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    MTD Important questions and answers

    (Short answer questions)

    1) State any three applications of thermodynamics in metallurgy

    Ans:

    In the construction of phase diagrams

    To understand various chemical reactions

    To predict the correct reducing agent in

    Ferrous and non ferrous extractive metallurgy

    To understand phase transformations during heat treatment of an alloy

    To predict new phases in the development of newer materials

    ) State a) the first law of thermodynamics !) its mathematical equation and c) la!el the terms

    Ans:

    Statement:

    The total energy of an isolated system remains constant, though there are changes from one form of energy to

    another.

    Mathematical equation:

    ! " # $

    $here ! %hange in internal energy of the closed system

    " ! &eat transferred to or from the system$ ! wor' done (y or on the system

    ") Define a) heat of reaction !) heat of reaction at constant pressure c) heat of reaction at constant #olume

    Ans:

    a) $eat of %eaction: The amount of heat evolved or a(sor(ed in a chemical reaction when the reaction is

    performed according to a (alanced chemical equation at a given temperature and pressure is called heat of

    reaction

    !) $eat of %eaction at constant &ressure: The heat of reaction at constant pressure at a certain temperature is

    defined as the amount of heat evolved or a(sor(ed at constant pressure when the reaction is performed according

    to a (alanced chemical equation. This is represented (y &

    c) $eat of %eaction at constant #olume: The heat of reaction at constant volume at a certain temperature isdefined as the amount of heat evolved or a(sor(ed at constant volume when the reaction is performed according

    to a (alanced chemical equation. This is represented (y or )

    ') rite any three statements of Second law of thermodynamics

    Ans:

    *tatement+:

    - It is impossi(le to construct a machine, operating in cycles, which will produce wor' continuously (y a(sor(ing

    heat from a single heat reservoir.

    - /elvin+0lanc' statement of the second law1

    *tatement+2:

    - &eat does not flow spontaneously from a cold to a hot (ody

    - %lausius statement of the second law1

    *tatement+3:

    - $henever a spontaneous process ta'es place, it is accompanied (y an increase in the total entropy of the universe

    That is *univ 45

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    ) hat information can !e drawn from a positi#e slope change of a line in the *llingham diagram+

    Ans:

    The slope of the llingham diagram is given (y +6*5

    7ut ,S-. S-products/ S-reactants

    ,S-(ecomes negative when S-reactants0 S-products then only the slope 8/,S

    -)(ecomes positive,

    This implies that the reactants are undergoing phase transformation which may (e either melting are (oiling

    depending upon the value of /,S-if +6*5 is low it is melting if +6*5 is high it is (oiling.

    ) hy the line of the reaction 2 3 4. 24 is down wards in the *llingham diagram.

    Ans:

    For the reaction

    234..0 24

    $e have a solid reacting with a gas to produce two moles of gas, and so there is a su(stantial increase in entropy

    so the line slopes sharply downward as shown (elow in the figure.

    ntropy change for the reaction is given as

    *5! 2*58%9# 2*5

    ;%4< *5

    8921

    &ere *5;%4 is assumed negligi(le since ;%4 is solid.

    *ince for every one mole of consumption of 92gas two moles of %9 gas is (eing produced

    Therefore 2*58%9 4 *5

    892

    7ut slope of the line is given as ! +*5; 5

    &ence the line of the reaction 2% < 52! 2%9 is down wards in the llingham diagram.

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    5) Slate Trouton6s rule and its use

    Ans:

    Trouton7s rule: It states that the molar entropy of vapori=ation of various liquids at their normal (oiling point is almost

    same and is a(out >?+>> @ /+mol+9r 2 cal.

    I.e. * ! &AT(.p! 2 cal

    8se:

    The rule is used to estimate the enthalpy of vapori=ation of liquids whose (oiling points are 'nown.

    9) Define: a) fugacity !) acti#ity

    Ans:

    a) ;ugacity: itis a measure of escaping tendency of a su(stance to prefer one phase 8liquid, solid, gas

    over another. It is represented (y BfC

    Fugacity appers as pressure (ut not really the pressure it is also 'nown as pseudo pressure or fictitious

    pressure (ut at low pressures fugacity approaches pressure mathematically it is expressed as

    !) Acti#ity: the activity of a su(stance is defined as the ratio of fugacity in the given state to the

    fugacity in the standard 8or pure state. Mathematically it is expressed as shown (elowa . f Magnesium alloy contains =1 atom? aluminium 2alculate the composition of the

    alloy in wt? The atomic weights of Al and Mg are =9 and '" respecti#ely

    Solution:

    ,55FF

    FF X

    MgXatwtMgatomAlXatwtAlatom

    AlXatwtAlatomofAlWt

    +

    =

    ,5532.2G:H.I,,558I>.2JH.I,

    I>.2JH.I,F X

    XX

    XofAlWt

    +

    =

    . ="

    Therefore $t of Mg ! 55+2.3

    !55

    1-) Define rate of a reaction and e@press it mathematically

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    Kate of a reaction can (e defined as the decrease in concentration of reactants or increase in concentration of

    products with respect to time. Mathematically it can (e expressed as shown (elow:

    Kate of a reaction !

    If concentration is expressed (y BxC then the change in concentration is expressed (y BdxC, Then dx is positive for

    the products since the concentration of products increases with time, whereas for reactants dx is negative since the

    concentration of reactants decrease with time.

    Then the rate of reaction can also (e expressed as

    Kate !

    $here dx +++ the change in concentration of products

    dT+++ the change in time9r

    Kate !

    $here dx +++ the change in concentration of reactants

    dT+++ the change in time

    11) Define a) System !) Surrounding c) !oundary with a neat setch

    Ans:

    System: L *ystem is that part of the universe, which is under thermodynamic study

    Surroundings: verything external to the system is called surroundings

    Boundary: The real or imaginary surface separating the system from the surroundings is called the

    (oundary

    *@ample: L gas entrapped in a cylinder fitted with piston as illustrated (elow

    %hange in concentration of reactants or products

    Time

    dx

    dT

    +dx

    dT

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    1) A System a!sor!s cal of heat from the surroundings at constant pressure then it performs a wor of '19-

    C calculate the change in internal energy of the system

    Solution:

    Lccording to First law of thermodynamics d8 . E >

    &ere ". 2.H ' cal ! 2H55 N G.> @ ! 5GH5 @

    $ ! ?H5 @ ,Oow d) ! 5GH5 # G>5

    . 5- C

    1") Define a) *@othermic reaction !) *ndothermic reaction and c) state sign con#entions

    Ans:

    a) *@othermic reaction: The reactions, which are accompanied (y the evolution of heat, are 'nown as exothermic

    reactions. In exothermic reactions the enthalpy of products is less than that of the reactants.

    !) *ndothermic reaction: The reactions, which are accompanied (y the a(sorption of heat, are 'nown as endothermicreactions. In endothermic reactions, the enthalpy of products is more than that of the reactants. For these reactions & !

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    That is *univ 45O.7: any one statement can (e sufficient

    ( " ! Tds

    $here " ! amount of heat transferred

    T ! Temperature in /elvin

    ds ! %hange in entropy

    1) State any three limitations and three uses of *llingham diagrams

    Ans:

    The limitations of llingham diagram are:

    It is applica(le only to su(stances in their pure states that is pure metals and pure oxides or sulphides, (ut pure

    su(stances are rare in metal extractions therefore the information derived from the diagram cannot (e directly applied

    to actual reactions

    2 %ompounds are assumed to (e stoichiometric which often is not true

    3 The diagram shows the direction in which equili(rium lies (ut it does not specify the conditions under which it would

    (e reached

    G Oo information on the 'inetics of the reactions can (e o(tained from the diagrams

    O.7: any three limitations can (e sufficient

    The uses of the *llingham diagram:

    To determine the reducing agent for reducing a given metallic oxide to metalP

    2 To determine the partial pressure of oxygen that is in equili(rium with a metal oxide at a given temperature, this

    will help us to predict the temperatures at which a metal is sta(le and the temperatures over which it will

    spontaneously oxidi=e

    3 Qetermine the ratio of car(on monoxide to car(on dioxide 8%9A%92ratio that will (e a(le to reduce the oxide to

    metal at a given temperature.

    G To determine the relative sta(ilities of oxides and sulphides etc.

    O.7: any three uses can (e sufficient

    1) a) Define *llingham diagram !) write its principle of construction and c) draw the line for the reaction 2 3 4

    . 24

    Ans:

    a)The llingham diagram is a plot (etween the standard free energy change 8R 5 of a reaction as a function ofthe temperature 8T

    !)The principle of construction is ,F- . ,$-/T ,S-

    $here 6R5 ! *tandard free energy change

    6&5 ! standard enthalpy change

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    6*5! standard entropy change

    c)

    15) The #apour pressure pG (measured in mm $g) of liquid arsenicG is gi#en !y log p . / '-

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    Ans:

    a L solution is a homogeneous mixture composed of two or more su(stances. In the study of solutions, it is

    customary to designate the component present in larger proportion as the solvent and the one in smaller

    proportion as the solute.

    ( quation for calculating wt from atom

    ,55FF

    FF X

    BXatwtBatomAXatwtAatom

    AXatwtAatomofAWt +=

    *imilarly,

    ,55FF

    FF X

    BXatwtBatomAXatwtAatom

    BXatwtBatomofBWt

    +=

    c quation for calculating atom from wt

    ,55AFAF

    AFF X

    atwtBBwtatwtAAwt

    atwtAAwtofAatom

    +=

    *imilarly,

    ,55AFAF

    AFF X

    atwtBBwtatwtAAwt

    atwtBBwtofBatom

    +=

    -) State the effect of temperature on rate of reaction with Arrhenius rate equation

    Ans:

    *ffect of Temperature:The rate of a reaction is increased considera(ly (y a little increase in temperature. The effect of

    temperature on the rate of the reaction can (e explained (y considering the Lrrhenius rate equation, which is given (elow

    Kate ! L e+"AKT

    $here L+++0re exponential constant

    "+++Lctivation energy

    K+++)niversal gas constant

    T+++Temperature in /

    From the a(ove equation, we can conclude that the rate of a reaction increases exponentially with temperature.

    1) a) Define $omogeneous systemG gi#e one e@ample !) Define heterogeneous systemG gi#e one e@ample

    Ans:

    a) $omogeneous system:L system, which consists of *ingle 0hase and )niform composition throughout, is called a

    &omogeneous system

    *@ample:L pure *olid or Siquid, L mixture of gases, $ater < salt

    O.7: any one example can (e sufficient

    !) $eterogeneous system:L system, which consists of More than one 0hases and %omposition, is not uniform is called

    &eterogeneous system

    *@ample:L Mixture two or more immisci(le solids,

    L Mixture of two or more immisci(le liquids,

    Ice in contact with water

    Ll+2 *i

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    Rrey cast Iron

    0earlite

    Metal # *lag + Ras

    ) Define a) $eat 2apacity !) $eat capacity at 2onstant &ressure c) $eat capacity at 2onstant olume

    Ans:

    a) $eat 2apacity

    - The quantity of heat required to raise the temperature of a (ody (y 5

    - If the temperature of the (ody is raised from Tto T2(y passing an amount of heat ", the heat capacity % of the

    (ody is given (y

    % ! " A 8T2+T

    89r

    dT

    QC=

    !) $eat capacity at 2onstant olume

    - It is the amount of heat required to raise the temperature of a (ody (y 5at constant volume

    - It is denoted (y %v

    - Lt constant volume heat capacity can (e given as

    - dT

    qVVC =

    - $here qvis heat a(sor(ed at constant volume

    c) $eat capacity at 2onstant olume

    - It is the amount of heat required to raise the temperature of a (ody (y 5at constant 0ressure

    - It is denoted (y %p

    - Lt constant 0ressure heat capacity can (e given as

    -

    dTC

    QP

    P=

    - $here "0is heat a(sor(ed at constant pressure

    ") calculate the heat of formation of J&2l0 at =9 K from the following data:J&0 3 "(2l). L&2l" N$

    -=9 . /19 calO L&2l" 3 (2l) . J&2l0 N$

    -=9 . /""9 cal

    Solution:

    The required reaction is: ;04 < HA2 %l 2! ;0%lH4

    Set

    2;04 < 38%l2! 2V0%l3W &5

    2> ! +H.> 'cal+++++++ 8

    V0%l3W < 8%l2 ! ;0%lH4 &52> ! +33.> 'cal++++++++ 82

    Multiply eqn 82 with B2C (oth sides

    2V0%l3W < 28%l2 ! 2;0%lH4 &5

    2> ! +J?.J 'cal+++ 83

    Oow add eqn 8 and 83

    2;04 < H8%l2! 2;0%lH4&52> ! +>3.G 'cal+++++++ 8G

    Qivide eqn 8G with B2C

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    ;04 < HA2 8%l2! ;0%lH4 &52> ! +G.? 'cal++++++ 8H

    The a(ove reaction is same as that of the required reaction

    Therefore the heat of formation of ;0%lH4 at 2> / is ! /'15 cal (Ans)

    ') a) write the mathematical relation !etween entropy and heat !) define spontaneous process and gi#e one

    e@ample

    Ans:

    a) Mathematical relation !etween entropy and heat: For a reversi(le change ta'ing place at a fixed

    temperature 8T, the change in entropy 8* is equal to heat energy a(sor(ed or evolved divided (y thetemperature 8T

    That is,

    S.

    ET

    !) Spontaneous process: L process, which proceeds on its own accord, without any outside assistance, is

    termed a spontaneous or natural process or irreversi(le process.

    *@amples Kolling stone from high level to low level

    2 &eat flow from high temperature to low temperature

    3 Ras flow from high pressure to low pressure

    O.7: any one example can (e sufficient

    ) a) Define *llingham diagram !) State any three uses of *llingham diagrams

    Ans:

    aThe llingham diagram is a plot (etween the standard free energy change 8R 5 of a reaction as a function of

    the temperature 8T( The uses of the llingham diagram:

    H To determine the reducing agent for reducing a given metallic oxide to metalP

    J To determine the partial pressure of oxygen that is in equili(rium with a metal oxide at a given temperature, this

    will help us to predict the temperatures at which a metal is sta(le and the temperatures over which it will

    spontaneously oxidi=e

    ? Qetermine the ratio of car(on monoxide to car(on dioxide 8%9A%92ratio that will (e a(le to reduce the oxide to

    metal at a given temperature.

    > To determine the relative sta(ilities of oxides and sulphides etc

    ) a) write the principle of construction of *llingham diagram and !) draw the lines for the reactions 2 3 4.

    24O 2 3 4. 24Ans:

    a)The principle of construction is ,F- . ,$-/T ,S-

    $here 6R5 ! *tandard free energy change

    6&5 ! standard enthalpy change

    6*5! standard entropy change

    !)

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    5) a) write the !asic idea that the phase equili!ria pro#ides !) write the 2lapeyron equation for solid > liquidequili!rium of a single component system and la!el the sym!ols

    Ans:

    a)0hase equili(ria, a (ranch of chemical thermodynamics. The (asic idea the phase equili(ria provide is to

    understand how we predict what phases will (e sta(le in a system at a given temperature, pressure, and

    composition, and how those sta(ility relations will change as the temperature, pressure and compositions are

    varied.

    !) ;or solid P liquid, equili(rium the %lapeyron eqn is:

    d&

    .

    N$fusion

    dT T N$here d0Xchange in pressure &fusionXSatent heat of fusion

    TXmelting point dTXchange in melting point

    U X change in volume of liquid and volume of solid 8 i.e., Ul# Us

    9) Define: a) fugacity !) acti#ityAns:

    a) ;ugacity: itis a measure of escaping tendency of a su(stance to prefer one phase 8liquid, solid, gas over another. It isrepresented (y BfC

    Fugacity appers as pressure (ut not really the pressure it is also 'nown as pseudo pressure or fictitious pressure (ut at low

    pressures fugacity approaches pressure mathematically it is expressed as

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    !) Acti#ity: the activity of a su(stance is defined as the ratio of fugacity in the given state to the fugacity in the standard

    8or pure state.

    Mathematically it is expressed as shown (elow

    a . f

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    Lccording to %harlesC Saw + U Y T 8Lt constant 0 and m

    %om(ining the a(ove two laws we can write U Y TA0

    9r 0UAT ! /

    $here / is a constant

    / value can (e found as follows:

    ++++++++++++++ 8

    $here05is atm pressure

    U5is 22.G lit

    T5is 2?35/

    If the a(ove values are su(stituted in the a(ove equation 8 we get / !5 .5>2 lit+atm

    This value of B/C i.e. 5.5>2 lit+atm is constant and is same for every gas. &ence B/C is replaced (y a new

    constant called )niversal gas constant. This is represented (y BKC Oow the ideal gas equation ta'es the following

    form for +mole of gas

    &

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    &K represents the sum of enthalpies of reactantsTherefore & ! &7+&L

    Oow differentiating 83 on (oth sides (y T

    7ut %p!

    There fore

    Oow

    The a(ove equation is called /irchoffCs equation

    ') Deri#e the free energy relationships

    Ans:

    Lccording to first law of thermodynamicsE. d8 3

    Lccording to second law of thermodynamics dS . E

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    9r

    Lgain according to the definition of free energy R ! &+T*

    Qifferentiating the a(ove equation dR ! d& + Td* + *dT

    9r dR ! d 8)

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    Que to translational, vi(rational and rotational motions of molecules, atoms and electrons etc of the

    system

    Internal energy is a state property and extensive in nature.

    !) &roof of ds . - for a re#ersi!le process

    Set sand s2denote the entropies of (loc' and (loc' 2 respectively, and let Tand T2denote the temperatures of

    the (loc's. Set a small quantity qof heat flows from (loc' to (loc' 2, and then the entropy change is given as

    under:

    E

    ,2 T

    Q

    T

    Qds

    =

    1

    ,,

    ,2 TTQds =

    12,

    2,

    TT

    TTQds

    =

    For a reversi(le process TT2that is the difference in T and T2 is negligi(le such that for practical purposes we

    can ta'e T !T2

    Then the entropy change d*rev! q 8T +T2A T T2! 5

    Therefore for reversi(le process entropy change dS . -

    ) Deri#e the 2lapeyron equation for Hiquid > apour equili!rium of a single component system

    Ans:

    %onsider the phase transformation of Siquid to Uapour of a one+component system such as pure metal

    Siquid 8S \ Uapour 8U 8

    *ince the S and U, phases are in equili(rium with each other. Therefore, the change in the Ri((s free energy for

    the transition given (y quation 8 is =ero.

    That is, R ! 5. 82

    RU] RS! 5 83

    RU ! RS 8G

    If the temperature and pressure are altered slightly such that the free energy changes slightly (y dR without

    distur(ing the equili(rium

    Then 8RU< dRU ! 8 RS< dRS 8H

    From 8G RU ! RS

    T1

    T2

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    dRU! dRS 8J

    7ut we 'now from dR ! ] *dT < Udp

    Therefore dRU ! ] *U dT < UU dp 8?

    dRS ! ] *SdT < USdp 8>

    From 8J ] *U dT < UU d0 ! ] *SdT < USd0 8

    Therefore 8*U] *S dT ! 8UU+ US d0 85

    9r * dT ! U d0 8

    9r d0AdT ! *A U 82

    7ut * ! &vapAT

    $here &vap is latent heat of Uaporisation

    T is the temperature where phase transformation ta'es place that is (oiling point

    There fore

    d&.

    N$(1")

    dT T NThe a(ove equation is called 2lapeyron *quation

    5) a) Define the equili!rium constant for the reaction aA 3 !B . c2 3 dD

    !) 2alculate the equili!rium constant for the reaction:

    JQi40 3 ($) . JQi0 3 ($-)

    At 5- 2 from the following data:

    JQi0 3 U (-) . JQi40O NF4. /''"1 3 =9'' T C

    ($) 3 U (-) . ($-)O NF4. /'- 3 '9T C

    Ans:

    a) aA + bB cC + dD

    The upper case letters L and 7 are the reactants and % and Q are the products. The lower case letters a, (

    are the no. of moles of reactants and c, d are the no. of moles of products

    The equili(rium constant 8/c is defined as given (elow

    !) 2alculation of equili!rium constant:

    Set

    ;Oi4 < ^ 852 ! ;Oi94P R9! +2GG32 < >.GG T @++++++++++++++++8

    8&2 < ^ 852 ! 8&25P R9! +2GJ252 < HG.>T @ +++++++++++++++++++++82

    9n *u(tracting 8 from 82

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    ;Oi94 < 8&2 ! ;Oi4 < 8&25P R9! +>>+G3.JGT @ @ ++++++++++++++++83

    The a(ove equation is the required reaction

    R9at T! ?H5>+G3.JG_523 @ ! +GJH2G.?2@

    7ut the relation (etween equili(rium constant and free energy change is given (y

    R9! +KTln/e

    $here /eis the equili(rium constant

    Therefore

    /e ! xp 8+R9AKT

    ! xp 8GJH2G.?2A>.3G_523

    ! xp 8H.G?

    ! "5

    9) a) Define %aoult7s law !) *@plain the factors causing a real solution to de#iate from ideal !eha#iourAns:a) Raoults law:

    The activity of a component of an ideal solution is equal to its mole fraction. That is in an ideal solution ofsolvent A and solute B,

    Where

    Activity of A

    ole fraction of A!imilarly aB" #BWhere aB Activity of B#B ole fraction of B

    ( The deviations of a real solution from ideal (ehaviour are of two types they are

    0ositive deviation and 2 Oegative deviation

    The factors, which cause the deviations, are the interactions (etween similar and dissimilar atoms of the

    components of the solution. %onsider a solution in which the solvent is BLC and solute is B7C and if the interactions

    (etween similar atoms are FL+Land F7+7 and the interactions (etween dissimilar atoms are FL+7then

    If FL+L! F7+7 ! FL+7 Then the solution (ecomes ideal

    2 If FL+L F7+7 Then the solution under goes negative deviation

    2 If FL+L>FL+7< F7+7 Then the solution under goes positive deviation

    The deviations are shown (elow in the diagram

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    = a) *@plain the terms: heat and wor with units State sign con#entions of heat and wor

    !) *@plain a!out wor of e@pansion Deri#e equation for wor of e@pansionAns:

    a) $eat and or:

    $hen a change in the state of a system occurs, energy is transferred to or from the surroundings. This energy may (e

    transferred as heat or mechanical wor'.

    $eat: &eat is defined as the energy in transit

    or: Mechanical wor' is defined as force N distance. 9r $ ! 0U

    $here 0 # is pressure U # is the change in volume

    8nits of $eat: In *.I units heat is measured in oules.

    8nits of or:In *.I units $or' is measured in `oules.

    Sign 2on#ention of $eat:

    The sym(ol of &eat is ".

    If the heat flows from the surroundings into the system to raise the energy of the systemP it is ta'en to (e positive,

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    Ras at constant pressure

    dl

    Therefore Wor$ " %ressure # chan&e in 'olume

    Qerivation of equation for wor' of expansion:

    %onsider a gas contained in a cylinder fitted with a friction less piston. The constant pressure acting is 0. If the gas

    expands at constant pressure, the piston moves through a distance let this distance is DdlE.

    Then $or' ! force N distance

    $ ! f N dl

    7ut f ! 0 N L 8here L!area of cross section of the piston

    Therefore $ ! 0 N L N dl

    ! 0 N U 8here U ! change in volume

    1-) Deri#e the relation !etween & (&ressure) and (olume) in an adia!atic re#ersi!le process for an ideal gas

    Ans:

    Deri#ation of %elation !etween & > in an Adia!atic %e#ersi!le&rocess for an Ideal gas

    In an adia(atic reversi(le process, the first law of thermodynamics ta'es the following form

    " ! d) < 0dU according to first law

    5 ! d) < 0dU since in an adia(atic process " ! 5

    d) ! +0dU

    Llso we have d) ! %vdT

    %vdT ! +0dU

    The state equation of an ideal gas is 0 U !

    K T 8for mole of gas

    Qifferentiating, 0dU < Ud0 ! KdT ++++++++++ 8(

    *u(stituting 8a in 8(

    0dU < Ud0 ! K 8+0dUA%v ++++++++++++++++++++++ 8c

    7ut we have for I mole of ideal gas %p# %v ! K +++++++++++++ 8d

    *u(stituting 8d in 8c

    0dU < Ud0 ! 8%p# %v 8+0dUA%v

    dT .+0dU

    ++++++++++ 8a%v

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    Ans:

    a) $ess7s law of constant heat summation

    This law can (e stated as: If a chemical change can (e made to ta'e place in two or more different ways, whether in one or

    several steps, the amount of total heat change is same no matter (y which method the change is (rought a(out.

    The law also follows as a mere consequence of the first law of thermodynamics.

    Set us suppose that a su(stance BLC can (e changed to BC directly

    L ! < "

    $here " is the heat evolved in the direct change. $hen the same change is (rought a(out through intermediate stages

    L < 7 ! "2

    7 < % ! "3

    % < ! "G

    The total evolution of heat ! "2

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    Fe293< 3%9 ! 2Fe ! +2?.2 /@++++++++ 8>

    qn 8> is the required eqn therefore N$-=9. /5 KC (Ans)

    1) 2alculate the standard enthalpy and entropy changes at 2 for the reaction:

    J2u0 + (4) . J 2u40

    NFo

    . /1==-/1' T log T 3 1"" T C

    Ans:

    2alculation ofstandard enthalpychange:

    The enthalpy change in terms of Ro is given as

    Therefore RoAT ! +J25AT+J.G log T < 23.3

    ! +J25AT+J.G ln TA2.353

    ! +J?J> @

    ! +J?.J> '@

    Therefore the standard enthalpy change ! /1519 C

    2alculation of standard entropy change:

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    The enthalpy change in terms of Ro is given as

    Therefore Ro! +J25+J.G T log T < 23.3 T

    ! +J25+?.2 T ln T < 23.3 T

    ! +?.2T_AT < ln T 1 < 23.3

    ! +?.2 < ln T1 < 23.3

    ! +?.2 < ln T1 < 23.3

    7ut

    so. ?.2 < ln T1+23.3

    ! ?.2 < ln 2>1 # 23.3

    ! ?.2

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    2 2p.3 T / T/

    $hereGG are constants%p +++ &eat capacity at constant pressure

    T +++ Temperature in /elvin

    ( For a process at %onstant Uolume, there is no xpansion and %ontraction i.e. 8dU ! 5 *o $ ! 5

    *o the equation " ! < $ will (ecome " U ! 6U&eat a(sor(ed at constant Uolume i.e., "Uis equal to the energy increase 6Uaccompanying the

    0rocess

    c Deri#ation of Fi!!s > $elmholtV equation

    The Ri((s free energy for a closed system R ! &+T*

    Qifferentiating the a(ove equation dR ! d& + Td* + *dT

    9r dR ! d 8)

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    +++++++++

    ++++ 8 dT T U

    $e can ma'e two approximations, which will give us the %lausius+%lapeyron equation.

    The first approximation is that the volume of a given amount of gas is much larger that the volume of an equivalent

    amount of solid or liquid. There fore

    U ! Ugas] Uliquidb Ugas

    9r2 The second approximation is that the vapour produced (ehaves as an ideal gas therefore we can use the state equation

    of the ideal gas and we can replace Ugas(y the following.

    0U ! KT 8ideal gas equation if n!

    That is, we set Ugas! KTAp.

    $ith these two approximations quation 8 (ecomes

    d0A0 ! 8 &A KT2 dT

    d ln0 ! 8 &A KT2 dT

    9r2

    lnRTH

    dTPd =

    The a(ove equation is called the 2lausius/2lapeyron equation

    The a(ove equation ta'es the following form if it is integrated (etween pressures 0 , 02and temperatures (etween T, T

    assuming & is constant

    2

    ,

    2

    ,

    2

    :8lnT

    dT

    R

    HPd

    T

    T

    P

    P

    =

    1,,

    .ln2

    22

    ,2

    ,

    TTR

    H

    P

    P

    =

    The a(ove equation is very useful for estimating vapor pressures at any temperature if the vapour pressure at another

    temperature and the heat of vaporisation over that temperature range are 'nown.

    1) Deri#e the ant $off7s *quation for Isotherm for the reaction of the type aA 3 !B . c2 3 dD

    Ans:

    %onsider the given reaction at a temperature 8T and a pressure 80

    aL < (7 ;++4 c% < dQ

    The upper case letters L and 7 are the reactants and % and Q are the products. The lower case letters a,( are the no. of

    moles of reactants and c,d are the no. of moles of products

    The free energy change for the a(ove reaction is written as

    6R ! R products# R reactants++++++++++++++++++ 8

    R products! cR% < dRQ++++++++++++++++++++++++++ 82

    R reactants! aRL < (R7+++++++++++++++++++++++++++ 83

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    7ut R ! R5< KT lna 8BaC is activity

    Lpplying the a(ove equation to 82 and 83

    cR%! cR5< cKT lna% 8 BacCis the activity of B%C

    *imilarly

    dRQ! dR5< dKT lnaQ +++++++++++++++++++++++ 8G

    aRL! aR5< aKT lnaL++++++++++++++++++++++++ 8H

    (R7! (R5< (KT lna7++++++++++++++++++++++++ 8J

    *u(stituting the a(ove 8G, 8H and 8J in 8

    6R ! 8cR%

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    $hen L+7 interactions are wea'er than L+L or 7+7 interactionsP then is 4 8positive deviation from ideality,

    system can then tend towards phase separation and mixing is endothermic.

    xample: Fe+%u liquid alloy

    ;actors causing de#iations from ideal !eha#iour

    ( The deviations of a real solution from ideal (ehaviour are of two types they are 0ositive deviation and 2 Oegative deviation

    The factors, which cause the deviations, are the interactions (etween similar and dissimilar atoms of the

    components of the solution. %onsider a solution in which the solvent is BLC and solute is B7C and if the interactions

    (etween similar atoms are FL+Land F7+7 and the interactions (etween dissimilar atoms are FL+7then

    If FL+L! F7+7 ! FL+7 Then the solution (ecomes ideal

    2 If FL+L; FL+74 F7+7 Then the solution under goes negative deviation

    2 If FL+L4 FL+7; F7+7 Then the solution under goes positive deviation

    15 a) rite any fi#e differences !etween re#ersi!le and Irre#ersi!le processes !) Define Thermodynamic

    equili!rium c) e@plain the criteria for the esta!lishment of thermodynamic equili!rium

    Ans:

    Keversi(le process Irreversi(le process

    It ta'es place with infinitesimally small change It ta'es place with a finite change

    2 These are ideali=ed and true in principle only 2 Lll actual processes which occur are Irreversi(le

    3 *ystem is at equili(rium at all stages of the

    process

    3 *ystem is at equili(rium only at the initial and

    final stages of the process

    G $or' done is maximum G $or' done is less than that of reversi(le process

    H fficiency of reversi(le process is maximum H fficiency is less

    J This process can (e retraced (ac' in the opposite

    direction

    J this process cannot (e retraced (ac' in the

    opposite direction

    ? xample: Ice can (e formed (y withdrawing

    heat from water at 55% under normal pressure.

    Further if heat is supplied at the same rate to ice at

    55% under normal pressure water is formed at the

    same rate.

    ? xample: when two gases are allowed to mix up

    into a given volume at a constant temperature, the

    pressure in the container increases. 7ut the two

    gases cannot (e separated (y `ust with drawing the

    pressure from the mixture.

    ( L *ystem is said to (e under Thermodynamic equili(rium if the system under goes infinitesimal change so that we

    L(le to determine the values of its state properties such as 0ressure, Temperature and Uolume.

    c For the esta(lishment of Thermodynamic equili(rium the system should (e in

    Thermal equili(rium

    Mechanical quili(rium and

    %hemical equili(rium

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    Thermal *quili!riummeans the temperature of the system must (e uniform throughout the system I.e. dT ! 5 $here

    BdTC represents change in temperature

    Mechanical *quili!riummeans the 0ressure of the system must (e uniform throughout the system I.e. d0 ! 5 $here

    Bd0C represents change in pressure.

    2hemical *quili!riumthe %hemical %omposition of the system must (e uniform throughout the system I.e. d% ! 5

    $here Bd%C represents change in temperature

    19 a) Define heat capacity at constant pressure (2p) and heat capacity at constant #olume(2#)

    !) Deri#e the relation !etween 2pand 2#for an ideal gas from first principles

    Ans:

    a)

    $eat capacity:The heat capacity is the amount of heat required to raise the temperature of the system (y one degree%elsius.

    The relationship (etween heat and temperature change is usually expressed in the form shown (elow

    $here % is the heat capacity

    $eat capacity at constant pressure82&:

    $here"p ! &eat content of the system at constant 0ressure

    $eat capacity at constant #olume82#:

    $here

    E#! &eat content of the system at %onstant Uolume.

    !) Deri#ation of the relation !etween 2pand 2#from first principles

    Lccording to the definition of heat capacity at constant 0ressure 8%p, %p is given (y the following equation

    2 . E

    ,T

    2p .Ep

    ,T

    2p .E#

    ,T

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    2p !

    Ep//////////////(1)

    NT

    9r

    2p !R$

    //////////////()NT

    $here Ep or R$is heat content of the system at constant pressure Tis change in temperature

    Lccording to the definition of heat capacity at constant 0ressure 8%v, %v is given (y the following equation

    2# !E#

    ///////////////(")NT

    4r

    2# !R8

    ///////////////(')NT

    $hereE or R8 is heat content of the system at constant volume. Tis change in temperature

    Lccording to first law of thermodynamics: E . d8 3

    That is E . d8 3 &d

    Lt constant pressure: Ep. d8 3 &d

    Qividing the a(ove equation with dT 8change in temperature)

    *u(stituting (1)and (')in () we get

    For one mole of an Ideal gas, the state equation is & . %T

    Qifferentiating on (oth sides: &d 3 d& . %dT

    Lt constant pressure: d& . -

    Therefore &d 3 - . %dT

    That is &d . %dT

    d.

    %dT &

    *u(stituting this in eqn ()

    Ep.

    d83

    &d////////////////()

    dT dT dT

    2p . 2# 3 p d ///////////////()dT

    2p . 2# 3& %

    &

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    Therefore

    9r

    (1=) ;e4"G &!4 and 2u4 present in a dead roasted tin ore react according to the following equationsG during

    leaching with 111 M $2l at -2

    ;e4"3 $2l (- $4) . ;e2l"(1- $4) 3 "$4

    &!4 3 $2l (- $4) . &!2l(1-- $4) 3 $4

    2u4 3 $2l (- $4) . 2u2l (- $4) 3 $4

    2alculate the heat of leaching process at -2 per 1-- Kg of oreG assuming that the Si4and Sn4present in the

    ore are unaffected during leaching

    Fi#en:

    (i) 2omposition of dead roasted tin ore is 1-? ;e4"G ? &!4G ? 2u4G 1? Si4and rest Sn4

    (ii) Standard heats of formation of ;e4"G 2u4G &!4G ;e2l"(1- $4)G 2u2l (- $4)G &!2l(1-- $4)G $4G $2l

    (- $4) at -2 are /1="G /'--G /'G /=-G /"-=G /9"G /9" and /15' cal3.2

    &29 +J>.32

    &%l 8H5 &29 +?.G

    Fe293< J &%l 8H5 &29 ! 2Fe%l38H5 &29 < 3&29 +++++++++++ 8

    &eat of leaching due to mole of Fe293is ! 2&f Fe%l3 8H5 &29< 3&f &291+&f Fe293< J &f&%l 8H5 &291

    ! 2_8+2.5 < 38+J>.321 # +J.3>.2J 'cal

    That is heat of leaching due to J5 g 8since mole of Fe293 ! J5 g Fe293 ! +>>.2J ' cal

    2p . 2# 3 %

    2p / 2# . %

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    Lmount of Fe293present in 55 'g of ore ! 5 of 55 'g ! 5 'g

    Therefore heat of leaching due to 5 'g or 5555 g of Fe293! 8+>>.2JAJ51_5555 ! +HHJ.2H ' cal

    0(9 < 2 &%l 8H5 &29 ! 0(%l2855 &29 < &29+++++++++ 82

    &eat of leaching due to mole of 0(9 is ! &f 0(%l2 855 &29< &f &291+&f 0(9< 2 &f &%l 8H5 &291

    ! 8+>3.2 < 8+J>.321 # 8+H2.G < 28+?.G1

    ! +JG.32 'cal

    That is heat of leaching due to 223.2 g 8since mole of 0(9 !223.2 g 0(9 !+JG.32 ' cal

    Lmount of 0(9 present in 55 'g of ore ! H of 55 'g ! H 'g

    Therefore heat of leaching due to H 'g or H555 g of 0(9 ! 8+JG.32A223.21_H555 ! +GG5. ' cal

    %u29 < 2 &%l 8H5 &29 ! 2%u%l 8H5 &29 < &29+++++++++ 83

    &eat of leaching due to mole of %u29 is ! 2 &f %u%l 8H5 &29< &f &291+&f %u29< 2 &f &%l 8H5 &291

    ! 28+35. < 8+J>.321 # 8+G5.5 < 28+?.G1

    ! +HH.32 'cal

    That is heat of leaching due to G3. g 8since mole of %u29! G3. g %u29! +HH.32 ' cal

    Lmount of %u29present in 55 'g of ore ! H of 55 'g ! H 'g

    Therefore heat of leaching due to H 'g or H555 g of %u29! 8+HH.32AG3.1_H555 ! +33 ' cal

    Therefore the heat of leaching process at 2H5% per 55 /g of ore

    ! +HHJ.2H ' cal < +GG5. ' cal < +33 ' cal

    ! /99=-1 cal (Ans)

    -) a) State the concept of ;ree energy and gi#e its mathematical equation !) 2hromium and 2ar!on present in

    stainless steel form 2hromium car!ide at ---2 Show !y thermodynamic calculations which of the metals among

    SiG TiG and should !e alloyed to stainless steelG so as to pre#ent the formation of 2hromium car!ide

    Fi#en:

    " 2r 3 2 . 2r"2O NF-. /-9--/'- T cal

    Si 3 2 . Si2O NF-. /155- 3 1 T cal

    3 2 . 2O NF-. /---- 3 1 T cal

    Ti 3 2 . Ti2O NF-. /'--- 3 9 T cal

    Ans:

    a The energy freely availa(le in the system for doing useful wor' is called free energy. It is represented (y BRC. This is

    also 'nown as thermodynamic potential or Ri((sCs free energy. The mathematical definition of free energy is given (y

    F . $ > TS

    $here &+enthalpy of the system

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    T+a(solute temperature

    *+entropy of the system

    The a(ove equation infers that the total energy availa(le in the system is B&C Lnd the energy which is not used for doing

    wor' is BT*C this is also called the (ound energy, the remaining energy that is BRC is only availa(le in the system for doing

    useful wor' which is the free energy.

    !) Solution:

    First find the numerical values of the R 5at T ! J55 < 2?3 ! >?3 / of the given reactions

    3 %r < 2% ! %r3%2P R5! +25>55+G.5 T cal

    R5of %r3%2 at >?3 / ! +25>55+G.5_>?3 ! +2G22 cal

    *i < % ! *i%P R5! +2??5 < .JJ T cal

    R5of *i% at >?3 / ! +2??5 < .JJ_>?3 ! +325.> cal

    U < % ! U%P R5! +25555 < .J T cal

    R5

    of U% at >?3 / ! +25555 < .J_>?3 ! +>J53.2 calTi < % ! Ti%P R5! +GH555 < 2.> T cal

    R5of Ti%at >?3 / ! +GH555 < 2.>_>?3 ! +G2HHH.J cal

    Oow %ompare the Ualues of R5of the a(ove compounds I,e 2r"2O Si2O 2O and Ti2O

    9n comparison, it can (e concluded that R5of Ti% is more negativeP this means the compound Ti% is more sta(le than

    %r3%2

    Therefore, to prevent the formation of %hromium car(ide Tishould (e added

    1) a) *@plain a!out the internal energy of a thermodynamic system !) Deri#e Fi!!s > $elmholtV equation

    Ans:

    a) Internal energy:

    The total of all the possi(le 'inds of energy of a system is called its internal energy. Internal energy of a system is

    made up of (y

    Internal potential energy

    Internal 'inetic energy

    Internal potential energy :

    This consists of the energy

    Que to electrostatic force of attraction (etween nucleus and electrons of the atom

    Que to inter molecular attractions

    Que to chemical (onds

    Internal 'inetic energy:

    This consists of the energy

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    Que to translational, vi(rational and rotational motions of molecules, atoms and electrons etc of the

    system

    Internal energy is a state property and extensive in nature.

    !) Deri#ation of Fi!!s > $elmholtV equation

    The Ri((s free energy for a closed system R ! &+T*

    Qifferentiating the a(ove equation dR ! d& + Td* + *dT

    9r dR ! d 8)

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    Lccording to this equation, the rate at which the natural logarithm of the vapor pressure of a liquid changes with

    temperature is determined (y the molar enthalpy ofvapori=ation of the liquid, the ideal gas constant, and thetemperature of the system.

    If we assume that Hvdoes not depend on the temperature of the system, the %lausius+%lapeyron equation can

    (e written in the following integrated form where Cis a constant.

    CRTHP V +=ln

    The plot of vapor pressure versus temperature is shown (elow

    (

    For solid \ liquid, equili(rium

    d0 ! &fusiondT T U

    $here d0Xchange in pressure &fusionXSatent heat of fusion

    TXmelting point dTXchange in melting point

    U X change in volume of liquid and volume of solid 8i.e., Ul# Us

    2 For liquid \ vapour, equili(rium

    d0!

    &vap

    dT T U

    $here d0Xchange in pressure &vapXSatent heat of vaporisation

    TX(oiling point dTXchange in (oiling point

    U X change in volume of Uapour and volume of liquid 8 i.e., Uv# Ul

    3 For solid \ Uapour, equili(rium

    d0!

    &su(limation

    dT T U

    $here d0Xchange in pressure &su(limationXSatent heat of su(limation

    TXsu(limation point dTXchange in su(limation point

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    U X change in volume of Uapour and volume of solid 8i.e., Uv# Us

    ") Deri#e the ant $off7s *quation for Isotherm for the reaction of the type aA 3 !B . c2 3 dD

    Ans:

    %onsider the given reaction at a temperature 8T and a pressure 80

    aL < (7 ;++4 c% < dQ

    The upper case letters L and 7 are the reactants and % and Q are the products. The lower case letters a,( are the no. of

    moles of reactants and c,d are the no. of moles of products

    The free energy change for the a(ove reaction is written as

    6R ! R products# R reactants++++++++++++++++++ 8

    R products! cR% < dRQ++++++++++++++++++++++++++ 82

    R reactants! aRL < (R7+++++++++++++++++++++++++++ 83

    7ut R ! R5< KT lna 8BaC is activity

    Lpplying the a(ove equation to 82 and 83

    cR%! cR5< cKT lna% 8 BacCis the activity of B%C

    *imilarly

    dRQ! dR5< dKT lnaQ +++++++++++++++++++++++ 8G

    aRL! aR5< aKT lnaL++++++++++++++++++++++++ 8H

    (R7! (R5< (KT lna7++++++++++++++++++++++++ 8J

    *u(stituting the a(ove 8G, 8H and 8J in 8

    6R ! 8cR%

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    ( Qepending on the (ehaviuor of solutions they are classified as Ideal solutions 2 Keal solutions and 3 Kegular

    solutions

    1) Ideal solutions: L *olution is called ideal solution if the interactions (etween similar and dissimilar atoms is the same

    that is in an ideal solution of L and 7, the L+L, 7+7 and L+7 interactions are identical

    9r FL+L! F7+7! F%+% $here F indicates the interaction

    xample: %u+Oi alloys

    For an ideal solution &mix,ideal ! 5

    *mix,ideal 5

    Umix,ideal ! 5

    ) %eal solution:L *olution is called real solution if the interactions (etween similar and dissimilar atoms is different

    that is in a real solution of L and 7, the L+L, 7+7 and L+7 interactions are different

    9r FL+L F7+7 F%+% $here F indicates the interaction

    For a real solution &mix,real 5

    *mix,real 5

    Umix,real 5

    xample: Fe+Oi, Fe+%u

    ") %egular solution:L special type of a non+ideal or real solution whose (ehaviour is in (etween the ideal and rea

    solutions for which the enthalpy of mixing is non+=ero (ut the entropy of mixing is identical to that of ideal solutions

    For a regular solution &mix,reg 5

    *mix,reg ! *mix,ideal

    Umix,reg 5