lecture
TRANSCRIPT
• 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