Monday, September 16, 2019

Unizor - Physics4Teens - Energy - Gravitational Field - Problems





Notes to a video lecture on http://www.unizor.com

Problems on Gravity

Problem 1
Gravitational potential of a spherical gravitational field around a point-mass M at a distance r from it is defined as the work performed by gravity to bring a probe object of a unit mass from infinity to this point and is expressed as
Vr = −G·M /r
Why is this formula independent of trajectory of a probe object or its exact final position relative to the point-mass M, but only on a distance itself from the source of gravity?

Solution
Any movement can be represented as infinitely many infinitesimal displacements, combined together into a trajectory.
In our three-dimensional world the force and an infinitesimal displacement of a probe object are vectors, so the infinitesimal work dW performed by the force of gravity F during the movement of a probe object, described by the infinitesimal displacement dS, is a scalar product of these two vectors:
dW = F·dS
Note that the vector of gravitational force F is always directed towards the source of gravity.
Since a displacement vector dS can be represented as a sum of radial (towards the source of gravity) dSr and tangential (perpendicular to radius) dSt components, the above expression for a differential of work can be written as
dW = F·(dSr + dSt) =
F·
dSr + F·dSt
The second component in the above expression is a scalar product of two perpendicular vectors and is equal to zero. That's why we can completely ignore tangential movements, when calculating the work done by a central gravitational field, as not contributing to the amount of work. The total amount of work will be the same as if our probe object moved along a straight line towards the source of gravity and stopped at a distance r from it.

Problem 2
Given two point-masses of mass M each, fixed at a distance 2R from each other.
Prove that the gravitational potential of a gravitational field produced by both of them at each point on a perpendicular bisector between them equals to a sum of individual gravitational potentials of these point-masses at this point, as if they were the only source of gravitation. In other words, prove that gravitational potential is additive in this case.

Solution
Let's draw a diagram of this problem (you can download it to display in a bigger format).

Our two point-masses are at points A and B, the probe object is at point D on a perpendicular bisector of a segment AB going through point C.
The force of gravity towards point A is a segment DE, the force of gravity towards point B is a segment DF.
We will calculate the potential of a combined gravitational field of two point-masses at point D, where the probe object is located.
Let's assume that the segment CD equals to h.
The magnitude of each gravitational force equals to
F = G·M·m /(h2+r2)
Represent each of these forces as a sum of two vectors, one (green on a drawing) going vertically along the bisector CD, another (red) going horizontally parallel to AB.
Vertical components of these two forces will add to each other, as equal in magnitude and similarly directed downwards on a drawing, while horizontal ones will cancel each other, as equal in magnitude and opposite in direction to each other. So, the combined force acting on a probe object is a sum of vertical components of gravitational forces with a magnitude
Ftot = 2·G·M·m·sin(φ)/(h2+r2)
Since sin(φ) = CD/AD,
sin(φ) = h /[(h2+r2)1/2]
Ftot = 2·G·M·m·h /(h2+r2)3/2
If the gravitational field pulls a probe object along the perpendicular bisector of a segment AB from infinity to a distance h from the segment, the magnitude of a combined force of gravity, as a function of a distance from the segment x is changing, according to a similar formula:
Ftot(x) = 2·G·M·m·x /(x2+r2)3/2
To calculate work performed by a gravitational field pulling a probe object from infinity to height h above the segment AB, we have to integrate
Wtot = [∞;h]Ftot(x)·dx
It's supposed to be negative, since the direction of a force is opposite to a positive direction of the coordinate axis, we will take it into account later.
Wtot = 2GMm·x·d/(x2+r2)3/2
(within the same limits of integration [∞;h])
This integral can be easily calculated by substituting
y=x2+r2,
2·x·dx = dy,
infinite limit of integration remaining infinite and the x=h limit transforming into y=h2+r2. Now the work expression is
Wtot = G·M·m·y−3/2·dy
with limits from y=∞ to y=h2+r2.
The indefinite integral (anti-derivative) of y−3/2 is −2·y−1/2.
Therefore, the value of integral and the work are
Wtot = −2·G·M·m·(h2+r2)−1/2
For a unit mass m=1 this work is a gravitational potential of a combined gravitational field produced by two point-masses on a distance h from a midpoint between them along a perpendicular bisector
Vtot = −2·G·M·(h2+r2)−1/2
At the same time, the gravitational potential of a field produced by each one of the point-masses, considered separately, equals to
Vsingle = −G·M·(h2+r2)−1/2
As we see, the gravitational potential of two point-masses equals to a sum of gravitational potential of each of them, considered separately.
IMPORTANT NOTE
With more cumbersome calculations this principle can be proven for any two (not necessarily equal) point-masses at any point in space (not necessarily along the perpendicular bisector). This principle means that gravitational potential is additive, that is the gravitational potential of any set of objects at any point in space equals to sum of their individual gravitational potentials.

Problem 3
Express mass M of a spherical planet in terms of its radius R and a free fall acceleration g on its surface.

Solution
Let m be a mass of a probe object lying on a planet's surface.
According to the Newton's 2nd Law, its weight is
P = m·g
According to the Universal Law of Gravitation, the force of gravitation between a planet and a probe object is
Fgravity = G·M·m /R2
Since the force of gravitation is the weight Fgravity = P,
m·g = G·M·m /R2
from which
M = g·R2 /G

Problem 4
Express gravitational potential VR of a spherical planet on its surface in terms of its radius R and a free fall acceleration g on its surface.

Solution
From the definition of a gravitational potential on a distance R from a source of gravity
VR = −G·M /R
Using the expression of the planet's mass in terms of its radius R and a free fall acceleration g on its surface (see above),
M = g·R2 /G
Substituting this mass into a formula for potential,
VR = −G·g·R2 /(G·R) = −g·R


Tuesday, September 10, 2019

Unizor - Physics4Teens - Energy - Energy of Gravitational Field - Gravitational Field



Notes to a video lecture on http://www.unizor.com

Gravitational Field

Studying forces, we have paid attention to a force of attraction, that exists between any material objects, the force of gravity.
For example, if a comet from outer space flies not far from a Sun, it is attracted by Sun and changes its straight line trajectory.

In Mechanics we used to see the force as something between the objects touching each other, like a man pushing a wagon. In case of gravity the force obviously exists, but it acts on a distance, in "empty" space.
In Physics this concept of force acting on a distance is described by a term field. Basically, field is the area in space where some force acts on all objects or only on objects that have specific property. The force in this case depends on a point in space and an object that experiences this force and, as a result of the action of force, changes its movement.

Gravitational field exists around any material object (the source object of a field) and acts as an attraction towards this source object, experienced by any other material object (probe object) positioned in this field.
As described in the "Gravity, Weight" chapter of "Mechanics" part of this course, the magnitude of the gravitational force F is proportional to a product of masses of a source object and a probe object, M and m, and it is inversely proportional to a square of a distance r between these objects:
F = G·M·m /
where G - a constant of proportionality, since the units of force (N - newtons) have been defined already, and we want to measure the gravitational force in the same units as any other force.

The direction of the gravitational force acting on a probe object is towards the source object.

Let's return to our example of a comet flying not far from the Sun and, being attracted to the Sun, changing its trajectory. Obviously, to change the trajectory, some energy must be spent. So, we conclude that gravitational field has certain amount of energy at each point that it spends by applying the force onto a probe object.

To quantify this, assume that the source of gravity is a point mass M fixed at the origin of coordinates. Position a probe object of mass m at coordinates {r1,0,0} and let it go. The force of gravity will cause the motion of this probe object towards the center of gravity, the origin of coordinates, so the movement will be along the X-axis. Let the ending position of the probe object be {r2,0,0}, where r2 is smaller then r1. Let x be a variable X-coordinate (distance to the origin).

According to the Universal Law of Gravitation, the force of attraction of a probe object towards the source of a gravitational field at distance x from the origin equals to
F = −G·M·m /
where minus in front of it signifies that this force is directed opposite to increasing the X-coordinate.
This force causes the motion and, therefore, does some work, moving a probe object from point {r1,0,0} to point {r2,0,0} along the X-axis. To calculate the work done by this variable force, we can integrate dx from x=r1 to x=r2:
W[r1,r2] = [r1,r2]dx =
= −[r1,r2]G·M·m·
dx /x² =
= G·M·m /x
|[r1,r2] =
= G·M·m /r2 − G·M·m /r1 =
= (G·M /r2 − G·M /r1
)·m

The expression
V(r) = −G·M/r
is called gravitational potential.
It's a characteristic of a gravitational field sourced by a point mass M at a distance r from a source.
It equals to work needed by external forces to bring a probe object of mass m=1 to a point at distance r from a source of the field from infinity.
Indeed, set m=1, r1=∞ and r1=r in the above formula for work W[r1,r2] and take into consideration that gravitational field "helps" external forces to move a probe object, so the external forces spend negative amount of energy.

Using this concept of gravitational potential V(r), we can state that, to move a probe object of a unit mass from distance r1 relative to a source of gravitational field to a distance r2 relative to its source in the gravitational field with gravitational potential V(r), we have to spend the amount of energy equal to V(r1)−V(r2).
For a probe object of any mass m this amount should be multiplied by m.
If r2 is greater than r1, that is we move a probe object further from the source of gravity, working against the gravitational force, this expression is positive, we have to apply effort against the force of gravity. In an opposite case, when r2 is smaller than r1, that is we move closer to a source of gravity, the gravitational force "helps" us, we don't have to apply any efforts, and our work is negative.

Therefore, an expression EP=m·V(r) represents potential energy of a probe object of mass m at a distance r from a source of a gravitational field with gravitational potential V(r).

A useful consequence from a concept of a gravitational potential is that the force of gravity can be expressed as the derivative of the gravitational potential.
F = G·M·m /r² = m·dV(r)/dr
which emphasizes the statement that the gravitational potential is a characteristic of a field itself, not its source.
We, therefore, can discuss gravitational field as an abstract concept defined only by the function called gravitational potential.

Sunday, August 18, 2019

Unizor - Physics4Teens - Energy - Energy of Nucleus - Fusion



Notes to a video lecture on http://www.unizor.com



Nucleus Fusion



Fusion is a nuclear reaction, when light nuclei are brought together and combined into a heavier ones.

The reason for this reaction to release the energy is the difference
between amount of energy needed to overcome the repulsion between nuclei
because they have the same positive electric charge (this energy is
consumed by fusion) and the potential energy released by strong forces, when the formation of a combined nucleus occurs (this energy is released by fusion).

The former is less than the latter.



When the light nuclei are fused into a heavier one, the excess of potential energy of strong forces, released in the process of fusion,
over the energy needed to squeeze together protons against their
repulsion is converted into thermal and electromagnetic field energy.



Analogy to this process can be two magnets separated by a spring.


The magnets represent two separate protons, the magnetic force of attraction between them represents the strong force
that is supposed to hold the nucleus together, when these particles are
close to each other, the spring represents the electrical repulsive
force between them, acting on a larger distance, as both are positively
charged.

It's known that magnetic force is inversely proportional to a square of a
distance between objects, while the resistance of a spring against
contraction obeys the Hooke's Law and is proportional to the length of
contraction.

On the picture magnets are separated. To bring them together, we have to
spend certain amount of energy to move against a spring that resists
contraction. But the magnetic attraction grows faster then the
resistance of the spring, so, at some moment this attraction will be
greater than the resistance of a spring. At this moment nothing would
prevent magnets to fuse.



As is in the above analogy, if we want to fuse two protons, we have to bring them together sufficiently close for strong forces to overtake the repulsion of their positive charges.



Consider the following nuclear reaction of fusion.

One nucleus of hydrogen isotope deuterium 1H2 with atomic mass 2 contains one proton and one neutron.

One nucleus of hydrogen isotope tritium 1H3 with atomic mass 3 contains one proton and two neutrons.

If we force these two nuclei to fuse, they will form a nucleus of helium 2He4 and releasing certain amount of energy:

1H2 + 1H3 = 2He4 + 0n1



It's not easy to overcome the repulsion of protons. High temperature and
pressure, like in the core of our Sun, are conditions where it happens.
On Earth these conditions are created in the nuclear bomb, using the
atomic bomd to achieve proper amount of heat and pressure, thus creating
an uncontrlled fusion.

Controlled nuclear reaction of fusion is what scientists are working on right now. So far, it's still in the experimental stage.

Monday, August 12, 2019

Unizor - Physics4Teens - Energy - Energy of a Nucleus - Fission



Notes to a video lecture on http://www.unizor.com

Nucleus Fission

Fission, first of all, is a nuclear reaction, when heavier nuclei are split into lighter ones.
The reason for this reaction to release the energy is the difference between amount of energy needed to break strong forces that hold the nucleus together (this energy is consumed by fission) and amount of potential energy in positively charged and repelling protons inside nucleus (this energy is released by fission).
The former is less than the latter.

When the heavy nucleus is broken into parts, the excess of potential energy of squeezed together protons against their repelling force over the energy of strong forces that keep nucleus together is converted into thermal and electromagnetic field energy.

Analogy of this is a spring squeezed tightly and held in this position by a thread. A thread plays the role of strong forces, while a potential energy of a squeezed spring plays the role of protons kept close to each other by a this force. When you cut a thread, the spring will release the potential energy, similarly to protons repelling from each other.

Electrically positively charged protons repel each other and, at the same time, are bonded together by strong forces inside a nucleus. At the same time neutrons are also bonded by strong forces among themselves and with protons without any repulsion.
So, the more neutrons the nucleus has - the stronger it is. Neutrons only add "bonding material" to a nucleus without adding any repelling forces that work against the nucleus' stability.

Uranium-238 with 92 protons and 146 neutrons (92U238) naturally occurs on Earth and is relatively stable.
Uranium-235 with the same 92 protons and 143 neutrons (92U235) has less "bonding material" (less neutrons) and is more susceptible to fission.

All it takes to break the nucleus of 92U235 is a little "push" from outside, which can be accomplished by bombarding it with neutrons. In the process of fission, caused by hitting a nucleus of 92U235 with a neutron, it can transforms into Barium-141 with 56 protons and 85 neutrons 56Ba141, Krypton-92 with 36 protons and 56 neutrons 36Kr92 and 3 free neutrons.
As we see, the numbers of protons is balanced (input: 92, output: 56 and 36), as well as a number of neutrons (input: 1 free hitting neutron and 143 in a nucleus of 92U235 total 144, output: 85 in a nucleus 56Ba141, 56 in a nucleus of 36Kr92 and 3 new free neutrons total 144).

Let's express this reaction in a formula (letter n denotes a neutron):
0n1 + 92U235 =
56Ba141 + 36Kr92 + 3·0n1


What's interesting in this reaction is that it not only produces energy because we break a heavy nucleus into lighter ones, but also that it produces 2 new neutrons that can bombard other atoms, causing a chain reaction and, potentially, an explosion (atomic bomb). However, if we absorb extra neutrons, it will allow to slowly release of nuclear energy (nuclear power stations).

Monday, August 5, 2019

Unizor - Physics4Teens - Energy - Energy of a Nucleus



Notes to a video lecture on http://www.unizor.com



Energy of Nucleus



In this lecture we will analyze the energy aspect of nucleus - the central part of an atom.



By now we have built a pyramid of energy types, related to the depth of our view inside the matter.



First, we analyzed the mechanical energy - the energy of moving macro-objects.



Our next view deep into the world of macro-objects uncovered the molecules - the smallest parts of macro-objects that retain their characteristics. The movement of these molecules was the source of thermal energy, which we often call the heat.



Next step inside the molecules uncovered atoms, as the molecules'
components. There are about 100 types of atoms and their composition
inside the molecules creates all the thousands of different molecules. Chemical reactions
change the composition of atoms in molecules, thereby creating new
molecules from the atoms of old molecules. This process broke some
inter-atomic bonds and created the new ones and is the source of chemical energy.



Now we look deep inside the atoms and find there 3 major elementary particles - electrically positively charged protons and electrically neutral neutrons inside a small but heavy nucleus and electrically negatively charged electrons,
circulating around nucleus on different orbits. For electrically
neutral atoms the numbers of protons and electrons are equal. Nuclear energy is hidden inside the nucleus and is the subject of this lecture.



The first question we would like to answer is "What holds nucleus, its
protons and neutrons, together, considering protons, as electrically
positively charged particles must repel each other?"



The answer is simple. There are other forces in the Universe, not only
electrostatic ones, that act in this case. These intra-nucleus forces
that hold the nucleus together are called strong forces. They are strong
because they are the source of attraction between the protons that is
stronger than electrostatic repelling. However, these strong forces act
only on a very small distance, comparable to the size of a nucleus
inside an atom. For example, at a distance 10−15m the strong force is more than 100 times stronger than electrostatic one.



If, by regrouping protons and neutrons, we will be able to create different atoms (inasmuch as regrouping atoms in chemical reaction we create new molecules), a new source of energy, based on strong forces, the nuclear energy, can be uncovered in the course of nuclear reaction.



There is another form of nuclear reaction related to
transformation of elementary particles. Under certain circumstance a
neutron inside a nucleus can transform into proton and, to keep the
total electrical charge in balance, it emits an electron. This reaction
is called beta-decay and it also produces energy in the form of electromagnetic waves of very high frequency (gamma-rays).



Nuclear reactions are a very powerful source of nuclear energy, which is
so much more powerful than other types of energy, that, if misused, it
might represent a danger for life on our planet.



There is a clear analogy between nuclear and chemical reactions.

What happens with atoms in the chemical reaction, happens with protons
and neutrons in nuclear reaction. Some atomic bonds break in a chemical
reaction, some are created. Some nuclear bonds between protons and
neutrons break in a nuclear reaction, some are created.



Sometimes the chemical reaction happens by itself, as long as
participating substances are close together, but sometimes we have to
initiate it, like lighting methane gas with a spark or a flame of a
match to initiate continuous burning.

Similar approach is valid for nuclear reaction. Sometimes it happens by
itself, but sometimes it should be started, like bombarding the nucleus
with neutrons, after which it continues by itself.



Here is an interesting fact.

Physicists have measured the masses of protons, neutrons and many
different nuclei that contain these protons and neutrons and have
discovered that the sum of masses of individual protons and neutrons is
greater than the mass of a nucleus that contain these exact particles.

For example,

mass of proton is 1.0072766 atomic mass units or 1.6726·10-27kg,

mass of neutron is 1.0086654 atomic mass units or 1.6749·10-27kg.

At the same time, mass of deuterium nucleus, that contains 1 proton and 1
neutron is 2.0135532 atomic mass units, which is smaller than the sum
of masses of proton and neutron (1.0072766 + 1.0086654 = 2.015942).

This so-called "mass defect" is directly related to nuclear energy - the energy of strong forces that hold the nucleus together.



A simplified explanation of this effect is based on the law of energy
conservation. Consider the force of gravity between a planet and an
object above its surface. The object has certain potential energy and,
if dropped to the ground, this potential energy transforms into other
forms, like kinetic, thermal etc.



Similarly, if we consider two independent neutrons (or neutron and
proton, or two protons) on a very small distance from each other, but
not forming a nucleus, there is a potential energy of the strong forces
acting between them. If we let these two particles to form a nucleus,
analogously to an object falling towards the surface of a planet, this
potential energy should be transformed into other forms, like thermal.



Now the Theory of Relativity comes to play, that has established the equivalence of mass and energy by a famous formula E=m·c².
According to this equivalence, if some energy is released during the
formation of a nucleus from individual protons and neutrons, there must
be certain amount of mass released associated with this energy. That is
the explanation of "mass defect".



It should be noted that to form a nucleus of deuterium from 1 proton and
1 neutron is easier than to form a nucleus that contains more than one
proton, because electrostatic repulsion between positively charged
protons prevents their bonding. So, to bring protons sufficiently close
to each other for strong forces to overcome the electrostatic
repulsion, we have to spend some energy. The net energy released by
forming a nucleus from protons and neutrons is the difference between
the energy released from strong forces taking hold of these particles inside a nucleus and the energy consumed to overcome repulsion of protons.



Actually, as we attempt to form bigger nuclei, the energy we have to
spend to overcome electrostatic repulsion forces become greater than
amount of energy released by forming a nucleus. This border line is
approximately around the nucleus of iron Fe. Forming iron
and heavier elements from protons and electrons is a process that
consumes more energy than releases. These heavier nuclei will produce
energy, if we reverse the procedure, breaking them into individual
protons and neutrons.



The mechanisms described above are used in nuclear reactors and atomic
bomb, where heavier elements are broken into lighter ones (fission),
releasing energy, and in hydrogen bomb, where lighter elements are
bonded together to release the energy (fusion).

Monday, July 29, 2019

Unizor - Physics4Teens - Energy - Atoms and Chemical Reactions - Interat...





Notes to a video lecture on http://www.unizor.com



Interatomic Bonds



Atoms in a molecule are bonded together to form a stable chemical substance or compound.

The mechanism of bonding is quite complex and different for different
molecules. In fact, the complexity of these bonds is outside of the
scope of this course. However, certain basic knowledge about molecular
bonding and molecular structure is necessary to understand the following
lecture, where we will make certain calculations related to energy
produced or consumed in chemical reactions.



The key to a mechanism of bonding atoms into molecules lies in an internal structure of atoms.

For our purposes we can consider the orbital model of atom as consisting
of electrically positive nucleus and electrically negative electrons
circulating on different orbits around a nucleus. This is only a model,
not an exact representation of what's really happening inside the atom,
but this model gives relatively good results that correspond to some
simple experiments.



Two different particles can be found in a nucleus - positively charged
protons and electrically neutral neutrons. The number of protons inside a
nucleus and electrons circulating on different orbits around a nucleus
should be the same for electrically neutral atoms in their most common
state.



For reasons not well understood by many physicists, each orbit can have
certain maximum number of electrons that can circulate on it without
"bumping" into each other. The higher the orbit - the more electrons it
can hold. The lowest orbit can hold no more than 2 electrons, the next -
no more than 8, the next - no more than 14 etc.



Consider a few examples.



1. Let's consider the structure of a simplest molecule - the molecule of
hydrogen, formed by two atoms of hydrogen. Each hydrogen atom has one
electron on the lowest orbit around a nucleus. The maximum number of
electrons on this orbit is two, in which case the compound becomes much
more stable. So, two atoms of hydrogen grab each other and the two
electrons, each from its own atom, are shared by a couple of atoms, thus
creating a stable molecule of hydrogen with symbol H2. The bond between two atoms of hydrogen is formed by one pair of shared electrons, so structurally the molecule of hydrogen H2 can be pictured as

H−H.



2. Atom of oxygen has 8 electrons - 2 on the lowest orbit and 6 on the
next higher one. The next higher orbit is stable when it has 8
electrons. So, two atoms of oxygen are grabbing each other and share 2
out of 6 electrons on the outer orbit with another atom. So, each atom
has 4 "personal" electrons, 2 electrons that it shares with another atom
and 2 electrons that the other atom shares with it. Thus, the orbit
becomes full, all 8 spots are filled. The bond between two atoms of
oxygen is formed by two pairs of shared electrons, so structurally the
molecule of oxygen O2 can be pictured as

O=O

(notice double link between the atoms).



3. Our next example is gas methane. Its molecule consists of one atom of
carbon (6 electrons, 2 of them on the lowest orbit, 4 - on the next
one) and 4 atoms of hydrogen (1 electron on the lowest orbit of each
atom). Obviously, having only 4 electrons on the second orbit, carbon is
actively looking for electrons to fill the orbit. It needs 4 of them to
complete an orbit of 8 electrons. Exactly this it finds in 4 atoms of
hydrogen that need to complete their own lowest orbit. Sharing
electrons, one atom of carbon and 4 atoms of hydrogen fill their
corresponding orbits, thus creating a molecule of methane CH4 with can be pictured as

     H

      |

H−C−H

      |

     H




4. Carbon dioxide molecule contains 1 atom of carbon, that needs 4
electrons to complete its orbit, and 2 atoms of oxygen, each needs 2
electrons to complete its orbit: CO2. By
sharing 2 electrons from each atom of oxygen with 4 electrons from atom
of carbon they all fill up their outer orbit of electrons and become a
stable molecule, pictured as

O=C=O

(notice double link between the atoms).



5. Ethanol molecule contains 2 atoms of carbon, 1 atom of oxygen and 6 atoms of hydrogen connected as follows

     H   H

      |     |

H−C−C−O−H

      |     |

     H   H


(notice single bond between atoms of carbon and oxygen in ethanol, while
the bond between them in carbon dioxide has double link)



6. Hydrogen peroxide molecule contains 2 atoms of hydrogen and 2 atoms of oxygen connected as follows

H−O−O−H

(notice single bond between atoms of oxygen, not like in a molecule of oxygen)



Numerous examples above illustrate that bonds between atoms can be
different, even between the same atoms in different molecules. That's
why it is important to understand the structure of molecules, how
exactly the atoms are linked and what kind of links exist between them.
This is the basis for calculation of the amount of energy produced or
consumed by chemical reactions that rearrange the atoms from one set of
molecules to another.



Obviously, bonds O−O and O=O are different.
The first one is facilitated by one shared electron, the second one - by
two. The amounts of energy, needed to break these bonds, are different
too. Therefore, when calculating the energy of chemical reaction, it's
important to understand the kind of bond between atoms in each separate
case.

Tuesday, July 23, 2019

Unizor - Physics4Teens - Energy - Chemical Energy of Atomic Bonds







Notes to a video lecture on http://www.unizor.com



Energy of Atomic Bonds

in Molecules




In this lecture we will analyze the energy aspect of chemical reactions.

Consider the reaction of burning of methane. This gas is used in regular
gas stoves, so the reaction happens every time we cook something.

A molecule of methane consists of one atom of carbon C and four atoms of hydrogen H, the chemical formula of methane is CH4.

You can imagine a molecule of methane as a tetrahedron, in its center is
an atom of carbon and on each of its four vertices is an atom of
hydrogen.

A molecule of oxygen, as we know, consists of two atoms of oxygen and has a chemical formula O2.



As a result of the reaction of burning of methane, water and carbon dioxide are produced, according to the following equation:

CH4 + 2O2 = 2H2O + CO2

So, during this reaction

(a) four atomic bonds between carbon and hydrogen in one molecule of methane are broken,

(b) one atomic bond in each molecule of oxygen (out of two) are broken,

(c) two atomic bonds between hydrogen and oxygen in each molecule of water (out of two) are created,

(d) two atomic bonds between carbon and oxygen in a molecule of carbon dioxide are created.



Amounts of potential energy of the different atomic bonds are
experimentally determined, which would lead to calculation of the amount
of chemical energy released (for exothermic) or consumed (by
endothermic) reaction.



To make experiments to determine potential energy of the bonds inside a
molecule, we have to make experiments with known amounts of components
in chemical reaction. The reaction above includes one molecule of
methane and two molecules of oxygen. Obviously, we cannot experiment
with one or two molecules. The solution is to experiment with proportional amounts of components, say, 1 million of molecules of methane and 2 million of molecules of oxygen.



To explain how to do this, we have to get deeper into atoms. Physics
models atoms as consisting of three kinds of elementary particles -
protons (electrically positively charged), neutrons (electrically
neutral) and electrons (electrically negatively charged). This is a
relatively simple model, that corresponds to most of experiments, though
the reality is more complex than this. For our purposes we can view
this model of atom as a nucleus, that contains certain number of protons
and neutrons, and a number of electrons circulating the nucleus on
different orbits.



Electrons are very light relatively to protons and neutrons, so the mass
of an atom is concentrated, mostly, in its nucleus. Protons and neutron
have approximately the same mass, which is called atomic mass unit. So, the mass of an atom in atomic mass units
("atomic weight") is equal to the number of protons and neutrons in its
nucleus. This mass is known for each element of the Periodic Table of
Mendeleev, that is for each known atom.

For example, it is determined that atom of hydrogen H has atomic weight of 1 atomic unit, atom of carbon C has atomic weight of 12, atom of oxygen O has atomic weight 16.



Knowing atomic weights of atoms, we can calculate atomic weight of molecules. Thus, the atomic weight of a molecule of methane CH4 is 12+4=16. Atomic weight of a molecule of oxygen O2 is 16+16=32. Atomic weight of water H2O is 2+16=18.



Now we can take components of any chemical reaction proportional to the
atomic weight of corresponding molecules, which will result in
proportional number of molecules. For example, not being able to
experiment with one molecule of methane CH4 and two molecules of oxygen O2, we can experiment with 16 gram of methane and 64 gram
of oxygen, and the proportionality of the number of molecules will be
preserved - for each molecule of methane there will be two molecules of
oxygen.



As you see, taking amount of any mono-molecular substance in grams
equaled to the atomic weight of the molecules of this substance (called a
mole) assures taking the same number of molecules, regardless of the substance. This number is the Avogadro Number and is equal to N=6.02214076·1023.

Thus, one mole of methane CH4 (atomic weight of C is 12, atomic weight of H is 1) weighs 16g, one mole of silicon Si2 (atomic weight of Si is 14) weighs 28g, one mole of copper oxide CuO (atomic weight of Cu is 64, atomic weight of O
is 16) weighs 80g etc. And all those amounts of different substances
have the same number of molecules - the Avogadro number (approximately,
of course).



The theory behind the atomic bonds inside a molecule is quite complex
and is beyond the scope of this course. Based on this theory and
experimental data, for many kinds of atomic bonds there had been
obtained an amount of energy needed to break these bonds, that is its
inner chemical energy.

Thus, chemical energy of atomic bonds inside a mole of methane CH4
is 1640 kilo-joules (because a molecule of methane has 4 bonds between
carbon and each atom of hydrogen, each bond at 410KJ), inside a molecule
of oxygen O2 - 494 kilo-joules (1 bond between 2 atoms oxygen at 494KJ), inside a molecule of carbon dioxide CO2 is 1598 kilo-joules (2 bonds between carbon and each atom of oxygen, each 799KJ), inside a molecule of water H2O is 920 kilo-joules (2 bonds between oxygen and each atom of hydrogen, each 460KJ).



Let's go back to methane burning:

CH4 + 2O2 = 2H2O + CO2

This chemical reaction converts 1 mole of methane (16g) and 2 moles of
oxygen (64g) into 1 mole of carbon dioxide (44g) and 2 moles of water
(36g).

The energy we have to spend to break the atomic bonds of 1 mole of methane and 2 moles of oxygen, according to above data, is

Ein = 1640 + 2·494 = 2628 KJ

The energy we have to spend to break atomic bonds of 2 moles of water and 1 mole of carbon dioxide, according to above data, is

Eout = 2·920 + 1598 = 3438 KJ

The net energy is

Enet = 2628 − 3438 = −810 KJ

This net energy is the amount of thermal energy released by burning 16g
of methane, using 64g of oxygen, obtaining as a result 44g of carbon
dioxide and 36g of water.