Classical mechanics formula

8/10/2019 Classical mechanics formula

1/20

Lecture 37

Last Lecture:

Equations of state. The ideal gas.Kinetic model of a gas. (Chap. 18 Sec. 1-4)

This Lecture:Maxwell-Boltzmann distribution.

Van der Waals Equation. (Chap. 18 Sec. 5)


2/20

Last time we connected the Macroscopic

and Microscopic pictures of an ideal gas

2total translational kinetic energy of molecules

3pV

23 1

2 2kT m v Average kinetic energy of one molecule

(for monoatomic ideal gas)

23

A

1.38 10 J/KR

kN

Boltzmanns constantwhere

when combined with the ideal gas law: PV = nRT

we found:


3/20

Two separate containers of gas are in thermal equilibrium with each other.One contains He and the other contains Ar. Which of the following

statements is correct?

1. .

2. .

3. .

2 2

He Ar

2 2

He Ar

2 2

He Ar

v v

v v

v v

3 They have the same average kinetic energy.

2K E kT

21 3 The heavier mass must have slower speeds.2 2

m v kT

Clicker Question: Molecular speeds


4/20

Root-mean-square velocities

The key quantity is mean of squared velocities

2

rmsv v

Take the square-root back to the units of velocity

2

rms

1 3

2 2mv kT

Example: 3 molecules at 400,500,600 ms-1

2 2 2

400 500 600average 500 m/s

3

400 500 600rms 507 m/s

3

Not the same!


5/20

v

Mean free path

Assume molecules are rigid spheres of radius r. Only one moves, with speed v.

It will collide with another molecule when their centers are at distance 2r.

2r

vdt

ie, in dtit collides if there are other molecules with centers inside this cylinder:

2

2 N

dN r vdt V

Collision per unit time:24

dN Nr v

dt V

If all molecules are moving, a similar calculation gives24 2

dN Nr v

dt V

mean 2

1

4 2

Vt

dN r vN

dt

Average time between collisions:

Average displacement between collisions

(mean free path)mean

24 2

Vvt

r N


6/20

Molecular speeds

Not all molecules have the same speed.

If we have Nmolecules, the number of molecules with speeds

between vand v + dv is:

( )dN Nf v dv

( )f v = distribution function (probability density)

= probability of finding a molecule with speed between v

and v + dv

( )f v dv


7/20

Maxwell-Boltzmann distribution

2

3/ 2

2 /(2 )( ) 42

mv kT mf v v ekT

Maxwell-Boltzmann

distribution

higher T

higher speeds aremore probable


8/20

Distribution = probability density

= probability of finding a molecule with speed between vand v + dv( )f v dv

Normalization:

0( ) 1f v dv

2

11 2

( ) probability of finding molecule with speeds between andv

vf v dv v v

= area under the curve


9/20

Most probable speed, average speed, rms speed

2mp

kTv

m

Most probable speed (where

f(v) is maximum)

00

0

( ) 8( ) ...

( )

vf v dv kTv vf v dv

mf v dv

Average speed

2 20

3( ) ... kTv v f v dv m

Average squared speed

2

rms

3kTv v

m rms speed


10/20

Beyond an ideal gas

We used this picture to derive the

Ideal Gas Equations

But real gases are more complicated


11/20

The Van der Waals state equation

The ideal gas equation is the first approximation to the behavior of gases.Now lets start to correct the idealization of a gas.

Reduce the assumptions by including:

molecule size reduces volume to move

molecular interactions (attraction) reduces pressure

2

2

np a V nb nRT

V

a, b are determined empirically.

Different for each gas

This model is used for more extreme conditions. If the gas is diluted

(ie, n/Vis very small), the corrections are negligible.

2

21n np a V b nRT

VV


12/20

pVdiagrams

Expansion at constant

pressure

(isobaric process)

Convenient tool to represent states and transitions from one state toanother (line represents a series of thermal states)

V

p

B

VA VB

A

states

process

If we treat the helium in the

balloon as an ideal gas, we can

predict T for each state:

A/BA/B

pVT

nR

Example: helium in balloon expanding in the room and warming up


13/20

ThispVdiagram can describe:

V

p

B

pA

pB

A

A. A tightly closed container cooling

down.

B. A pump slowly creates a vacuuminside a closed container.

C. Either of the two processes

In either case, volume is constantand pressure is decreasing.

In case A, becauseT decreases.

In case B, because n decreases.

(isochoric process)

Clicker Question: Constant volume


14/20

Isothermal curves

For an ideal gas,nRT

pV

(For constant n, a hyperbola for each T;Boyles Law )

1 2 3 4

T T T T Each point in apVdiagram

is a possible state (p, V, T )

Isothermal curve = all states with the same T


15/20

A container is divided in two by a thin wall. One side contains an ideal

gas, the other has vacuum. The thin wall is punctured and disintegrates.

Which of the following is the correctpVdiagram for this process?

Initial state

Final state

2

Initial state

Final state

Initial state

Final state

3

Initial state

Final state

4

1A B

C D

Clicker Question: Free expansion


16/20

Final state has larger V, lowerp

During the rapid expansion, the gas does NOT uniformly fill Vat a uniformp

hence it is not in a thermal state.

hence no states during process

hence this process is not represented by line

Initial state

Final state

2

Initial state

Final state

Initial state

Final state

3

Initial state

Final state

4

1A B

C D


17/20

Further beyond the ideal gas

When a real gas is compressed, it eventually becomes a liquid

Decrease volume at constant

temperatureT2:

At point a, vapor begins to condense

into liquid.

Between a and b: Pressure and Tremain

constant as volume decreases, more of

vapor converted into liquid. At point b, all is liquid. A further

decrease in volume will require large

increase inp.


18/20

The critical temperature

For T >> Tc, ideal gas.

critical temperature = highest temperature where a

phase transition happens.

T

p

solidliquid

gas

Triple

point

Critical point

Supercritical fluid

Critical point for water: 647K and 218 atm


19/20

pVT diagram: Ideal gas

States are points on

this surface.


20/20

pVT diagram: Water

Phase transitions

appear as angles.

Classical mechanics formula

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