driving force for solidification this learning object will introduce the concept of undercooling...

Driving force for solidification

This Learning object will introduce the concept of undercooling and driving force for phase transformations

Subject: MEMS

Mentor: Prof. Gururajan

Authors:

Learning Objectives

After interacting with this Learning Object, the learner will be able to:• calculate the change in entropy of the system• calculate the driving force of the system during phase

transformation

Definitions:1) Gibbs free energy: The part of the energy of a system that is available as work during

chemical change is known as free energy of the system. The Gibbs free energy of a system is defined by the equation G = H – TS where H is the enthalpy, T the absolute temperature and S the entropy of the system.

2) Undercooling Cooling a material below the temperature of an equilibrium phase

change fast enough to not allow the occurrence of the transformation.

3) Driving force: The gain in free energy of a system during transformation from one

phase to another. It is given by the formula ∆G ∆Hf – T

4) Melting point of ice: The temperature at which free energy of water (liquid) is equal to the

free energy of ice (solid)

≈

mT

H fΔ ≈mT

TH ΔΔ f

Definitions:5) Boiling point of water The temperature at which free energy of water vapour (gaseous

state) is equal to the free energy of water (liquid)

6) Transformation temperature/ Temperature of transformation: The temperature at which free energy of state is equal to free

energy of state of transformations.

7) Specific heat of a substance: It is the quantity of heat (in joules) required to raise the temperature of 1 gram of pure substance by one degree Kelvin.

8) Latent heat of fusion/vaporisation: The quantity of heat absorbed or released when a substance

changes its physical phase at constant temperature.

9) Entropy of fusion: It is the change in entropy at the time of melting of a material

It is defined by the equation

αβ

int=

Δ=Δ

pomelting

heatLatent

T

HS

m

Derivation:The free energies of the liquid and solid at temperature T are given by

GL = HL – TSL

GS = HS – TSS

Therefore, at a temperature T,

∆G = ∆H – T ∆S ………………. (1)

Where ∆H = HL – HS and ∆S = SL – SS

At the equilibrium melting temperature Tm the free energies of the solid and liquid are equal i.e. ∆G = 0

Therefore, ∆G = ∆H – Tm ∆S = 0

Derivation:Therefore at Tm

∆S = ……………. (2)

This is known as “Entropy of fusion”.

For small undercoolings (∆T) the difference in the specific heats of the liquid and solid (CL

p - CSp ) can be ignored.

Therefore, ∆H and ∆S are approximately independent of temperature.

Combining equations (1) and (2), we get

∆G ΔHf – T ( for small ΔT)

m

f

T

HΔ

≈m

f

T

HΔ

mT

TH ΔΔ f≈

Diagram for reference:

For the animator – Draw this image

• As temperature increases, enthalpy increases with slope Cp i.e specific heat at constant pressure

2) As temperature increases, Gibbs free energy decreases at a rate given by (-S)

Diagram for reference:

For the animator–Draw this image and the let the text come below the image

Variations of the free energies of solid and liquid phases with temperature at constant pressure

solid

liquid

solid liquid

Tm

• The Gibbs free energy of the liquid decreases more rapidly with increasing temperature than that of the solid• For temperatures upto Tm, the solid phase has the lowest free energy and is therefore the stable equilibrium phase• For temperatures above Tm, the liquid phase has the lowest free energy and is therefore the stable equilibrium phase• At both the phases have the same value of G and both solid and liquid can exist in equilibrium.

Master Layout

Difference in free energy between water (liquid) and ice (solid) close to melting point

The curvature of the Gw and Gi lines have been ignored

w

i

Title of the graph

Footnote

Tm - Ice point

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2

4

Step No: 1

Instructions for the animator Audio narration/ text to be displayed

Display the image of water in glass first then the thermometer

Can we keep water below its freezing point without allowing it to turn into ice?

The red line of the thermometer goes down from 300 C to - 300 C

Click on the NEXT button to find out the answer.

NEXT

Images copyrighted

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Step No: 2


Display the animation of green box. If we keep water in the refrigerator eventually it will turn into ice.

Display master layout as above If we plot the graph of Gibbs free energy for cooling of water as a function of temperature it will look like this

Master layout do not show - Gi line, title of the graph and footnote

Show water in vessel being kept in the refrigerator for cooling.

Vessel is taken out and show ice in the vessel.

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Step No: 3

Master layout http://en.wikipedia.org/wiki/File:Melting_icecubes.gif


Display the animation of green box. Now if we keep ice at room temperature to melt eventually it will turn into water.

Display master layout as above If we plot the graph of Gibbs free energy for melting of ice as a function of temperature it will look like this

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Step No: 4

Master layout

Instructions for the animator

Audio narration/ text to be displayed

The master layout remains on screen as in previous slide

Observe the graph of Gibbs free energy v/s temperature as shown above.• the two lines intersect at a point called Tm

• below Tm ice has lower free energy hence it exists in stable equilibrium• above Tm water has lower free energy hence it exists in stable equilibrium• at both water and ice have the same free energies and co-exist in equilibrium.

Master Layout 1



w

i

T3 T2 T1

T2

Tm - Ice point

Master Layout 1



w

i

T1

∆T

∆G

Tm - Ice point

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4

Step No: 5

Master layout 1



Display master layout as explained above

If water is cooled at temperature T1 below Tm, the driving force ∆G is small and the barrier for water to form ice is large.

Master Layout 2



w

i

T1

∆T

∆G

T2

Tm - Ice point

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5

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2

4

Step No: 5

Master layout 2




If water is cooled at temperature T2 below Tm, the driving force ∆G is higher than that at T1

It means that the gain of free energy by the system during the transformation of water to ice is more.

Master Layout 3



w

i

T1

∆T

∆G

T2

Tm - Ice point

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5

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2

4

Step No: 5

Master layout 3




If water is cooled at temperature T3 below Tm, the driving force ∆G is much higher than that at T1 and T2

It means that the gain of free energy by the system during the transformation of water to ice is very large i.e the driving force ∆G is very large.

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Step No: 5

Master layout 3




You must have observed that higher the undercooling ∆T, the driving force is large.In this case, undercooling of water is how much below Tm can we keep water without allowing it to turn into ice

The driving force is given by ∆G L - T≈

mT

L

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Step No: 5

diagram on slide 7



Display diagram as explained above

In general, the graph of Gibbs free energy as a function of temperature for any material (in solid and liquid state) looks like this.

If the driving force is large the system will move to its stable state quickly. If the driving force is small then the system will take a longer time to move to its stable state. Thus water can be kept in its liquid state below 00 C.

Want to know more…(Further Reading)

Graphs/Diagram(for reference)

Animation Area

Test your understanding (questionnaire)

Lets Learn!

Lets Sum up (summary)

Instructions/ Working area

Radio buttons (if any)/Drop down (if any)

Interactivity options

Sliders(IO1)/ Input Boxes(IO2)/Drop down(IO3)

(if any)

Play/pause Restart

Output result of interactivity (if any)

What will you learn

Credits

Definitions

Derivation

INSTRUCTIONS SLIDE

• Please make sure that the questions can be

answered by interacting with the LO. It is

better to avoid questions based purely on

recall.

Questionnaire for users to test their

understanding

Questionnaire1) NPTEL link will be provided by Prof. Gururajan

2) Questions:

(Raghavan, V. , Material Science and Engineering, Prentice

Hall of India, Vth edition)

a) The free energy change during melting of ice at 0o C is equal

to

i) enthalpy of melting – entropy of melting

ii) 0 iii) 273 iv) can’t say without more data

b) Calculate the entropy increase when one mole of ice melts into

water at 0o C (Hint: Use formula for ∆S)

Answers: ∆S = 22.04 J mol-1K-1

c) Calculate the entropy change and the free energy change

during the melting of gold at its melting point. The enthalpy

of fusion for gold is 12.6 kJmol-1

Answers: ∆S = 9.43 J mol-1K-1 , ∆G = 0

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Links for further reading

Books:1) Raghavan, V. , Materials Science and Engineering, a first

course, Second edition, Prentice Hall of India, New Delhi2) Jena, A. K., Chaturvdei, M.C., (1992), Phase transformation in materials, Englewoods Cliffs:

Prentice Hall3) Porter, D., Easterling, K., Phase Transformation in metals

and alloys, 3rd edition

Weblinks:NPTEL link will be provided by Prof. Gururajan

Summary• Driving force is the gain in free energy of the system during transformation from one state to another.

• Undercooling is cooling the material below its melting point (transformation temperature) without allowing it to convert into the stable state.

• The larger the undercooling the associated driving force is also large.

• The rate of transformation depends upon the driving force.

• Driving force is given by ∆G ∆Hf - T≈mT

HfΔ

driving force for solidification this learning object will introduce the concept of undercooling...

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