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H= U+PV  By plotted the equation ''2'' in ''1''

   Differential equation (3)results

 واف ه شغم وخت نهخ حج غط راج  حم  ح ست  هت نخل    ض: ح ىن ه ناث نات  -ىىا شخاق 

 

dU=  dU  نات ن هت ف  غخن ا

dU= dU=TdS- PdV

 اف ر dUتنا ح   

( ز تنا  فisothermal stat) ح0''=''dT(  تنان  ناخ5)

  و 

d G= VdP   but for Ideal gas '' RT=PV'' so that V=RT/P by plugging

in equation (6)

 

d G= RTdP/p = RTdlnP

 

Q2-Find the change in free energy associated with the isothermal

expansion of 1 mole of ideal gas from 2 to 1 atm pressure at 298kº.?

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n=1moles , p1=1atom ,p2=2atom for Idial gas and T= 298 kº 

3290.573 J/mole

 

Q3- For a reaction, Calculate S   and  H  for the reaction at 400k°?

  at T=400k°

  -

1572583.32J/mole

 

-

S

 

-S= -16.6lnT + 16.6  -16.6ln + 16.6

   ol 

Q 4 - For the reaction Fe   = Fe (l) ,   

  Why two temperatures are dependent term present

in the,    equation. Although   is independent of

temperature?

Do both of these terms drop out if   = 0 for a reaction?

 

(

 

 

(

 

From above relation shown that the temperatures are dependent term

 present in the,    equation. Although   is independent of

temperature.

Yes drop out because   

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Q 5- The reaction of the chemical potential   to the termed

Functions U, H, A, AND G are given by the equation

Is there a corresponding useful relation between   and the entropy

S of a phase .

 

G the is function of p,T and the number of molecules in the system

G= (T,P,ni ,n j,nk )

ni ,n j,nk  = number of molecules where the system is fixed at number of

molecules constant .

  

   

 (1)

At the number of molecules for the all system content constant

equation (1) simplified by the equation (2)

 

dG =SdT+v dp2

 

at differential (2) 

  (3)

  (4)

Plutting (3)&(4) in (1)

 

∑  (5)

where ∑  equal the sum of all boundary of system at

differential the molecules where constant  

 (6)

Pl utting (6) in (5)

 

dG =SdT+v dp+∑  (7)

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ن تن ت ن ة  رث نغط   ى ةن تان ف  غخن  ح تنان ينهىظاث  نك  ن ة ع نط  وظت نخت ذ د حال نهاة    ان

ص هن انزم   . نخب ف  غح د ذ تهغن

∑   (8)

∑   (9)

∑   (10)

From (8),(9),(10) can shown the entropy(S) does not has same relation

with chemical potential

.

()  

But has this relation

  

Q 6 - Would it possible for to equals zero? Explain. ? Yes ,equals zero

Explain :-

standard Gibbs free energy of formation of a compound is the change

of  Gibbs free energy that accompanies the formation of 1 mole of a

substance in its standard state from its constituent elements in their

standard states (the most stable form of the element at 1 bar of pressure

and the specified temperature, usually 298.15 K or 25 °C).

This table includes several Standard Gibbs function of formation

evaluated at 298 K. Note that all values are in kJ/mol.

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SpeciesPhase

(Matter)

Chemical

FormulaΔGf 

o (kJ/mol)

Aluminum  

Aluminum Solid Al 0

Aluminum Chloride   Solid AlCl3  -628.8

Aluminum Oxide  Solid Al2O3  -1582.3

Barium  

Barium Chloride Solid BaCl2  -806.7

Barium Carbonate Solid BaCO3  -1134.4

Barium Oxide Solid BaO -520.4

Barium Sulfate Solid BaSO4  -1362.2

Beryllium  

Beryllium Solid Be 0

Beryllium Hydroxide  Solid Be(OH)2  -815.0

Boron  

Boron Trichloride  Solid BCl3  -388.7

Bromine 

Bromine Liquid Br 2  0

Bromine Trifluoride  Gas BrF3  -229.4

Hydrobromic Acid  Gas HBr -53.4

Calcium  

Calcium Solid Ca 0

Calcium carbide Solid CaC2  -64.9

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Calcium

Carbonate (Calcite) Solid CaCO3  -1129.1

Calcium Chloride Solid CaCl2  -748.8

Calcium Chloride Aqueous CaCl2  -816.05

Calcium Hydride  Solid CaH2  -142.5

Calcium hydroxide   Solid Ca(OH)2  -897.5

Calcium Oxide  Solid CaO -603.3

Calcium Sulfate Solid CaSO4  -1309.1

Calcium Sulfide  Solid CaS -477.4

Carbon  

Carbon (Graphite)  Solid C 0

Carbon (Diamond)  Solid C 2.900

Carbon Dioxide Gas CO2  -394.39

Carbon disulfide Gas CS2  67.1

Carbon Monoxide Gas CO -137.16

Carbonyl

Chloride (Phosgene)Gas COCl2  -204.9

Caesium  

Caesium Solid Cs 0

From table above all elements in their standard states

(oxygen gas, graphite, etc.) have 0 standard Gibbs free energy change

of formation, as there is no change involved.

Δr G = Δr G˚ +  RT  ln Qr ; Qr  is the reaction quotient. 

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At equilibrium, Δr G  = 0 and Qr  = K  so the equation becomes Δr G˚ =

− RT  ln K ; K  is the equilibrium constant. 

 

 = -RT

 

 

IF    = 1

Then    = Zero.

 

Q7 – apply the phase rule to a system consisting of molten iron in equilib.

With δ-Fe and iron vap.does the state of this

System represent a triple point ?

  In this case, C=1, P=3

F=c-p+2

C=1 , p=3

F=1-3+2= 0    8  – The vap. Pressure of zinc is 0.1atm at 990k and 1 atm at 1180k.

Calculate the heat of vaporization of zinc?

 

 ,  , , =1 & R=8.314 j/mole.k o 

 

 

 

 

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9 –  For second –  order transition , drive

 

 For the Ideal gases

 

PV=RT 1

 

At constant V= constant 

By differential equation

 Vdp=RdT

 

 =

   2

 

 

 

   

  ()  

 

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Q10 –  the following pressure of zinc have been determined for copper-

zinc alloys at the 1060cº.

xzn  1.0 0.45 0.3 0.2 0.5 0.1 0.05

pzn,mmH 3040 970 456 180 90 45 22

thi t ht

1-  Pi=xi = constant for the system obey Raoult''s

2-    = xi  & xi0 for the system obey Raoult's law

3-   xi , xi1 Pi=k   xi for the system obey Hennery's law

where k=constant .

4-  For this system Xi= xzn ,   where

Xi= volume fraction ,

 pure vapour component pi= vapour pressure of

ideal solution component .  By excel calculation can shown the system obey to Rault's or

Hennery's law see Table below and Figure.

xzn      1 3040 3040 1

0.45 970 436.5 0.45

0.3 456 136.8 0.3

0.2 180 36 0.2

0.5 90 45 0.50.1 45 4.5 0.1

0.05 22 1.1 0.05

 

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Calculation

  xzn    from Table .

  azn has liner relation with xzn 

  xzn .vs. pzn Near liner

From all calculation I think the system oby to Raoult's

 

-500

0

500

1000

1500

2000

2500

3000

3500

0 0.2 0.4 0.6 0.8 1 1.2

   p   z   n 

xzn 

0

0.2

0.4

0.6

0.8

1

1.2

0 0.2 0.4 0.6 0.8 1 1.2

   a   z   n

 

xzn 

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Q11 –   If a2=kx2(1+bx2) , find a1 by the G.D. eq.?

From Gibbs-Duhem eqs. , we get

 

     

   

  

  

  

 By integrations

   

   

Q12  –   from the eq. for the activity coff of zinc in copper- zinc alloys

TγZn = - 4600x2cu , find the acti. Coff.of copper at 1500k by mean of

the G.D eq.

Activity coefficient of zinc

 (1)

From (1) 

 = -0.368856

 (2)

From

 

 (3) 

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Deferential equation (2)     (4)

By plutting (4) in used .G.D.eq.

  G.D.eq. 

    (6)

 By integration (6)

∫ ∫   (7)

 

[ ]  (8)

   [ ]   [ ] 

Supposed the alloy content 40% zn ,60%cu  The the equation (7) will come to

 

   

 

At  

 

0.059017 

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  = 1.060793

 

Q13  –   the following eqs. Have been proposed to excrescent activity

coff. Data for a system at fixed T and P .

       

 G.D. eq .

   

 

   By putting in (2)& (3) in (1)     

Where   &  putting in (4)

 

        

 

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   From(6) the system are not satisfy the G.D.eq .

 Q14  –   if a solute lowers the surface tension , do you expect to be

present at the surface in larger or in smaller conc. Than in the

interior .?

In the case of solutions of two or more substances , the surface

minimization process is different and more complex, because different

molecules present in a solution have different intensities of attractive

force –  fields and also have different molecular volumes and shapes as a

result , the molecules that have the greater fields of force tend to pass into

the interior, and those with the saller force- fields remain at the surfaceand surface tension has relation below with force.  Where the force is low the surface tension is low , In this case I expect

the solute concentrated at the surface as result low force that is cused by low surface tension . The solutes dispersing in a liquid instead of

concentrating on it is surface, rise the surface tension while solutes

concentrating on the liquid surface lower it is surface tension (Gibbs  –  

Thomson principle) . If C is the overall solute concentration in a dilute

solution , s is the excess of solute concentration on the surface over it isconcentration in deeper layer   is the change in  for a small change

( in solute concentration , R is the molar gas constant , and T is the

absolute Temperature , then the Gibbs adsorption equation is As follows

S = -

 ( 

Thus, a positive S.

-  i.e. a higher concentration of the solute on the liquid surface than

in the deeper layers, is related to a negative  and would lead to a

fall in the surface tension .

-  i.e. a lower concentration of the solute on the liquid surface than in

the deeper layers, is related to a positive and would lead to a

rise in the surface tension .

  from Biophysics &Biophysical chemistry

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  By D.Das

-  For a solute that lowers the surface tension .the surface excess

concentration is positive

-  For a solute that rises the surface tension .the surface excess

concentration is negative.

Q15 –  if a soap film is extended adiabatically , do you expect its temp

. to rise or fall ?( the surface tension of the film may be assumed to

decrease with a rise in temp.)?

At adiabatically process the temperature is not change of soap film in the

first but can be soap absorbed the energy from surrounding system ifso that the energy of film equaled = E+H where, E= energy of & H =

Energy of soap film . where the energy absorbed grater from the energy

soap in all time as result as I expected the soap will be rise heat in most

time.

Q16  –   given a liquid phase which has a surface tension of 0.5j/m2,

calculate the work done by the system if a mole of liquid is converted

f th f f ph 01 i dit t thi fi 1μ

thick.)?

For thin film soap

   

   

   

     

 A =(       

   

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we have produced A in 2filmBut for thick