T1_Kinetic Molecular Theory

Thermal and Magnetic Properties of Materials

THERMAL PROPERTIES

1

Introduction

Chinese porcelain kiln

Useful Invaralloys for clock making

Allow room for thermal expansion

Glass-Metal seal

Bimetal switchThe incredible Ceramic Hobs


THERMAL PROPERTIES

2

Kinetic Molecular TheoryThe ideal gas

Heat and Heat Capacity

Boyle’s law (1662): A fixed amount of gas at T=Cons obeys ��Charles’ law (1787): At constant pressure a gas obeys ��Gay-Lussac’s law (1809): A fixed volume of gas obeys ��

��

�� and��

��=8.314�J/mol�K

��

Heat= Thermal energy in transit

HotCold

How much heat needs to be supplied to an object to raise its temperature by dT? � !��"��!�

��Cis the Heat Capacity in J/K

Heat Capacity per unit of mass = Specific Heat =!#$%&##�'(�)*+�',-+.)in J/(kg·K)

Heat Capacity per unit of mol= Heat Capacity of 1 molof the substance in J/(mol·K)

In a gas,!/� ��/

!0� ��0


THERMAL PROPERTIES

3

Kinetic Molecular TheoryMicrostates, macrostates

A particular configuration is a microstateof the system

If the coins are distinguishable, each of the 210possible states have the same probability

If the coins are IN-distinguishable, we can only distinguish when the number of tails and heads is different.

A particular configuration here is a macrostateand they do not have the same probability

Number of states with 4 heads and 6 tails = 1234353= 210


Number of states with 8 heads and 2 tails = 12373�3= 45

Number of states with 10 heads and 0 tails = 1

-The system could be described by a very large number of equally likely microstates-What we measure is the property of a macrostateof the system and they are not equally likely


Total 252

210�245�210�2784


THERMAL PROPERTIES

4

Kinetic Molecular TheoryStatistical definition of temperature

91:1;91<

9�:�;9�<

The system with energy 9=is in one of the :=;9=<microstates for that energy

When both are in thermal contact, the whole system can be in any one of the :1;91<:�;9�<microstates

In thermal equilibrium the system will appear to choose a configuration that maximizes number of microstates

��91:191:�9��

-Each one of the possible microstates of a system is equally likely to occur.

Assumptions that we have to make:

-The system’s internal dynamics are such that the microstates of the system are continually changing.

-Given enough time, the system will explore all possible microstates and spend an equal time in each of them.

-Therefore, the system will be most likely found in a configuration represented by the most microstates. For large number of particles, the phrase “most likely” becomes “absolutely, overwhelmingly likely”


THERMAL PROPERTIES

5

Kinetic Molecular TheoryStatistical definition of temperature

91:1;91<

9�:�;9�<

The system with energy 9=is in one of the :=;9=<microstates for that energy

When both are in thermal contact, the whole system can be in any one of the :1;91<:�;9�<microstates

In thermal equilibrium the system will appear to choose a configuration that maximizes number of microstates

��91:191:�9��

:�9��:191�91>:191�:�91

�9��9��91�

As the total energy is constant 991>9�, then �91?�9�and �9��91 @?��:1�:1�91�

:��:��9�

�AB:1�91�AB:�

�9�This condition defines the most likely division of energy

or “being at the same temperature” ��AB:�

�9�

NoteAs the entropy is defined as 1CDE

DF, this quickly leads to

GHI�JKL


THERMAL PROPERTIES

6

Kinetic Molecular TheoryThe Boltzmann distribution

9?M :;9?M<NOPONQRSN�TU��

MPVPUOW

The “system” is so small that :M�microstate for each MXM�:;9?M<��

Since Tis related to AB:microstate and MY9we can Taylor expand around E, or M�,

AB:9?MAB:9?�AB:9��9M>ZAB:9?M

��>Z

��AB:

�9

Neglecting higher terms of the expansion (see problem 6),

:9?M:9[�\]^C @

So the probability function describing the system is then,

XM�[�\]^C @Boltzmann distribution

The probability has to be normalized with all the possible microstates,

XM[�\]^C @_[�à]^C @ =

Where _[�à]^C @ =�bis called the partition function


THERMAL PROPERTIES

7

Kinetic Molecular Theory

The Boltzmann distributionExercise for computer freaks

20 x 20 matrix with random changes between locations,

Initial state is unlikely as it has only 1 microstate associated

�=1

Number of possible microstates like this�=400*399=19600

The Bloltzmanndistribution is simply a matter of probability


THERMAL PROPERTIES

8


The Boltzmann distributionExercise for the computer freaks

20 x 20 matrix starting with 2 quanta per site(i.e. more energy in the initial state) Final distribution of a

1000 x 1000 matrix


THERMAL PROPERTIES

9


The Maxwell -Boltzmann distribution

Assume a monoatomic gas where the only energy available to the particles is the kinetic energy,

c;defdg,dh)9]1�ide�>1

�idg�+�1�idh�1�id�

For a given direction, say x, the number of particles between deand de>�dewill be proportional to,

jde�[�klmn�]^C @

And we need to normalize this function so ojde p�p�de�

(see appendix at the end of the lesson),

q[�klmn�]^C @�dep

�pris�� @sr��

ijdeisr��[�klmn�]^C @

The 3 directions are indistinguisable, so the number of particles between cand c>�cis proportional to,

jde�dejdg�dgjdh�dh�[�klmn�]^C @�de[�kltn�]^C @�dg[�klun�]^C @�dh[�kln�]^C @�de�dg�dh


THERMAL PROPERTIES

10

Kinetic Molecular TheoryThe speed distribution(see Appendix with solutions to the different integrals)

Fraction of molecules between dcand d>�d

v;d<�d�;wrd��d<�;[�kln�]^C @<O course we have to normalize this function by making ov;d< p

2�d�

qd�[�kln��]^C�d p

2�w

ris�� @�

So that,

v;d<�dwr

is��

��@d�[�kln�]^C @�d

Maxwell-Boltzmann speed distribution

v;d<�d��xiy[z��{z|}~�[��}|��d�y[|�[[��d�{��d>�d�;�z�y��v�y[}�j�{|�|�{|�d<

v;d<


THERMAL PROPERTIES

11

Kinetic Molecular TheoryThe speed distribution

The maximum of v;d<comes easily by differentiating,

�v;d<�d�dk&es��

iThe experimental justification of the Maxwell-Boltzmann velocity distribution,

We can calculate easily dfd�and d�k#d�dqdv;d<�d

p

2��ri @

d�qd�v;d<�dp

2��i

and d�k#��i


THERMAL PROPERTIES

12

Number of molecules travelling in a certain directions at a certain speed

�:wr

Fraction of particles whose trajectories lie in an elemental solid angle �:is,

The solid angle corresponding to particles travelling between �and d�is the shaded area,�:sr��}��

So,�:wr�

s�}��

And the number of particles in that direction with speeds between vand v+dv,�v;d<��d��s�}��

Number of molecules hitting a wall

Volume sweptin a dt,��d��|�RP�So the number of particles hitting a wall of area A,

��d��|�RP��v;d<��d�1��}��And the number of particles hitting a wall per unit of time and unit of area, having speeds between vand v+dv,and travelling at angles between �and d�,

d�RP��v;d<��d�1��}��



THERMAL PROPERTIES

13


Deduction of the equation from the kinetic molecular theory.

��sid�~��

The ideal gas law

Linear momentum change perpendicular to the wall transferred by every particle collision

Adding the contribution of each of the particles we can calculate the pressure on the container,

Xq��0&�)=.�+�;X{z|}~�[��|z{d[�}�j��}|��d�}��|�[��}z[~|}��<�

Xq�dp

2q;sid~��<d~��vd�d�s�}��

��

2�iq�dp

2d�v;d<q~��}��

2

Using the integral o~��}�� 21

�we have that,

X��id�

Using the total number of molecules ��, and that d��i@,

��

Proportion of particles per unit of energy v;9<


THERMAL PROPERTIES

14

Molecular velocity and energy distribution

For a monoatomic gas, 91�id�� 9id�d

{Number of particles between Eand E+dE}= v;9<�9{Number of particles between vand v+dv}= v;d<�d� v;9<v;d<Dl

D`Substituting f(v)by Boltzmann distribution function, 91

�id�and �9id�d,

v;9<sr

��

��@9�[�`]C @

Boltzmann factor

{Probability of being between Eand E+dE} =v;9<�9If the gas is not monoatomic (translational + rotational + vibrational energy)

v;9<!;9<��[�`]C @

For virtually all practical purposes, activation energies EA>>kTand [�`]C @dominates the function


THERMAL PROPERTIES

15

Thermally Activated ProcessesThe role of EA

Example:If a molecule bond breaks in collisions giving 1eV of energy and another breaks more easily, needing only 0.8 eV. What are the relative chances of each kind to break at 300 K?

��f��f��6[��For T=300 K��f�s��[�

v;9<!;9<��[�`]C @�!�[�`]C @

The term in 9changes much slower than [�`]C @so lets assume it constant

v;9��[�<!�[�1+/2�2�5+/ @!�s��1� For 1eV

v;9��[�<!�[�2�7+/2�2�5+/ @!�w��14 For 0.8eV

A decrease of 20% implies that now it is more than 2000 times more likely to break the bond

Example:Compare the number of atoms in a gas having E>1eVat 1000 K and 300 K

9�� @��s

9�� @��w

v;9��[�<!�[�1+/2�2�5+/ @!�s��1� For 300K

v;9��[�<!�[�1+/2�275+/ @!��5 For 1000K

What it takes 15000 years at room temperature is done in 1 second at 1000K


THERMAL PROPERTIES

16

Thermally Activated Processes

Even if EA>>3/2kT, there is a probability of jumping the barrier

Probability of jumping ��Number of times an atom tries to jump per sec., �2Probability of having energy > EA

Probability of having energy > EAXz�y9�9�qv;9<p

`��9�2�[�`�]C @

Rate of jumps = Frequency of jumps ��2�2�[�`�]C @��[�`�]C @

Arrheniusrate equation (rate of a thermally activated process)


THERMAL PROPERTIES

17

Thermally Activated ProcessesThe role of EA

��[�`�]C @��>9�

��[�`�]C @��>9�

��

��>��s��>�OTU

kT(arbitrary units)

EA=1

EA=5

EA=10

��[ �`�]C@


THERMAL PROPERTIES

18

Thermally Activated ProcessesUsing Arrhenius plots

��[�`�]C @��?9��

�>��i��>~Example:A hot-curing single-part epoxy glue called TickYTackA, sets in 6 minutes at 137ºC, 10 minutes at 127ºC or 30 minutes at 100ºC. Calculate the activation energy EAand the proportionality rate A.

The slope of the ��vs. 1/Tis,

?9��?��??��

s��?s��?s��

Taking k=86·10-6eV·K-1, EA=0.49 eV

And A can be calculated from the line equation,

P��[?s��?�2

�?�2�?;?��<�?s��

�� ¡¡¡¡�¢£K��


PROBLEMS

19

1.-The world’s oceans contain approximately 1021kg of water. Estimate the total heat capacity of the world’s oceans. cwater=4.18·103J/(kg·K). Calculate the increase in its T if we could use 10 years of the total energy burned by the entire world (World Energy Consumption 13·1012J/s)

2.-With the energy that Spain burns in 1ms (Spanish power consumption 2,67·108MW·h/year) you want to increase the temperature of all the gold in the US federal Fort Knox reserve (4000 tons of gold). What would be the temperature increase of all that gold? Cm(Au)=25.4 J/(mol·K) and M(Au)=197.

5.-Two bodies, with heat capacities C1and C2(assumed independent of temperature) and initial temperatures T1and T2respectively, are placed in thermal contact. Show that their final temperature Tfis given by �(;!1�1>!��<;!1>!�< @6.-In the overhead 5 we neglected the second term onwards of the Taylor expansion, which is Dn¤¥¦

D`nM�. Show that this term equals ?\n

]^CnDCDànd hence show that it can be neglected compared with the first

two terms if the reservoir is large.

3.-Calculate the rmsspeed of hydrogen (H2), helium (He) and oxygen (O2) at Room Temperature (300 K). The atomic masses of H, He and O are 1, 4 and 16 respectively. Compare these speeds with the escape velocity on the surface of i) the Earth (11.2 Km/s), ii) the Sun (617 Km/s). Proton Mass 1.67·10-27

Kg and kB=1.38·10-23J/K. Discuss the results.

4.-Why does the Moon have no atmosphere? During the 13 days of sunlight, the surface of the Moon can reach a temperature of 500K. Knowing that the escape velocity of the Moon is 2410 m/s, calculate if a molecule of Hydrogen (m=3.33·10-27 kg) escapes after hitting the Moon surface. Explain what would happen to heavier molecules and why there is no atmosphere. kB=1.38·10-23J/K.


PROBLEMS

20

10.-A chemical reaction has an activationenergy 9�1�[�. Compare the rates of the reaction when you

pass from T=300K to T=310K

9.-How many molecules of gas fit inside a 30 liters “ultra high vacuum” (UHV) chamber at 10-10torrand room temperature. 1 torr= 133.3 N/m2

8.-Find the average energy §9�for:a) A n-state system, in which a given state can have energy �fMfsMf¨f�M.b) A harmonic oscillator, in which a given state can have energy �fMfsMf¨f�Mf¨©

Hint: Assume Mvery small so a discrete sum can be solved like an integral.

7.-In a very simple two state-system where there are only two states, one with energy 0 and the other with energy M��. What is the average energy of the system?


APPENDIX

21

The Gaussian integral:q[�ªen�� p

�pr«

It can be proved by evaluating the 2-D integral,q��

p

�pq��[�ª;en¬gn< p

�pq[�ªen�� p

�p�q[�ªgn�� p

�p�

Using polar coordinates,�q��

2q�z�z[�ª�n p

2

Which with the substitution ®«z�,�q��

2��s«q�®�[�h

p

2r«

..and more

Differenciate the Gaussian integral by «, as xdoes not depend on «,so ��« @[�ªen?��[�ªenand ��« @r«@?rs«��@ @so that,q��[�ªen�� p

�p�s

r«�

Repeating the process you could reach the general formula,q��¯[�ªen�� p

�ps�3�3s�¯

r«�¯¬1

As all these functions are even, the integral 0 to ©is 1�of the one from ?©to ©

To integrate ��¯¬1[�ªenfrom ?©to ©is easy as the function is odd and so the integral is zero. To integrate from 0 to ©, start off with o�[�ªen�� p

2, which can be evaluated by noticing that �[�ªenis almost what you get when you differentiate [�ªen. The following s�>�powers differentiate with respect to «as before

q��¯¬1[�ªen�� p

2�3s«¯¬1

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T1_Kinetic Molecular Theory