Assign 3 Solutions

Thermodynamics of Multicomponent Systems – 3C04 Assignment 3 – Solutions

1. The molar volumes of solid and liquid lead at the normal melting temperature of lead are respectively, 18.92 cm3 and 19.47 cm3. Calculate the pressure which must be applied to lead in order to increase its melting temperature by 20 centigrade degrees. ANSWER: Thermodynamic Data: ∆H ( s → l ) = 4810 J / mol

Ttrans = 600 K ∆V ( s → l ) = 19.47 − 18.92 = 0.55 cm3 mol

∆H dP = dT T ∆V ∆H  T2  ln   ∆V  T1   4810   620  =  ln    0.55   600 

P2 − P1 =

= 286.76 J / cm3  82.057  = 286.76 *    8.314  = 2830 atm

2. Measurements of the saturated vapor pressure of liquid NdCl5 give 0.3045 atm at 478K and 0.9310 atm at 520K. Calculate the normal boiling temperature of NdCl5.

ANSWER: The vapor pressure is given by the following equation: A ln p = + B T For measurement 1 we get: A ln ( 0.3045 ) = +B 478 For measurement 2 we get: A ln ( 0.9310 ) = +B 520 Solving for A and B yields:

A = −6614

B = 12.65 Therefore, at 1 atm pressure the normal boiling point of NdCl5 is:

6614 + 12.65 T T = 523K

ln (1) = −

3. One mole of solid Cr2O3 at 2500K is dissolved in a large volume of liquid Raoultian solution of Al2O3 and Cr2O3 in which XCr2O3 = 0.2 and which is also at 2500 K. Calculate the changes in enthalpy and entropy caused by the addition. The normal melting temperature of Cr2O3 is 2538 K, and it can be assumed that the ∆S m , Al2O3 = ∆S m ,Cr2O3 .

ANSWER: Thermodynamic Data: ∆H m, Al2O3 = 107500 J

Ttrans = 2324 K 107500 = 46.25 J / K 2324 = 46.25* 2538 = 117400 J

∆Sm, Al2O3 = ∆Sm,Cr2O3 = ∆H m,Cr2O3

Cr( s ) → Cr(l )

∆H = ∆H M = 117400

Cr(l ) → Cr( solution )

∆H = ∆H MID = 0

∆H Total = 117400 J Cr( s ) → Cr(l )

∆S = ∆S M = 46.25

Cr(l ) → Cr( solution )

∆S = ∆S MID = − R ln X Cr = − R ln 0.2 = 13.38

∆STotal = 46.25 + 13.38 = 59.63 J / K

4. The variation, with composition, of GXS for liquid Fe-Mn alloys at 1863 K is listed below. a. Does the system exhibit regular solution behaviour? XS XS and GMn at XMn = 0.6. b. Calculate GFe

c. Calculate ∆G M at XMn = 0.4. d. Calculate the partial pressures of Mn and Fe exerted by the alloy of XMn = 0.2.

XMn

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

XS

395

703

925

1054

1100

1054

925

703

395

G ,J

ANSWER: a)

∆G = ∆G ID + ∆G EX

∆G EX = ∆G M − ∆G ID = ΩX A X B Solving for Ω using the data given: XMn

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

XFe

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

XMnXFe

0.09

0.16

0.21

0.24

0.25

0.24

0.21

0.16

0.09

GXS, J

395

703

925

1054

1100

1054

925

703

395

Ω

4389

4394

4405

4392

4400

4392

4405

4394

4389

Taking the average of Ω above gives 4395J. Therefore, the system exhibits regular solution behaviour. b) EX ∆GFe = RT ln γ Fe = ΩX Mn X Mn = 4395 * 0.6 * 0.6 = 1582 J EX ∆GMn = RT ln γ Mn = ΩX Fe X Fe = 4395 * 0.4 * 0.4 = 703 J

c)

∆G = ∆G ID + ∆G EX = RT ( X Fe ln X Fe + X Mn ln X Mn ) + RT ( X Fe ln γ Fe + X Mn ln γ Mn ) = RT ( X Fe ln X Fe + X Mn ln X Mn ) + ΩX Fe X Mn

= 8.314 *1863* ( 0.6 ln 0.6 + 0.4 ln 0.4 ) + 4395 * 0.4 * 0.6 = −9369 J

d) At XMn = 0.2

 4395 * 0.82   = 1.2  8.314 *1863 

γ Mn = exp 

aMn = 1.2 * 0.2 = 0.24 At XMn = 0.2

γ Fe aFe

 4395 * 0.22  = exp   = 1.011  8.314 *1863  = 1.011* 0.8 = 0.809

The vapor pressure of Mn and Fe are determined as (Gaskell appendix):

 33440  o pMn = exp  − − 3.02 ln (1863 ) + 37.68  = 0.0493 atm  1863   45390  o pFe = exp  − − 1.27 ln (1863) + 23.93  = 4.55 x10−5 atm  1863  Using this data the partial pressures can be determined:

aMn =

pMn o pMn

pMn = 0.24 * 0.0493 = 0.0118atm aFe =

pFe o pFe

pFe = 0.809 * 4.55 x10−5 = 3.68 x10 −5 atm