• The concept of an atom as an indivisible

component of matter goes back to Indian and

Greek philosophers. The word comes from

átomos (Greek: ἄτομος), which means

"uncuttable“.

• The idea of atoms was used in 17th and 18th

century to explain chemical properties of various

substances.

• However, people thought that even if

atoms really existed they were far too

small to see even under the most powerful

microscopes. Speculating about them was

not really of much use.

• Thus, their existence was debatable till the

mid 19th century.

However, somewhere around this time (mid 19th

century) whether or not the atom is real became a

question of utmost importance.

REASON?

Image Sources:

http://campus.udayton.edu/~hume/Steam/Tacoma2.jpg

http://www.twmuseums.org.uk/engage/blog/wp-content/uploads/2013/06/Turbiniaspeedblog4-300x235.jpg

http://www.crudem.org/wp-content/uploads/2011/09/Steam-Engine-Rum-Factory-2144-x-1424.jpg

The reason was STEAM & Steam powered

machines: Industrial revolution.

It became necessary to understand the dynamics

of the constituents of steam (or a gas in general)

so as to improve the machinery and hence

efficiency.

Image Source: http://i1-news.softpedia-static.com/images/news2/Steam-Now-Produced-from-Almost-Freezing-Water-2.jpg?1354010321

Ludwig Eduard Boltzmann

(1844-1906)

Ludwig Eduard Boltzmann paved the way for

development of Statistical Mechanics, which

explains and predicts how the properties of

atoms (microscopic) determine the physical

properties (macroscopic) of matter.

Boltzmann's kinetic theory of gases seemed to

presuppose the reality of atoms and molecules.

He had to face strong opposition from the philosophers and

physicists of that era who argued that atoms were a mere

mathematical convenience rather than real physical

objects.

Image Source: http://upload.wikimedia.org/wikipedia/commons/thumb/a/ad/Boltzmann2.jpg/225px-Boltzmann2.jpg

In 1827, the botanist Robert Brown, looking

through a microscope at particles found in pollen

grains in water, noted that the particles moved

through the water but was not able to determine

the mechanisms that caused this motion.

Robert Brown(1773-1858)

Albert Einstein in 1905 successfully explained

the phenomenon by considering the atoms as

real objects and was also able to estimate their

size. Albert Einstein (1879-1955)

Image Sources:

http://upload.wikimedia.org/wikipedia/commons/thumb/3/32/Robert_Brown_%28botanist%29.jpg/220px-Robert_Brown_%28botanist%29.jpg

http://upload.wikimedia.org/wikipedia/commons/thumb/6/66/Einstein_1921_by_F_Schmutzer.jpg/220px-Einstein_1921_by_F_Schmutzer.jpg

• Gases are composed of a large number of

particles that behave like hard, spherical objects

in a state of constant, random motion.

• These particles move in a straight line until they

collide with another particle or the walls of the

container.

• These particles are much smaller than the

distance between particles. Most of the volume

of a gas is therefore empty space.

• There is no additional interaction (other than

when they collide) between gas particles or

between the particles and the walls of the

container.

• Collisions between gas particles or collisions

with the walls of the container are perfectly

elastic.

• Consider a box with a frictionless

piston filled with some gas.

• We are interested in finding out the force on the piston

due to the particles (atoms/molecules) constituting the

gas.

• This force, however, is not localized at a single point

but rather distributed over the entire area of the piston.

• A convenient way to measure it would be to talk about

force per unit area, i.e., Pressure.

𝑃 =𝐹

𝐴

Consider a particle which has a mass 𝑚 and

velocity 𝑣. If the 𝑥-component of the velocity is 𝑣𝑥,

then when the atom hits the piston (elastic

collision), this component gets reversed. The

change in momentum is

∆𝑝 = 𝑚 (−𝑣𝑥) − 𝑣𝑥 = −2𝑚𝑣𝑥.

(-ve sign represents the loss in momentum)

• Momentum delivered to the piston because of

this single collision = 2𝑚𝑣𝑥

• For simplicity let us assume that all the atoms

have the same velocity. (We will generalize this

to the case of unequal velocities soon).

• Let us consider a small time interval ∆𝑡. In this

interval, only the particles which lie within the

distance 𝑣𝑥∆𝑡 from the wall will be able to hit the

wall. Others won’t be able to reach the wall in ∆𝑡.

If 𝐴 is the area of the piston, then the particles

which lie within the volume 𝐴𝑣𝑥∆𝑡 will be able to hit

the piston.

If 𝑛 is the number of particles per unit volume,

𝑛 =𝑁

𝑉,

the number of particles that hit the wall in time ∆𝑡 is

𝑛𝐴𝑣𝑥∆𝑡

Thus, total momentum imparted to the piston in this

interval= (𝑛𝐴𝑣𝑥∆𝑡)(2𝑚𝑣𝑥).

In other words, the force on the piston,

𝐹 =(𝑛𝐴𝑣𝑥∆𝑡)(2𝑚𝑣𝑥)

∆𝑡= 2𝑛𝑚𝑣𝑥

2 𝐴.

(We are free to take the limit ∆𝑡 → 0, and still we

get the same result).

The pressure, therefore, is

𝑃 =𝐹

𝐴= 2𝑛𝑚𝑣𝑥

2.

• Now let us generalize to arbitrary velocities for

the particles. However, we are considering

identical particles, so masses are same for all.

• For that we need to replace 𝑣𝑥2 by the average

velocity in the x-direction. So,

𝑣𝑥2 →

1

2𝑣𝑥

2 .

• The factor of half has to be introduced because

𝑣𝑥2 counts contribution from both 𝑣𝑥 and −𝑣𝑥 ,

whereas we are focusing on 𝑣𝑥 only.

Thus,

𝑃 = 2𝑛𝑚𝑣𝑥2 → 𝑛𝑚 𝑣𝑥

2 .

But now there is nothing special about the x-direction, we

might as well consider y and z directions. Since there is no

preferred direction for the particles, for the averages we must

have

𝑣𝑥2 = 𝑣𝑦

2 = 𝑣𝑧2

Now if 𝑣2 is the velocity squared of the particles (in general

different for all), then 𝑣2 = 𝑣𝑥2 + 𝑣𝑦

2 + 𝑣𝑧2.

Image source: http://www.chem.ufl.edu/~itl/2041_f97/matter/FG10_014.GIF

Take the average,

𝑣2 = 𝑣𝑥2 + 𝑣𝑦

2 + 𝑣𝑧2 = 3 𝑣𝑥

2 .

Thus, finally we have

𝑃 = 2𝑛𝑚𝑣𝑥2 → 𝑛𝑚 𝑣𝑥

2 →1

3𝑛𝑚 𝑣2 .

𝑣2 is the mean-square velocity, i.e.,

𝑣2 =1

𝑁 𝑣𝑗

2 =1

𝑁 𝑣𝑗,𝑥

2 + 𝑣𝑗,𝑦2 + 𝑣𝑗,𝑧

2𝑁𝑗=1

𝑁𝑗=1 .

Here 𝑁 represents the total number of particle and

𝑗 the 𝑗th particle.

Now

𝑃 =1

3𝑛𝑚 𝑣2 =

2

3𝑛

1

2𝑚𝑣2 .

Clearly 1

2𝑚𝑣2 represents the average kinetic

energy for the particles. In other words, the kinetic

energy of the center of mass motion. Now using

𝑛 =𝑁

𝑉, we obtain

𝑃𝑉 =2

3𝑁

1

2𝑚𝑣2 .

𝑃𝑉 =2

3𝑁

1

2𝑚𝑣2

This is a truly remarkable relation. It relates the

average of microscopic property, the velocity of the

gas particles, to the macroscopic observable, the

pressure exerted by the gas on the piston/wall.

For a monatomic gas, e.g. Helium or Argon, i.e. molecules

with just single atom in them, it is reasonable to assume

that there is no other internal motion (rotation, vibration).

Thus the kinetic energy as obtained in the previous slide

will represent the total energy. We will represent it by

𝑼, the total internal energy of the gas.

Thus we have

𝑈 = 𝑁1

2𝑚𝑣2 ,

giving

𝑃𝑉 =2

3𝑈.

We might have a situation for a gas with complex

molecules, then one will have contributions from

internal motion also, rotation, vibration etc.

Therefore for generality, we write

𝑃𝑉 = 𝛾 − 1 𝑈.

Thus for a monatomic gas like helium we have

𝛾 =5

3, giving 𝑃𝑉 = (2/3)𝑈.

Take the differential of 𝑃𝑉 = 𝛾 − 1 𝑈,

We obtain

𝑃 𝑑𝑉 + 𝑉 𝑑𝑃 = 𝛾 − 1 𝑑𝑈

Let us examine the compression of the gas when

we apply force on the piston. Assume that the

process is adiabatic: No heat energy is added or

removed. Then change in internal energy,

𝑑𝑈 = −𝐹𝑑𝑥 = −𝐹

𝐴𝐴𝑑𝑥 = −𝑃 𝑑𝑉

𝑃 𝑑𝑉 + 𝑉 𝑑𝑃 = 𝛾 − 1 (−𝑃 𝑑𝑉)

Rearranging we obtain,

𝛾𝑑𝑉

𝑉+

𝑑𝑃

𝑃= 0.

Assuming that 𝛾 is a constant, as it is for a

monatomic gas, we can integrate the above

equation. We obtain

𝛾 ln 𝑉 + ln 𝑃 = ln 𝐶

Where ln C is the constant of integration.

Exponentiating both sides we obtain

𝑃𝑉𝛾 = 𝐶 (𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡)

Consider a photon gas. We will avoid talking in

terms of mass in this case as we are dealing with a

relativistic system, and it has a very different

behavior in the relativistic domain. However,

𝐹 = 𝑑𝑝/𝑑𝑡 still holds. Redoing the analysis again

and working with 𝑝 we arrive at

𝑃 = 2𝑛 𝑝𝑥𝑣𝑥

Introducing the averaged quantities and

considering the three directions we obtain

𝑃 =1

3𝑛 𝑝 ∙ 𝑣 ⇒ 𝑃𝑉 =

1

3𝑁 𝑝 ∙ 𝑣

• The momentum 𝑝 and velocity 𝑣 are in the same

direction and so 𝑝 ∙ 𝑣 = 𝑝𝑣. Now for a photon

𝑣 = 𝑐, the speed of light.

• Thus 𝑝 ∙ 𝑣 = 𝑝𝑐. Special theory of relativity tells

that 𝑝𝑐 for a photon is actually its total energy 𝐸.

• Thus 𝑁 𝑝 ∙ 𝑣 = 𝑁𝐸 = 𝑈, the internal energy of

the photon gas. Thus

𝑃𝑉 =1

3𝑈

Comparing this with

𝑃𝑉 = 𝛾 − 1 𝑈

We conclude that 𝛾 = 4/3, and therefore the

photon gas (radiation in a box) obeys

𝑃𝑉4/3 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡

Thus we know about the behavior of the radiation.

This can be applied to radiation of hot stars!

Image Source: http://1.bp.blogspot.com/-cB8lS60iMvM/ToXwJOy3_EI/AAAAAAAAAEw/goJQbNPE8Ic/s1600/Star_5.jpg