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Each object around us has certain shape and size. When we study motion of such objects, we will have to study motion of all the different particles of the object. Practically speaking, this is not possible and we will have to consider alternative ways to study motion of the object. Scientists have defined a concept called `Centre of Mass’ to take care of this situation. To consider motion of an object having finite size, they consider motion of the centre of mass of the object. The basic objective of this chapter is to understand this concept.

OBJECTIVE

We will attempt to define the concept of a point where the mass of the whole body can be assumed to be concentrated. This helps us in analysis of problems that involve application of forces that result in subsequent motion. We will study systems which have a non uniform distribution of mass. And finally we will understand the motion of centre of mass as the body moves.

PRE-REQUISITE

The following pre-requisites apply

Þ         Vector Representation.

Þ         Integration in one variable.

Þ         Force & Acceleration.

CORE CONCEPTS

Classically each and every object has some finite size, but we have assumed them as particles i.e. having mass but no size so far.

In translatory motion each point on an object undergoes the same displacement as any other point as time goes on, so that the motion of one particle represents the motion of whole object. Thus, the translatory motion of an object of finite size can be studied by analysing the motion of any constituent particle.

But, in cases where the motion is not translatory, rather than considering any point on the object, a point known as the centre of mass is defined, and the motion of this point is studied. The motion of the centre of mass leads to the analysis of the object as a whole

Conceptually, the point where the whole mass of body or system can be assumed to be concentrated for simplified study of its motion is called the centre of mass.

For a discrete system of particles the positions (see figure given below) of the centre of mass is

In chemistry, we deal with processes, is which are invariably associated with transfer of energy between the system under study and its surroundings. For example, heat is evolved when an acid is neutralised by a base. The heat transfer is basically due to the conservation of energy or first law of thermodynamics. It is of prime importance to a chemist to understand these energy changes & use this knowledge in his study of the subject.

Objective

The branch of physical chemistry, which deals with the study of heat changes, accompanying a chemical reaction is termed as Thermochemistry. Thermo (heat) Dynamics (work) is the study of those interactions among various materials which involve the transfer of heat, and the performance of work. Our aim in this chapter will be [to mathematically & conceptually understand] the changes that the system & surrounding undergo when exchange of energy takes place.

PRE-REQUISITE

Þ    Since this subject is incomplete without the use of mathematical relationships, it is important that we must understand the basic operations of Logarithms, Ratios etc.

Þ    The units & dimensions come in extremely handy in dealing with unknown variables & constants.

Þ    Basic stoichiometry & Mole concept are also important.

CORE CONCEPTS

The heat transfer is basically due to the conservation of energy or first law of thermodynamics. Thermo (heat) Dynamics (work) is the study of those interactions among various materials which involve the transfer of heat, and the performance of work.

Thermochemistry basically deals with the transfer of heat between a chemical system and its surrounding.

A system is defined as a specified part of the universe or specified portion of the matter which is under experimental investigation and the rest of the universe i.e. all other matter which can interact with the system, is surrounding.

Note

Þ      To calculate the heat transferred, the reactants and the products must be at the same temperature.

Depending on the heat transferred the reactions can be classified as

Exothermic Reaction

The reaction in which heat is transferred to the surroundings from the system.

Endothermic Reaction

The reaction in which heat is transferred to the system from the surroundings.

By SI convention, the heat transferred is taken as negative and positive for exothermic and endothermic reactions, respectively. In other words, the process which increases the energy of the system is taken as +ve and which decrease the energy of the system is taken as –ve.

The molar enthalpy H_m of any substance is a function of temperature and pressure, i.e. H_m = H_m(T, p). The pressure dependence is removed by defining the standard molar enthalpy H^°_m, which is the enthalpy of the substance at the standard pressure of 101.325 k Pa.

Note

Þ      It is impossible to determine the absolute value of enthalpy.

It is impossible to determine absolute value of enthalpy. The values we observe are based on the SI convention. However relative enthalpies of substances can be determined if the enthalpy of free elements at 25 °C and 1 atmosphere pressure are taken arbitrary as zero or in other words, the enthalpy of every element in its stable state of aggregation at 101.325 k Pa (or 1 atmosphere pressure) and at 25 °C is assigned a zero value.

At 101.325 kPa and 298.15 K, the stable state of aggregation of Nitrogen is the gaseous state, hence H^°_m (N2_g) = 0.

If an element exists in more than one allotropic forms, the most stable allotrope is assigned zero value.

Example

Solid sulphur (rhombic) and solid carbon (graphite) are assigned a zero standard molar enthalpy. i.e., H°.

Example:

To find out the standard molar enthalpies of various substances, the above conventions are used. For example, consider the following reaction:

According to the Graham’s Law of diffusion, the rate of diffusion of a gas is inversely proportional to the square root of it’s density or molecular weights. If r1 and r2 are the rates of diffusion of two gases, whose densities under the given conditions are d1 and d2 respectively, then we have

(M₁ and M₂ are the respective molecular weights of the two gases) and d₁ and d₂ are their respective vapour densities.)

Knowing the experimental gas laws, it is of interest to develop a theoretical model based on the structure of gases, which can co-relate the experiment. Fortunately, such a theory has been developed and is known as kinetic theory of gases.

Kinetic Theory of Gases

The word ‘Kinetic’ means ‘Motion’. Gaseous molecules are assumed to be in constant motion. A theory with the help of which the various gas laws can be derived mathematically is known as the kinetic theory of gases.

The main postulates of the theory are as follows:

Þ      A gas is made of extremely tiny particles called molecules. The molecules of any given gas are identical and have the same mass,

and the molecules are assumed to be dispersed in a lot of vacant space.

Þ      The individual molecules are relatively far apart from each other and they exert very little attraction for each other except under collision of molecules and near the liquification point. The real volume of the gas molecules at ordinary temperatures and pressures is very small in comparison to the total volume of the gas. Here we are talking about real gases or non-ideal gases since ideal gases cannot be liquefied.

Þ      The gaseous molecules are in continuous random, straight line motion with very high speeds in all directions. They collide frequently and this may bring about a change in the direction of movement and a redistribution of energy between the colliding molecules. The collisions are perfectly elastic (i.e. no loss of energy) but only redistribution of energy may occur.

Þ      The force of gravity has negligible effect on the speed of the gas molecules.

Þ      The pressure exerted by a gas is due to collisions made by gas molecules on the walls of the container. Gases not only distribute themselves throughout the total volume of the container but also exert uniform pressure on every point of the container.

Þ      The average K.E. of the molecules is directly proportional to the absolute temperature of the gas.

To find the unit of a given physical quantity in a given system of units:

ü      By expressing a physical quantity in terms of basic quantities we find its dimensions. In the dimensional formula replacing M, L, T by the fundamental units of the required system, we get the unit of physical quantity. However, sometimes we assign a specific name to this unit.

Example

Force is numerically equal to the product of mass and acceleration

i.e. Force = mass × acceleration

Its unit in SI system will be Kgms^-2 which is given a specific name “Newton (N)”.

Similarly, its unit in CGS system will be gm cms^-2 which is called “dyne”.

To find dimensions of physical constants or coefficients

The dimensions of a physical quantity is unique because it is the nature of the physical quantity and the nature does not change. If we write any formula or equation incorporating the given physical constant, we can find the dimensions of the required constant or co-efficient.

Example

From Newton’s law of Gravitation, the force exerted by one mass upon another is

To convert a physical quantity from one system of units to another

This is based on the fact that for a given physical quantity,

magnitude × unit = constant

So, when unit changes, magnitude will also change.

Example

1.         Convert one Newton into dyne

Sol. Dimensional formula for Newton  = [ML/T²]

or 1 N = 1 Kg m/s²    But 1 kg = 10³g

and 1 m = 10²cm

1N = (10³ g) ( 10²cm)/s² = 10^5 g cm/s² = 10^5 dyne

To check the dimensional correctness of a given physical relation

This is based on the principle that the dimensions of the terms on both sides of an equation must be same. This is known as the ‘principle of homogeneity’. If the dimensions of the terms on both sides are same, the equation is dimensionally correct, otherwise not. It is not necessary that a dimensionally correct equation is also physically correct but a physically correct equation has to be dimensionally correct.

Þ        Consider the formula, T = 2Π √(1/g)

where T is the time period of oscillation of a simple pendulum in a simple harmonic motion, l and g are the length of the pendulum and gravitational constants respectively. Check this formula, whether it is correct or not, using the concept of dimension.

Thus the above equation is dimensionally correct (homogeneous) and later you will come to know that it is physically also correct.

Þ        Consider the formula

Check this formula, whether it is correct or not, using the concept of dimension.

Dimensionally

In this case also the formula is dimensionally correct but, you know that it is physically incorrect as the correct formula is given by

Rutherford placed thin sheets of metal in the path of a-particles in order to see how various metals would affect the a-particle trajectory.

a-particles are actually helium atoms from which electrons have been removed. Each a-particle consists of a mass equal to about 4 times that of hydrogen atom and carries a positive charge of 2 units. It is represented by symbol .

If Thomson’s explanation were correct, a-particles would have been deflected at very small angles only

from a straight line path. But Rutherford found that maximum a-particles go straight, some get deflected at small angles, a few at large angles and in rare cases the deflection is 180° as shown in fig. 1.4. He hypothesised that deflection at 180° can arise only if an intense positive electric field is present inside atoms. Observations showed that a positive charge spread throughout a sphere of radius 10^-8 cm would be incapable of producing this field. Calculations showed that this radius should be of the order 10^-13 cm to account for scattering data. Based on these observations Rutherford presented following model for atom.

Þ        Atom consists of a nucleus which contains protons making it positively charged & mass being centered here in a small space of radius 10^-13 cm.

Þ        There is a lot of empty space around nucleus in which electrons are present. The total size of the atom is of radius 10^-8 cm.

Þ        Electrons can’t be stationary as they would be pulled by nucleus. Instead they are revolving around nucleus, the necessary centripetal force for revolutions is provided by attractive forces between nucleus & electrons

Bohr also argued the same that the electron  (being a charged particle) should also lose energy while moving in a circle (i.e. with an acceleration). As a result its orbit should become smaller and smaller and finally it should drop into the nucleus. But the fact is that atom is stable.

Niel’s Bohr supplied a solution to this problem by applying Planck’s quantum theory. Let us first study the Planck’s quantum theory.

Planck’s quantum theory (1901)

It states

Þ        Radiant energy is emitted or absorbed discontinuously in the form of tiny bundles of energy called Quanta.

Þ    Each quantum is associated with a definite amount of energy E which is proportional to frequency of radiation.

where,             h = Planck’s constant = 6.626 * 10^-34 Joule-sec.

v = Frequency of the light radiation

Þ        A body can emit or absorb radiations only in whole multiples of quantum i.e. E = nhv where
n = 1, 2, 3, …….

Bohr’s atomic model

The postulates of Bohr’s atomic theory stability of an atom are as follows

Electron revolves in only allowed stationary orbits. Energy of different stationary states vary. An electron can be excited from a lower state to higher state with the absorption of a quantum of energy, or can come down from a higher to lower state with emission of a radiation of energy (as shown in figure 1.5) equal to energy to quantum ΔE = E2 - E1 = hv. E2 & E1 are energies of the electron associated with stationary orbits.

The stability of the circular motion of an electron requires that the electrostatic force (due to the attraction between the nucleus and the electron) provides the necessary centrepetal force for the motion of electron.

where Z – atomic number

e – charge on electron

ε0 - permittivity of the charge in vaccum

r – distance between positive charge & electron

Angular momentum of electron is quantised i.e. electron can revolve only in those orbits where its angular momentum is an integral multiple of h/2Π.

where, v – velocity of electron

m – mass of electron

h – Planck’s constant

n = 1, 2, 3, …. are known as Principal quantum number.

Self adjusting Properties

Then x = sin θ                                                  …(2)

From (1) putting the value of θ in (2), we get,

Illustrations

Equal number of molecules of different gases under identical conditions of temperature and pressure occupy the same volume.

Hence, the volume occupied by one mole of an ideal gas at standard temperature (273.15 K) and pressure (101.325 K Pa) has a fixed volume (22.414 dm³). This indicates that the number of molecules contained in one mole of any real gas should be a constant quantity. This number is found to be 6.023 × 10²³ and is known as Avogadro number.

Important

Ideal gas is a gas which follows all the above gas laws under all conditions of temperature and pressure.

Real gases generally do not obey the gas laws, exactly, under all conditions of temperature and pressure.

The Ideal Gas Equation

Combination of Boyle’s and Charle’s laws. When temp. (T1) is kept constant and pressure is changed from p1 to p2, Let the new volume be V ^‘.

Important

Value of the proportionality constant k depends on:

(a)      Quantity of gas and

(b)         Units, in which p, V and T, are expressed.

On the basis of Avogadro’s hypothesis, 1 mole of all gases under similar conditions of temp. and  pressure occupies the same volume. Hence k will have the same value for 1 mole of any gas taken.

pV = kT. …(5)

(k is replaced by R called the molar gas constant).

For n moles of gas considered  (5) becomes

PV = nRT. …(6)

Eq. (6) is called the ideal gas equation showing the effect on the third variable when two of the three variables are changed simultaneously for a given amount of a gas. The units of R varies with the units of the other parameters (p, V, T).

e.g. R has the following values

0.0821 litre atm/ K /mole

5.28 × 10¹⁹ ev / K / mole

8.314 Joules / K / mole

1.99 cal / K / mole

0.002 k cal / K / mole

8.314 × 10⁷ erg / K / mole

Illustrations

1.   A two litre flask, containing O2 at 1 atm pressure is at a constant temperature at 27°C. The gas pressure is reduced to 10^–6 atm by attaching the flask to a vacuum pump. Assuming ideal behaviour, answer the following:

(a) What will be the volume of the gas which is left behind?

(b) What will be the no of molecule given in the problem?

Sol. Given that V₁ = 2 l, p₁ = 1 atm, T = 27°C = 300 K

We have the following results

(a)         The volume of oxygen left behind will be the same i.e. 2 l.

(b)         The number of moles of oxygen left behind is given by

Concept of friction:

Component of normal reaction

We have seen that a ball rolling on a floor stops after some time. When we switch off the engine of a car, it stops after traveling some distance. Similarly when we apply brakes, our bicycle comes to rest after traveling some distance. The above examples show that some invisible force is opposing the motion of one body over the other. This opposing force is called friction. Friction is an opposing force that comes into play when one body actually moves (slides or rolls) or even tries to move over the surface of another body.

Cause of friction

Roughness of surfaces is the cause of friction.

When two bodies are in contact with each other, the irregularities on the surfaces get interlocked and oppose any relative motion.

Normal reaction

When a body of mass m is lying on a horizontal surface, it presses the surface due its weight. Contact force F_c = R in case no external force is acting on the body. Also in this case friction is zero and F_c is perpendicular to the surface. In the diagram below, R is the normal reaction.

Static and kinetic friction

The opposing force that comes into play when one body tends to move over the surface of another (but the actual motion has not yet started) is called static friction. Limiting friction is the maximum opposing force that comes into play when one body is at the verge of moving over the surface of another body. It is denoted by f_s, and is called maximum force of static friction or limiting friction.

Laws of friction

The following are the laws of limiting friction:

(a) The magnitude of the force of limiting friction f_s, is directly proportional to the normal reaction R.

(b) The direction of the force of limiting friction is always opposite to the applied force.

(c) It is independent of the apparent area of contact.

(d) It depends on the nature and material of the surfaces in contact.

For example, when two polished metal surfaces are in contact,  = 0.2

When these surfaces are lubricated,  gets reduced. Hence, it depends on the nature of the surfaces.

Rolling friction

Rolling friction is always less than dynamic and static friction. When a body rolls on a level track, the area of contact is very small. This causes a depression in the surface below. This causes rolling friction.

The velocity of the point of contact of the wheel with respect to the floor remains zero all the time. Thus, rolling wheel constantly climbs a hill and has to simultaneously get itself detached from the road. Rolling friction is also directly proportional to the normal reaction and is inversely proportional to the radius of rolling body. Combining the two, we get,

F is directly proportional to R/r

F = μx (R/r)

Where μx is called coefficient of rolling friction.

Advantages and Disadvantages of friction

Friction is a necessary evil. It is necessary because we cannot do any work without it. At the same time, it is also an evil because it involves unnecessary wastage of energy.

There are advantages of friction. It can be understood with the help of the following examples:

Walking will not be possible without friction.

Brakes of vehicles will not work without friction.

Writing on a blackboard or a piece of paper is possible only due to friction between the blackboard and the chalk or the paper and the pen. Cleaning with sand paper will not be possible without friction.

There are disadvantages of friction. Friction is an evil and this can be understood with the following   examples:
Extra energy is required to overcome the friction between moving parts.

ü       Friction causes wear and tear of different parts of machinery.

ü       Frictional forces result in the production of heat, which causes damage to the machinery.

Methods of reducing friction

By polishing, by lubrication, By streamlining, By using ball-bearings,

Angle of friction

It is the angle between normal Reaction and the resultant of the normal reaction and limiting frictional force. i.e. we have a body of mass m which is placed on a table and we say that the body and the surface of the table have a coefficient of friction m between them. (fig. 3.7)

If we apply a small force F, the body will not move. Let us gradually increase the force until the body starts moving. At one stage the applied force will be equal to the frictional force. The coefficient of static friction ms  = F/N where F is the applied force and N = mg is the normal force.

Orthocentre is the point of concurrence of the altitudes of a triangle. We denote it by H. The triangle obtained by joining the feet of the altitudes is called pedal triangle. The position vector of the orthocentre is given by

It is possible to find distance of orthocentre from vertices and sides of a triangle.

Let H be the orthocentre

In the figure, we observe that

HK = BK tan ∟HBK = BK tan ∟LBC

HK = BK tan (90°-C) (in ΔBHK)

= AB cos B cot C

=c cos B cos C/ sin C = (c/sin C) cos B cos C

= 2 R cos B cos C

Similarly, we can prove that

HL = 2R cos A cos C

and       HM = 2R cos A cos B

Now, AH = AL sec (90 – C)

= AB cos A cosec C

= (c/sin C) cos A = 2 R cos A

Similarly we get BH = 2R cos B, CH = 2R cos C

Thus, distances of orthocentre from sides a, b, c, are given by
2 R cos B cos C, 2R cos A cos C, 2 R cos A cos B respectively and from vertices A,B,C are given by 2R cos A, 2R cos B and 2R cos C  respectively.

ΔKLM is pedal triangle for ΔABC

In quadrilateral BKHM,

∟BMH +∟ BKH= 90° + 90° = 180°

Þ         BKHM is a cyclic quadrilateral

Þ         ∟BMK = ∟BHK = 90° – ∟HBK

= 90° – ∟LBC

= 90° – (90° – C) = C

=> ∟KMH = 90° – C

Similarly, we can prove that

∟HML = 90° – C

=> ∟KML = 180° – 2 C

Similarly

∟KLM = 180° – 2B and ∟MKL = 180° – 2A

i.e. angles of pedal triangle are supplement of double of opposite angles of the original triangle

Now, in Δ AML,

Similarly, it can be shown that KM = b cos B, KL = c cos C

i.e. sides of pedal triangle are a cos A, b cos B, c cos C.

Illustration

1. MKL is the pedal triangle of ABC; prove that its area is

2 S cos A cos B cos C, where S is area of ΔABC.

Various kinds of substances that constitute matter can be roughly divided into three categories namely- gases, liquids and solids. The existence of matter in either of these three forms is a result of the competition between two opposing intermolecular forces

(1)     The forces of attraction, which hold the molecules together.

(2)     The thermal energy of these molecules which tend to increase the intermolecular distances.

If the thermal energy of the molecules is much greater than the forces of attraction, the state of matter that result is called the gaseous state. On the other hand, if the forces of attraction are greater than the thermal energy, we have the matter in the liquid state. When these forces of attraction are much more greater than the thermal energy compared to the liquid state, we have matter in its solid state condition. However, on the application of heat, the thermal energy of the molecules can be increased and as a result the intermolecular forces of attraction would relatively decrease simultaneously.

Molecules in the gaseous state possess high energy and have almost no force of attraction. They are far apart and show a great uniformity in behaviour, irrespective of their chemical nature, colour or odour. They are highly compressible and can also be expanded without limit. They also produce pressure on the walls of any container uniformly in all directions. They diffuse rapidly through one other to form a homogeneous mixture, and their separation is also not very easy.

Gas laws

Boyle’s law

At constant temperature, the volume of a sample of gas of definite mass varies inversely with its pressure.

i.e    when temperature is kept constant for a given mass of gas

where V = volume and  p = pressure

Introducing a constant k, we have

pV = k = constant                                …(1)

The value of the proportionality constant k depends upon the following factors:

(1)           Nature of a gas

(2) Temperature of the gas, and

(3)           The mass of the gas

Hence, at constant temperature, for a given mass of a gas, Boyle’s Law states

p₁V₁ = p₂V₂ = k = constant                  …(2)

Eq. (1) can be represented graphically as shown in figure given below.

The general term isotherm (i.e. at const. Temp.) is used to describe the above curves.