Thursday, July 31, 2008

Lecture notes -1(Thermodynanics)

THERMODYNAMICS

Some Important common thermodynamic terms

1. System :
The part of the universe selected for thermodynamic study is called system. For example boiling of water in a beaker.

2. Surroundings :

The remaining part of the Universe around the system, which is not under study, is called surroundings. For example, boiling of water in a beaker is an example of system and everything else around the beaker is the surroundings.

3. Types of System :

Generally system may be classified into three categories:

(i) Open system: A system, which can exchange matter as well as energy with its surroundings, is called an open system. For example, boiling of water in an open beaker.

(ii) Closed System: A system, which may exchange energy but not matter with surroundings, is called a closed system. For example, system remains inside the vessel.

(iii) Isolated system: A system which can neither exchange matter nor energy with the surroundings is called an isolated system.

4. Homogeneous and Heterogeneous System :

(i) Homogeneous system: A system is said to be homogeneous if it is uniform throughout. Such type of system consists of only one phase. For example, a system containing only a pure solid or a pure liquid or a pure gas or completely miscible liquids or gases etc.
(ii) Heterogeneous system. A system is said to be heterogeneous if it is not uniform throughout. Such type of system consists more than one phase i.e., Polyphase. For example, a system containing a mixture of two immiscible liquids or gases or solids etc.

5. Thermodynamic Properties:

(i) Extensive property, and (ii) Intensive Property.

(i) Extensive property:

The properties, which depend on the amount of the material in the system, are called ‘Extensive properties’. For example, volume, internal energy, heat capacity, free energy, mass heat content, entropy, etc.

(ii) Intensive property:

The Properties, which do not depend (independent) on the amount of the material but depend upon the nature of the material in the system, are called intensive properties. For example, viscosity, surface tension, thermal conductivity, boiling point, freezing points, refractive index, Vapour pressure of as liquid, temperature, density, specific heat etc.


6. Thermodynamic Equilibrium

(i) Thermal Equilibrium.
(ii) Mechanical equilibrium.
(iii) Chemical equilibrium.

(i) Thermal Equilibrium: A system is said to be in thermal equilibrium if there is no flow of heat from one portion of the macroscopic system to the another portion. This is possible only if the temperature remains the same throughout in all parts of the system.

(ii) Mechanical Equilibrium: A system is said to be in mechanical equilibrium if no mechanical work is done by one part of the macroscopic system to the another part of the system. This is possible only if the pressure remains constant throughout in all parts of the system.
(iii) Chemical Equilibrium: A system is said to be in Chemical equilibrium if the composition of the various phases in the system does not change with time.


7. Thermodynamic Process

It is the operation, which brings about the changes in the state or system. It is of the following types:
(i) Isothermal Process(dT=0)
(ii) Adiabatic Process(dq=0)
(iii) Isobaric process(dp=0)
(iv) Isochoric process(dv=0)
(v) Cyclic process(dE=0)

Path function: It is the property whose change depends upon the path followed by it and not upon the initial & final state of the system.eg. work

State function: It is the property whose change depends upon the initial & final state of the system and not upon the path followed by it


(i) Reversible process: A process is said to be reversible if it is carried out infinitesimally slowly so that the driving force is only infinitesimally greater than the opposing force. Such types of processes are ideal and cannot be realized in practice. This is because a reversible process requires infinite time for its completion.

(ii) Irreversible process: A process is said to be irreversible if it is not carried out infinitesimally slowly but the change is produced rapidly so that the system slowly does not remain in equilibrium condition. Both the reversible and irreversible process may be understood as follow.


LAWS OF THERMODYNAMICS

1. First Law is simply known as the ‘Law of conservation of energy’. The law may be stated in any one of the following ways:

(i) Energy can neither be created nor destroyed by any physical or chemical change. However, it can be changed into equivalent amount of the other form.

(ii) The total energy of isolated system remains constant.
(iii) It is impossible to construct a perpetual motion machine i.e., a machine, which works without consuming energy.
(iv) The total sum of mass and energy is constant in the Universe though these are interconvertible consuming and are related to each other by the expression
(E=mc2)
`
Where E = energy produced by destruction of mass ‘m’
c = velocity of light.
Therefore, the first law is modified and stated as “The total mass and energy of an isolated system remains unchanged”.

Mathematical Formulation
Let us consider a system having internal energy E1 in a particular set of condition. Suppose the q amount of the heat supplied to the system. Supplying the heat increases the internal energy of the system. Let the internal energy of the system in the final state is E2 and work done on the system is w, then:

E2 = E1 + q + w or E2-E1 = q + w or ΔE = q + w ------- (1)
Here ΔE = change in internal energy


Limitations

(i) It is unable to explain why do the physical and chemical processes occur in a natural direction. For example, ice melts at room temperature but water does not freeze at room temperature.
(ii) It does not predict the feasibility of a process in a particular direction.
(iii) It does not put any restriction on the direction of flow or heat. For example, without any external aid, ice may be used to heat water by extracting heat from ice at lower temperature and supplying it to the water at higher temperature.
(iv) It does not tell about the extent of convertibility of one form of energy into another. Due to the above limitations second law of thermodynamics came into existence.

Second Law of Thermodynamics

It may be stated in following ways:

(i) Heat cannot be completely converted into work without leaving changes either in the system or in the surroundings.

(ii) It is impossible for a self-acting machine, unaided by any external agency; to convey heat from a body at a lower temperature to one at higher temperature i.e., heat cannot pass from a colder body to a warmer body. Some work must be expanded to achieve this.(R. Clausius statement)

(iii) It is impossible for a cyclic process to take heat from a cold reservoir and convert it into work without at the same time transferring heat from a hot to a cold reservoir.(Kelvin statement)

(iv) All spontaneous processes or naturally occurring processes are thermodynamically irreversible; therefore, the entropy of the Universe is increasing continuously.

(v) The entropy of the Universe remains constant in a reversible process but it increases in an irreversible process.

ENTROPY:

The term ‘entropy’ is introduced by R.J.E.Clausius. Entropy is a Greek word (‘trope’ means change and ‘en’ stands for energy). Entropy is the property of substance, which measures the disorder or randomness in a system and is represented by ‘S’. Entropy is larger in case of more randomness. It is also a measure of the unavailable energy. In general for any substance
SVapour >SLiquid >SSolid

ΔS = qrev / T


Units:

Entropy change is expressed by a heat term divided by the absolute temperature.
In S.I. units entropy is expressed in JK-1 mol-1 (1 cal=4.186J)

Physical Significance

(i) Entropy is a state function: It does not depend upon the path by which the change takes place.
(ii) Entropy is a measure of randomness (disorder) of the system. Greater the randomness, higher is the entropy of the system.

SGas > SLiquid > SSolid


ENTROPY CHANGES FOR AN IDEAL GAS

Since entropy is a state function of system. Its value for a pure substance depends upon any two of the three variables P, T and V. Since temperature (T) is considered generally as one of the variables, the second variables to be considered may be either P or V.

To derive ideal gas equation for entropy change is considered over following cases:
Case I: Temperature (T) and volume (V) variations
Case II: Temperature (T) and Pressure (P) variations

CaseI: consider an ideal gas having volume V at temperature T and pressure P. Let the system absorbs an infinitesimally small amount of heat dqrev reversibly at temp. T, then the entropy of the system increases.
dE = dq –dW (1st law of thermodynamics) ------------------(1)
dE = CvdT ------------------(2)
dq = TdS ---------------(3)
dW = -PdV -------------(4)

Using (1),(2),(3) & (4) we have:
dS = CvdT / T + RdV / V

On integrating above equ. within limits S1 to S2, T1 to T2 and V1 to V2.

ΔS =nCV ln T2/T1 + nR ln V2/V1 for n moles of an ideal gas. --------------- (5)

Case II:
P1V1 / T1 = P2V2 / T2 ------------------ (6)
Cp –Cv = R ---------------- (7)

From (5), (6) & (7) we have:

ΔS= CP ln T2/T1 - R ln P2/P1
Special cases:
Isothermal Process: ΔST = R ln V2/V1 or - R ln P2/P1

Isobaric Process: ΔSP = CP ln T2/T1

Isochoric Process: ΔSV = CV ln T2/T1

Entropy Change in Reversible and Irreversible Process

Entropy change for a reversible process is always zero i.e. ΔSTotal = 0 and the entropy change for an irreversible process is always +ve. i.e. there is no net change in entropy for a reversible process and entropy always increases in an irreversible process.

Physical significance of work function
Decrease in change in work function (Helmholtz free energy function) is equal to max. work obtained from the system during change i.e. –ΔA = Wrev. = Wmax.

Physical significance of Gibb’s free energy
Decrease in change in free energy is equal to max. work other than pressure- volume work that can be obtained from the system during change i.e. –ΔG = Wrev. - P ΔV.



Gibb’s Helmholtz Equation in terms of free energy:

G = H-TS (1)

∆G = ∆H - T∆S (2)

Also dG = VdP- SdT (3)

(∂G / ∂T)p = -S (4)

from equ. 1& 4

G = H + T (∂G / ∂T)p

Also, - ∆S = S1 – S2

= [∂(G2-G1) / ∂T]p (from equ. 4 )

= [∂(∆G ) / ∂T]p (5)

from equ. 2 & 5

∆G = ∆H + T [∂ (∆G) / ∂T]p (6)

Gibb’s Helmholtz Equation in terms of work function (Helmholtz function):

A = E-TS (1)

∆A = ∆E - T∆S (2)

Also dA = PdV- SdT (3)

(∂A / ∂T)v = -S (4)

from equ. 1& 4

A = E + T (∂A / ∂T)v

Also, - ∆S = S1 – S2

= [∂(A2-A1) / ∂T]v (from equ. 4 )

= [∂(∆A ) / ∂T]v (5)

from equ. 2 & 5

∆A = ∆E + T [∂ (∆A) / ∂T]v


Alternative form of equ. (6)

[∂ / ∂T(∆G / T)]p = - ∆H / T2 (on differentiating ∆G / T w.r.t. T at constant P)

Significance:

1) Idea of spontaneity of a reaction
∆G < 0 reaction will be non spontaneous.
∆G > 0 reaction will be spontaneous.
∆G = 0 reaction will be in equilibrium.

Since ∆G = ∆H - T∆S
Consider exothermic & endothermic reactions for both +ve & -ve values of ∆H & ∆S with increase and decrease of temperature.
2) In the study of Galvanic cells as:

nFE = - ∆H + TnF[(∂E / ∂T)p (as ∆G = -nFEcell for an electrical cell)

Chemical Potential (Partial molal / Molar free energy)
It is defined as the change in free energy of the system when 1 mole of that particular substance is added at constant temp. & pressure, to such a large quantity of the system so that there is no appreciable change in overall composition of the system.
μi = (∂G / ∂ni)T,P,n1,n2----------


Clausis – Clapeyron Equation:
Consider 1 mole of a substance in two phases A & B in equilibrium at constant Temp & Pressure.
For this,

∆G = 0 or GB – GA = 0 (1)

For small change in pressure as P+dP, temperature as T+dT
Relative change in Gibb’s free energy for the two phases is as;
For phase A, GA+dGA & for phase B, GB+dGB

For such a small change, the system is still considered to be in equilibrium, so

GA+dGA = GB+dGB (2)

From (1) & (2)

dGA = dGB (3)

Also, dG = VdP-SdT (4)
From (3) & (4)

dP / dT = SB-SA / VB-VA = ∆S / ∆V

Also, ∆S = q / T

So dP / dT = q / T ( VB-VA )

For liquid – vapour equilibrium,

dP / dT = ∆HV / T( Vg-Vl ) (5)
∆HV = molar heat of vaporization.

Similarly for solid – liquid & solid –vapour equilibrium.

Integrated form of Clausis – Clapeyron Equation:

In comparison to gas (Vg), the Vs & Vl may be neglected and hence
Vg-Vs = Vg
Vg-Vl = Vg

Now equ. (5) may be written as,

dP / dT = ∆HV / T Vg
(6)
If vapours are assumed to behave ideally then,
Vg = RT / P
(7)
From (6) & (7)
dP / P = (∆H / RT2) dT
(8)
Assuming ∆H remains constant(condition for integration) & then integrating the above equ. Within limits P1 to P2 & T1 to T2 we have;

2.303 log P2 / P1 = ∆H / R [T2-T1 / T1T2] (9)

Applications:
1) Determination of molar heat of vaporization.
2) Determination of boiling points of a liquid at different pressures.
3) Determination of vapour pressures of a liquid at different temperatures.

Numericals:

1) Calculate the change in entropy for 1 mole of an ideal gas when its temperature rise from 300K to 600K under i) Isochoric condition & ii) Isobaric condition. (Cv = 2.5R)
Soln: T1 = 300K T2 = 600K n = 1
ΔS= CP ln T2/T1 - R ln P2/P1
ΔSV = nCV ln T2/T1 + nR ln V2/V1
For Isochoric condition
Using ΔS = nCV ln T2/T1 ΔS = 14.40J/K/mole
For Isobaric condition
Using ΔS = nCp ln T2/T1 & Cp –Cv = R ΔS = 20.16J/K/mole

2) The energy change (ΔG) in a process are found to be -138KJ at 303K & -135KJ at 313K. Calculate the enthalpy change at 308K.
Soln:
ΔG at (303 +313/2) K = -138-135/2
ΔG at308K = -273/2
= -136.5KJ
[∂ (∆G) / ∂T] = ΔG at 313 - ΔG at 303/T2 – T1
= -135 + 138/10
Now using ∆G = ∆H + T [∂ (∆G) / ∂T]p
∆H = - 228.9KJ

3) Calculate the change in volume for melting of 1 gm. mole of phenol. Given latent heat of fusion = 3250 cal/gmole.Melting point of phenol = 330K dP/dT = 287atm./K
Soln:
Using dP / dT = ∆HV / T( Vl-Vs )
( Vl-Vs ) = 0.034 ml

4) The latent heat of vaporization of water is 540 cal/gm at 1000C. Calculate the pressure at which water must be heated to produce superheated steam at 1500C.
Soln:
∆Hvap. = 540cal/gm
= 540 Χ 18 cal/mole
T1 = 1000C = 373K T2 = 1500C = 423K
P1 = 1 atm. P2 = ?
Now using 2.303 log P2 / P1 = ∆H / R [T2-T1 / T1T2]
P2 = 4.71 atm.

Lecture notes -1

Lecture notes -1

Assignment-4

Assignment-4

1. What is Lambert’s Law?
2. What is Lambert-Beer Law?
3. Define absorbance and transmittance of a solution.
4. Describe briefly the process which takes place when the solution of a metallic salt is sprayed into a flame?
5. Name the important components of a flame –photometer.
6. Write short notes on the following:
(i) Oileness (ii) Neutralization number (iii) Saponification number
(iv) Cloud point and pour point of lubricant.(iv) Synthetic lubricants
7. Define Viscosity Index. How is it determined for a lubricant?
8. Why graphite and MoS2 preffered as solid lubricants?
9. What do you understand by wear ? Give precise definition.
10. What is friction and what are its consequences?
11. Give a brief account of the classical laws of frictions
12. Give the Principle and Applications of I.R.& U.V. spectroscopy.










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Assignment-3

Assignment- 3


1. What are the differences between Galvanic cell and Electrolytic cell?
2. What is Galvanic series?
3. What do you mean by corrosion? How does it differ from erosion?
4. Explain how corrosion can be controlled by proper designing?
5. What are the effects of temperature, pH, over voltage and reactivities of metals influences the corrosion?
6. Write preparation properties and applications of following polymers
(a) Polyvinyl Chloride (PVC) (b) Polyvinyl acetate (PVA)
(c) Bakelite (d) Urea formaldehyde (UF) (e) Styrene rubber (GR-S / Buna-S / SBR)
(f) Nitrile rubber (GR-N / Buna-N / NBR) (g) Butyl rubber (GR-I)
7. What do you understand by crystallinity of a polymer?
8. What is the difference between thermosets and thermoplastics? Give their charcterisics and suitable examples.
9. Explain the addition and condensation polymerization reactions with one example for each.
10. What do you understand by Co-ordination Polymerisation? Explain with example.

Assignment-2

Assignment- 2

100 ml. of a sample of hard water neutralizes exactly 12 ml. of 0.12 N HCl using methyl orange as indicator. What kind of hardness is present? Express the same in terms of an equivalent of CaCO3.
2. 100 ml. of a water sample required 12.4 ml. of N/50 H2SO4 for neutralization to phenolphthalein end point. Another 15.2 ml. of same acid was needed for further titration to methyl orange end point. Determine the type and amount of alkalinity.

50 ml. of water sample requires 10 ml. of 0.01 N EDTA when titrated using buffer solution (pH=10) to attain the end point. Calculate the total hardness of a sample in terms of ppm equivalent of CaCO3 per litre.

4. 200ml. of hard water sample require 30 ml. of 0.02 M EDTA with NH4 Cl-NH4OH buffer and EBT indicator. Another 200 ml. of the sample is boiled for about half an hour and after filtering the ppt. the volume of filtrate is made 200 ml. again by the addition of distilled water.20 ml. of this boiled water sample requires 5 ml. of 0.01 ml. EDTA following the same procedure. Calculate the temporary and permanent hardness of the sample.

5. A water sample on analysis gives the following data : Ca2+ = 20 ppm, Mg2+ = 25 ppm, CO2 = 30 ppm, HCO3- = 150 ppm, K+ = 10 ppm. Calculate the lime(87%pure) and soda (91% pure) required to soften 1 million litres of water sample.

6. Calculate the quantity of lime and soda required for softening 60000 liters of water containing CO2 = 20mg/L, Ca(HCO3)2 = 20 mg/L, Mg(HCO3)2 = 25mg/L, HCl= 8.4 mg/L, Al2(SO4)3 = 40mg/L and Mg Cl2 = 12 mg/L.

7. What is the principal involved in reverse osmosis process of desalination of water? Describe the process and its advantages.

8. What is Electro dialysis and how is it carried out? Discuss the process, its advantage and limitations.

Assignment-1

Assignment-1

1. Define Thermodynamics and mention its scope. State and explain the important laws of thermodynamics. What are its limitations?
2. State and explain the first law of thermodynamics. Derive a mathematical formulation of the law. Justify the law on the basis of some common observations and discuss its limitation.
3. Five moles of an ideal gas expand isothermally and reversibly at 27 C from an initially volume of 5dm to 50 dm against a pressure that is greatly reduced. Calculate G & S for the process. (R= 8.314 J/K/mol)
4. Calculate the change in entropy for one mole an ideal gas when its temperature rises from 300k to 600k (i) Isochoric condition (ii) Isobaric condition
Cv = 2.5R
5. Water boils at 373 K at one atmospheric pressure. At what temperature will it boil when atmospheric pressure becomes 528 mm of Hg at some space station? Latent heat of water = 2.28KJ/g
6. At 373.6 K and 372.6 K the vapour pressure of water are 1.018 and o.982 atm. resp. Calculate the heat of vaporization of water.
7. The free energy change ( G) accompanying a given process is – 85.77 and -83.68 kj at 25 C and 35 C respectively. Calculate the change in enthalpy for the process at 30 C.
8. Calculate the number of phases in the following:
(a) SR SM

(b) I2 (s) I2 (g)

(c) MgCO3(s) MgO(s) + CO2 (g)

(d) N2O4 (g) 2NO2(g)

9. How many degrees of Freedom are present in the following system?

a) A gas in equilibrium with its solution in a liquid
b) Two partially miscible liquids in the absence of vapour

10. What do you understand by condensed system.