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Thermodynamics Class 3

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Previous class: the microscopic view of ideal gas “Temperature” !1, = . 3/2 kT = , proportional to the average molecule kinetic energy !2, Ideal gas law, PV=µRT !3, Van der Waals law for real gas: [P+a(µ/V)2][V-µb]=µRT. !4, dP = - nmg dh => n = n0 exp(- P.E. / kT) Boltzmann’s law !

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Questions from the last class: !a): = 0? = sum_{ijk} wi vcmj cosθk / N. θk is independent of wi and vcmj. And

sum_k cosθk = 0.

!

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The distribution of molecular speeds

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It’s important to know the velocity distribution. For example, we can calculate the average speed, thermal velocity, etc from the distribution function. !Now let’s consider the distribution of molecules in an atmosphere. !At height h, the number of molecules that pass upward though the lower plane in a vertical direction with velocity component greater than u is exactly the same as the number which pass through the upper plane with any upward velocity. !nvz>=0(h) = nvz>=u(0), with 1/2mu2 = mgh !So nvz>=u(0) / nvz>0(0) = exp(-1/2 mu2/kT) or !n >u is proportional to exp(-K.E./kT) ! Z 1

uuf(u)du = C exp(�1/2mu2/kT )

f(u)du = C exp[�(1/2)mu2/kT ]

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The distribution of molecular speeds (cont’d)

To get the normalisation coefficient, we have C = sqrt(m/2pi kT), using

Z 1

�1e

�x2dx =p⇡

So the distribution of velocity is f du = C exp(-K.E./kT). !If written in momentum which is also valid in relativity, f(p) dp = C exp(-K.E./kT) dp. !Note that both the spatial and the velocity distribution has the form of exp(-energy / kT). !Generalize to three dimensions, we have

f(vx

, vy

, vz

)dvx

dvy

dvz

/ e�mv2x

/2kT e�mv2y

/2kT e�mv2z

/2kT

A more formal way of showing the velocity distribution in thermal equilibrium.

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Maxwell assumed that in thermal equilibrium, f is a function of v only. f(vx), f(vy) and f(vz) are independent of each other. We can derive the form of f(vx, vy, vz) like this. !f(v) is independent of the direction of v.

f(vx

, vy

, vz

) = f(v2) = f(vx

)f(vy

)f(vz

)

@f(v2)

@vx

=df(v2)

dv2@v2

@vx

= 2vx

df(v2)

dv2=

df(vx

)

dvx

f(vy

)f(vz

)

df(v2)

dv2=

df(vx

)

dv2x

f(vy

)f(vz

)

1

f(v2)

df(v2)

dv2=

1

f(vx

)

df(vx

)

dv2x

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1

f(v2)

df(v2)

dv2=

1

f(vx

)

df(vx

)

dv2x

= ��

df(vx

)

f(vx

)= ��dv2

x

f(vx

) = C1e��v2

x

f(v) = Ce��(v2x

+v2y

+v2z

)

beta can be obtained from the definition of temperature.

C =

✓�

⇡

◆3/2Z 1

�1f(v)dv = 1

Degrees of freedom

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The degrees of freedom: the minimum number of independent coordinates that can specify the position of the system completely. !Monoatomic molecules: H, O, 3 degrees of freedom Diatomic molecules: 3 translational motion (x,y,z) + 2 rotational motion (ignoring the rotation along the line connecting the two atoms) + 1 vibrational motion (the distance between two atoms) = 6 degrees of freedom !In general, for a n-atom molecule: 3n degrees of freedom. Each atom needs three coordinates. !In these 3n: 3 translational motion of the molecule as a whole. 3 rotational motion as a whole if not all the atoms are collinear. So there are 3n-6 degrees of freedom for vibrational motion. !If all the atoms are collinear (e.g., a diatomic molecule), then there are 2 rotational degrees of freedom. and 3n-(3+2) = 3n-5 degrees of vibrational freedom.

Equipartition of energy (Equipartition Theorem)

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Suppose the kinetic energy of a molecule is K.E. =NX

i=1

1

2mv2i

f(v1, v2, v3, ..., vN ) / exp(�K.E./kT ) =NY

i=1

exp(�12

mv2i /kT )

exp(�12mv

2i

kT) as exp(��v2i ) with � =

12m

kTwrite

r�

⇡

Z 1

�1

1

2

mv2i exp(��v2i )dvi =1

2

kT

Z 1

�1x

2exp(��x2)dx = 1

2

r⇡

�

3

Z 1

�1exp(��x2)dx =

r⇡

�

So we have that the average kinetic energy of every degree of freedom is equal to 1/2 kT.

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For vibrational motion, we also have potential energy. If it’s a harmonic oscillator, then = !For a monoatomic gas: we have U = N (3/2) kT (3 degrees of translational motion. !For a diatomic gas: we have U = N * (3/2 + 2/2 + 1/2 + 1/2) kT = N 7/2 kT. !PV = NkT = (ɣ-1)U !So for a monoatomic gas: NkT = 2/3 U => ɣ = 5/3 ~ 1.667 For a diatomic gas: NkT = 2/7 U => ɣ = 9/7 ~ 1.286.

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Fig. 40–6.Experimental values of γ as a function of temperature for hydrogen and oxygen. Classical theory predicts γ=1.286, independent of temperature.

Failure of classical physics

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For diatomic gases like H2 or O2, we assumed that the vibrational motion is like a harmonic oscillator. What if that’s not true? If the coupling between two atoms is very rigid, and there is no vibration, then we will have U = (3/2+2/2) kT = 5/2 kT and ɣ = 7/5 =1.4。This looks quite close to experimental results. But ɣ varies with temperature for both H2 and O2. !It seems like as the temperature falls, certain kinds of motion “freeze out”. !This represents the failure of classical physics.

The resolution is in Quantum Mechanics. !In QM, the energy of a harmonic oscillator is not continuous, but is quantized. While in classical physics, it is continuous. !For a harmonic oscillator, its energy levels are

PnP0

=e�En/kT

e�E0/kT= e�n~!/kT =

nnn0

Failure of classical physics (cont’d)

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n = n0e�E/kT

with E varies from 0 to 1

hEi =R10 EndER10 ndE

= kTSo that

kT ⌧ ~!If , then all atoms are in the lowest energy level.

The contribution of the harmonic oscillator to the total energy of the molecule is negligible.