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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:

IIT is achievable, just by following the simple path…

Regularity in work: Get regular in your work right from day one, because courses and problems will increase with time and if you delay your work for one day also, then it gets piled up for days ahead.

Take some time to relax: Do not follow “only study” module. Play for some time to refresh your brains and then get back to work.

Take tests: Weekends give you some extra time so gear you to attempt some tests from IIT JEE Coaching Course along with stop watch. You will understand your strengths and weaknesses to improve them right in the beginning.

Limit your friendship: One or two friends are O.K., but not more than that. Avoid parties and minimize distractions. Discipline over you is a must.

Complete syllabus in time: there is a set schedule, drafted by instructors of IIT JEE Coaching Institute, which should be strictly followed. If you are not finished, then don’t panic. Devote some extra time to get it done in next few days.

Think positive: Positive attitude keeps you moving while negative attitude pulls backwards. Think positive then only you can achieve your goals.

Exercise for fifteen minutes: Meditation for five minutes and brisk walk in open for ten minutes is sufficient to exercise your brain and body to make you fit for exam.

Concentrate on Lectures: Don’t daydream of passing out during IIT Lectures. Maximize your concentration to devote minimum effort in understanding the topic, afterwards.

Study without stressing you: Overstress diminishes your ability to do work. It hinders in normal functioning of brain. Stay cool and devote as much time as possible to complete the work in time. If possible set a schedule for different subjects so that you complete your work in time.

No wasting of time: Greatest enemy is “hanging out” here and there. It eats away your time leaving you gasping at the last moment.

Don’t Worry Be Happy: when you worry then your brain concentrates on worrying rather than studying, then you have to devote extra time to that particular subject. So bury your worries to earn scores in IIT JEE. Take a break if you can’t overcome your worries. It relaxes your mind and makes you feel refreshed. Don’t compromise on sleep and nutrition. Take small and regular meals to feel full and satiated without feeling sleepy.

Last but not the least, neither lose confidence, nor get over-confident, and then attempt your course. You will certainly achieve your defined goals. Join www.smartlearnwebtv.com as online coaching Institute and get started.

IIT Entrance exams are special exams, conducted for those classes of students who are hard working, sincere, result oriented, and ambitious. These students have set goals and work towards achievement of such goal.

IIT Entrance exam requires lots of efforts. Students should have proper mind set to prepare him/her for such prestigious exam. A proper framework of mind, strong determination, thorough understanding of goal, tremendous will power and lots of effort makes a student achieve IIT ranking amongst students all over India.

When a student undergoes IIT JEE coaching, then Professors also extend all possible help for making their students achieve their goal.

IIT JEE preparation syllabus includes an extensive training program where students are prepared for IIT JEE exam exclusively. Therefore the IIT entrance exam syllabus is designed in such a way that a student prepares him/her thoroughly. The syllabus for IIT JEE preparation is equipped with exact syllabus needed for preparation. The syllabus for each subject gives extensive knowledge of that subject. All possible topics are explained with proper example and there are different question papers for students to assess them properly. This is not all. There are proper questionnaire to judge students for their strengths and weaknesses.

Question papers are also designed to prepare students on regular intervals, so that the students remain regular with their syllabus. Those students who lag behind start gear up their preparations to cover up for studies. IIT entrance exams’ preparation has to be very regular. It’s like the job of a postman, who has to deliver the mail in times of adversities, as well. Same way, whatever happens, IIT JEE preparation should never be omitted, at all.

Some IIT JEE entrance exams’ preparation syllabus is equipped with Professors’ Lectures as well. These lectures make the student understand that particular topic better. Students can listen and understand lectures at their convenience and with utmost concentration so that they can understand the subject. And if in case, the child is not able to understand the subject then he/she can refer to that CD again. However, these lectures are so understandable that any qualified student can understand the topic with ease.

http://www.smartlearnwebtv.com is one of the most prestigious coaching Institute which is preparing thousands of students all over India and helping them achieve their goals.

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.

SmartLearn’s academic team is providing you the best way to study mathematics. Learning Mathematics is like learning the art of solving problems and not just knowing formulae and concepts. You need to focus more on problem solving instead of just reading theories, formulae and solutions, if you want to improve your mathematics.

Following are some instructions/tips that would definitely help you learn mathematics better:

Make a habit to study Mathematics with a paper and pen: More you work out, better you remember. Even If you are learning formulae and concepts, have a habit of writing it down the paper. Concentration is the key word to study mathematics. Whether you are solving a problem or reading mathematical steps of a solution you need better concentration and focus. So my suggestion would be to sit on a table chair with no disturbance around. If your room is noisy, you can put cotton balls in your ears.

Spend more time in solving problems instead of reading solutions/theories/formulae: More you practice, better you would learn. It is very important that you solve problems to learn topics in mathematics. Just understanding concepts and learning formulae would not be sufficient to be able to solve questions in exam. In mathematics more than 50% of the knowledge comes through tricks/methods involved in solving problems. If you don’t practice questions, you don’t acquire this knowledge. In fact learning in Mathematics starts the day you start solving problems with pen and paper.

Learning Procedure: Priority should be given to learning formulae and concepts before working out the problems. Write them and see how confident you are. You should start working out solved-examples only after learning the concepts and formulae. This should be the procedure of study for easy understanding of the solved examples as in every questions we use multiple formulae. Formulae play a vital role in mathematics, as you will not be able to proceed with the problems once you forget the formulae. Then proceed with the next level of questions that is average level questions.

We can decide the level of the questions based on the ability to solve the questions.

Level 1 – If you can understand the solution just by a glance.

Level 2 – If you can understand the solution after going through the complete solution step-by-step.

Level 3 – If you find it hard to understand the solution.

You can start with 35-45 level 1 problems and make sure that the formulae and concepts of these problems are clear so that you can move to the next level of problems. Then practice at least 35-45 level 2 problems and make sure that the concepts and solutions are clear. Then try to solve level 3 problems. When compared to level 1 and level 2 problems, level 3 problems are less important.

Give importance to revision: In case you are not able to solve any of the problems, you try to solve it or else you see their solutions. Generally when you see the solution and know the problem solving methodology, you do nothing after that. In a weeks time you forget the solution. In case you are asked to solve the same problem again, you might not be able to solve it.

I strongly feel that you should solve the problems by yourself with paper and pen after seeing the solution. This helps you retain the solutions in your mind for a longer time when compared to seeing the solutions alone. When you try a problem mark the difficulty level of the question so that it will be easy for future reference. Select a few questions which are quite difficult and take up a test with a fixed time. This procedure increases your confidence and makes your mathematical skills more powerful.

At the time of revision you will have confusion as to what to revise at the time before the exams. Just take up the level 2 questions and start revising. This will be helpful only if you mark the questions as level 2 or level 1.

Trying out the problems before looking into the solutions: As far as we have reviewed, students have the tendency to look into the solution page before trying out the problems. They sit with the books and study materials around which makes them easy to look into the solutions before trying out the problems. We strongly fight back this kind of attitude in some students. Once you continue this, you will become a dependent on solutions and you tend to surrender yourself to the solutions. The students should have a fighting attitude when it comes to the problem solving in subjects. Not for all questions you can follow this, at least for 70%-80% questions you can try it for yourself before looking into the solutions.

Interest makes you perform better: yes, interest makes you work better and perform better. As mathematics involves puzzles, tricks, problems, you can perform better only if you like the subject. When you attempt a problem and you succeed in that, feel very proud of you. In the same way, don’t get frustrated when you fail to solve the problem. Instead feel thrilled when you take help from solutions. Try to solve questions which you took help from the solutions once again so that make sure that you don’t repeat the same mistake.

Creative learning process: Learning formulae is a tiring work for the students. When things are made creative it makes learning fun for you. You can make paper cards of the formulae and tricks and carry it with you. You can learn it at any place, this helps you build your knowledge in a creative way.

Help your friends: If your friends have some problem in solving a particular problem, help them in solving if you know how to solve. This will surely generate confidence in you and builds your confidence too.

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