (2 π mKT )32 V

h3

49. The order of magnitude of vibrational partition function of diatomic gas is

a. 1 -10 b. 10-100

c . 10 25- 1030 d. 0 -1

50. The degeneracy of the ground state nuclear energy level is given by--- partition function

a. Translational b. Rotational

c. Nuclear. d. vibrational

51. The nuclear partition function is

a. (I + 1)( 2I + 1) b. 8 π 2 IKTσ h2

c. (I + 1)( 3I + 1) d. (2π mKT )32 V

h3



52. The translational partition function of H2 gas confined in a box of 81 m3 is 3× 10 30 at 3000

K. The thermal de Broglie wavelength is

a. 3 A b. 3 nm

c. 27 nm d. 59 nm

53. The translational partition function of a gas confined in a box of 1 m3 with thermal de

Broglie wavelength 10 × 10 -11 m is

a. 3× 10 30 b. 1× 10 30

c. 3× 10 33 d.3× 10 3

54. If the value of 8 π 2 Kh2 is 2× 10 38 .The rotational partition function of H 2 gas at 100 K ,with

moment of inertia 1.5 × 10 -38gKg/m2 is

a. 165 × 10 -2 b. 1.65 × 10 -2

c... 165 × 10 -8 d. 3 × 10 2

55. The rotational partition function of a gas at 100 K ,with moment of inertia 1.5 × 10 -38Kg/m2

is 3 × 10 2 and if the value of 8 π 2 Kh2 is 2× 10 38 the gas is

a. Mono atomic b. homo nuclear diatomic

c. hetero nuclear diatomic d. can not be predicted

56. The rotational partition function of a diatomic gas at 100 K is 150 × 10 -2 and if the value

of 8 π 2 Kh2 is 3 × 10 38 , the moment of inertia of the gas is

a. 165 × 10 -2 b. 2 × 10 -2

c...1 × 10 -40 d. 165 × 10 2


The value of 8 π 2 Kh2 is 3× 10 38 . The moment of inertia of the gas is 2 × 10 -40 Kg/m2

a. 100 K b. 2 × 10 2 K

c... 3 × 10 3 K d. 50 K

58: If the value of 8 π 2 Kh2 is 1.5× 10 38 Calculate the rotational partition function of H 2 gas at


a. 1 × 10 2 b. 2 × 10 2



c. 3 × 10 3 d. 165 × 10 -2

59: The rotational partition function of a gas at 100 K ,with moment of inertia 2.2 × 10 -

40gm/cm2 is 165 × 10 -2 and if the value of 8 π 2 Kh2 is 1.5× 10 38 .The atomicity of the gas

a. 2 b. 1

c. 3 d 4

60: The rotational partition function of a diatomic gas at 100 K is 150 × 10 -2 and if the value

of 8 π 2 Kh2 is 1.5× 10 38 Find the moment of inertia of the gas

a. 1 × 10 2 b. 2 × 10 2

c.. 3 × 10 3 d. 2 × 10 -40


The value of 8 π 2 Kh2 is 1.5× 10 38 . The moment of inertia of the gas is 2 × 10 -40 gm/cm2

a. 200K b. 100K

c. 300K d 400K


0.5× 10 -4 Find the rotational constant.

a. 1 × 10 2 b. 2 × 10 2

c.. 3 × 10 3 d. 600 × 10 -2


0.5× 10 -4 what will be the spacing between the lines of microwave spectrum of the gas.

a. 2 cm-1 b. 6 cm-1

c. 12 cm-1 d 24 cm-1

64. The characteristic rotational temperature of H2 gas at 3000 K is 150 K. Its rotational partition function is a. 2 b. 1

c. 3 d 2065. What is the characteristic rotational temperature of a gas whose rotational partition function at 3000 K is 20. a. 200K b. 100K c. 300K d.150 K



66.. Calculate the vibrational partition function of a gas at 1000 K whose vibrational frequency

is such that e−h ϑKT = 0.5

a. 2 b. 1c. 3 d 20

67. .Calculate the vibrational partition function of a gas at 1000 K whose vibrational frequency

is such that hϑKT = 2.3 given e−2.3 = 0.100

a. 2 b. 1.11c. 3 d 20

68. The electronic partition function of Hydrogen and Helium isa. 2,2 b. 1,2c. 2,1 d 20,3

69.The nuclear partition function of ortho H2 and ortho D2 molecules.isa. 2,2 b. 1,2c. 3,6 d 20,3

70 The statistics applicable for distinguishable particles is

a. Bose- Einstein b. Maxwell

c. Fermi- Dirac d. none of the above

71. Most probable distribution of Maxwellons is

a. gi

eα+β ∈ i+1 b.

gi

eα+β ∈ i−1

c. g i

eα+β ∈ i d.gi

eα+β ∈ i+2

72. The number of ways of arranging 2 distinguishable particles in two states is

a. 2 b. 4c. 3 d 20

73. The number of ways of arranging 2 distinguishable particles in three states isa. 2 b. 4c. 9 d 20

74.The number of ways of arranging 3 distinguishable particles in two states is a. 2 b. 4c. 9 d 8

75 The number of ways of distributing two Maxwellons among four energy levels is

a. 10 b.64

c. 16 d. 15

76. The number of ways of arranging ‘ni ‘ Maxwellons in ‘gi’ states is



a. ∏ [¿+gi−1]!(gi−1) !×∋! b. N!∏

1

i (gi)¿

¿!

c.∏ [ g]!(gi−1) !×∋! d. ∏ [g ]!

¿¿ ¿

77. Most probable distribution of proton is

a. gi

eα+β ∈ i+1 b.

gi

eα+β ∈ i−1

c.1

eα+β ∈ i+1d.

g i

eα+β ∈ i

78. The statistics applicable for photon gas is



79. The most probable distribution by photons is

a. gi

eα+β ∈ i+1 b.

gi

eα+β ∈ i−1

c. g i

eα+β ∈ i d.gi

eα+β ∈ i+1

80. The number of ways of arranging ni Bosons in gi states is

a. ∏ [¿+gi−1]!(gi−1) !×∋! b. N!∏

1

i (gi)¿

¿!

c.∏ [ g]!(gi−1) !×∋! d. ∏ [g ]!

¿¿ ¿

81. Bose- Einstein statistics is applicable to molecules which have

a. integral spin b. half integral spin


82. Bosons are particles having

a. half integral spin b. integral spin


83. The number of ways of distributing two Bosons among four energy levels is

a. 10 b.20

c. 30 d. 50

84. Planc’s energy distribution formula can be derived from





85. The statistics applicable for protons is



86. Pauli’s principle is followed in



87. Most probable distribution of electron is

a. gi

eα+β ∈ i+1 b.

gi

eα+β ∈ i−1

c. g i

eα+β ∈ i d.gi

eα+β ∈ i−1

88. The number of ways of distributing two Fermions among four energy levels is

a. 2 b. 4

c. 6 d. 5

89 .The statistics applicable for H2 is



90. The most probable distribution by helium ion is

a. gi

eα+β ∈ i+1 b.

gi

eα+β ∈ i−1

c. g i

eα+β ∈ i d. gi

eα+β ∈ i−191. Fermions are particles having

a. zero spin b. half integral spin c. integral spin d. negative spin92. The number of ways of arranging ni Fermions in gi states is

a. ∏ [¿+gi]!(gi−1) !×∋! b. N!∏

1

i (gi)¿

¿! c.∏ [g ]!(gi−¿)!×∋! d. ∏ [g ]!

¿¿ ¿

93.Thermionic emission is satisfactorily explained by





94.The statistics applicable for electron gas is



95. .The specific heat capacity of sold remains constant at all temperature. This is stated by

a. Einstein b. Debye

c. Dulong and Pettit d. Boltzmann

96. Which is true according to Einstein

I. As temperature increases , the atoms vibrate simple harmonically.

II. A crystal can be considered as a system of non – interacting particles.

III .Each atom vibrates independently .

a. I ,II only b. I,II, III only c. II, III, only d. all

97. The expression for heat capacity according to Einstein is

a. 3R θE

Te

θE

T

(e¿¿θE

T – 1)2

¿ b. 3R

eθE

T

(e¿¿θE

T – 1)2

¿

c. 3R( θE

T¿¿2 e

θE

T

(e¿¿θE

T – 1)2

¿ d.3R(

θE

T¿¿2

eθ E

T

(e¿¿θE

T )2

¿

98. According to Einstein’s theory, the heat capacity of solid at low temperature is

a. zero b. Proportional to T3

c.. infinity d. 3R

99. According to Debye’s theory, the heat capacity of solid at low temperature is

a. independent b. Proportional to T3

c. zero d. 3R

100. ASSERTION(A) : vibrational partition function has very low value .

REASON(R): It depends on temperature

a. Both A and R are true and R is the correct explanation of A

b. Both A and R are true but R is not correct explanation of A

c. A is true but R is false

d. Both A and B are false



101. The stationary phase and mobile phase of Paper chromatography: is

a. solid- liquid b. liquid- solid

c. liquid – liquid d. liquid – gas

102. Principle is used in column chromatography is

a. diffusion b. adsorption

c. selective adsorption d. osmosis

103. Which is not adsorbent in column chromatography

a. Magnesium oxide b. Magnesium carbonate

c. Calcium carbonate d. Xylene

104. Eluotropic series is the series of solvents according to their increasing

a. eluting power b. adsorbing power

c. dissolving power d. polarity

105. The component which has greater adsorbing power is adsorbed in the -----of the column

a. lower part b. upper part

c. middle part d. lowest part

106. The position of migrated spots on the paper chromatograms are indicated by

a. RF b. Rx

c. RM d. all the above

107.. The Migration parameter RF and RM are related as

a. RM = log ( 1

RF) b. RM = log (

1RF

+1)

c. RM = log ( 1

RF -1) d. RM = log ( RF+1)

108. .Stahl’s triangle relates

1. adsorbant activity 2. nature of solute 3. nature of solvent

a. I and II only b. II and III only c. I and III only d. I,II and III

109.. The carrier gas not used in Gas-Liquid Chromatography is

a. Hydrogen b. Helium

c. Nitrogen d. Oxygen

110. Hydrogen is not used as carrier gas because

I. It may react with unsaturated compound



II. It creates fire and explosive hazard

III .It has low thermal conductivity

a. I only b. II only c. III only d. both I and II

KEY – UNIT-5 – FULL- THERMO / STATISTICAL / ANALYTICAL

1 2 3 4 5 6 7 8 9 10

B B C A D D C C D B

11 12 13 14 15 16 17 18 19 20

B C C D C B A B B D

21 22 23 24 25 26 27 28 29 30

D B D D B C B A B B

31 32 33 34 35 36 37 38 39 40

A C B A B C B C A C

41 42 43 44 45 46 47 48 49 50

A B B C C B B A D C

51 52 53 54 55 56 57 58 59 60

A A B D A C D D A D

61 62 63 64 65 66 67 68 69 70

B D C D D A B C C B

71 72 73 74 75 76 77 78 79 80

C B C D C B B C B A

81 82 83 84 85 86 87 88 89 90

A B A C C B A C B A

91 92 93 94 95 96 97 98 99 100

B C B B C D C A B B

101 102 103 104 105 106 107 108 109 110

C C D A A D C D D D


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2019- PG –TRB - UNIT 5- FULL MATERIAL, TESTS & KEY · Web viewPartition functions – Sackurtetrode equation –statistical approach to the third law of Thermodynamics – Maxwell