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NATIONAL LEVEL ONLINE TEST

Physics is the science that studies matter, its motion and behavior and the related entities. KTG & Thermodynamics both are in the Chemistry and Physics syllabus. And it is the scoring chapter as every year 2 questions of 8 marks are asked which are not less if we see the weightage.

To get full marks in this chapter, students must revise all the important concepts and formulae. Thermodynamics includes some of the important topics like First law of thermodynamics, Adiabatic process, Isothermal Process, Efficiency of Carnot engine, etc. In article we have provided quick notes covering the major topics of KTG and Thermodynamics, solved sample questions, Tips to remember, etc.

**Must Read:**

**Kinetic Energy: **Energy possessed by the atoms or molecules by virtue of their motion is called kinetic energy.

**Solids: **Matter which has got fixed shape and volume. The force of attraction between any two molecules of a solid is very large.

**Liquids : **Matter which has got fixed volume but no fixed shape. Force of attraction between any two molecules is not that large as in the case of solids.

**Gases: **Matter which does not have any fixed shape or any fixed volume.

**Ideal Gas:** An ideal gas is one which has a zero size of molecule and zero force of interaction between its molecules.

**Ideal Gas Equation:**

PV/T = Constant or

PV = nRT

Here, n is the number of moles and R is the universal gas constant.

**Gas Constant**

Universal gas constant (R)

R= P0 V0/T0

=8.311 J mol-1K-1

Specific gas constant (r)

PV= (R/M) T = rT,

Here,r = R/M

**Real Gas : **The gases which show deviation from the ideal gas behavior are called real gas.

**Avogadro’s number (N):** Avogadro’s number (N), is the number of carbon atoms contained in 12 gram of carbon-12.

N = 6.023×1023

** (A) To calculate the mass of an atom/molecule**

Mass of one atom = atomic weight (in gram)/N

Mass of one molecule = molecular weight (in gram)/N

** (B) To calculate the number of atoms/molecules in a certain amount of substance**

Number of atoms in m gram = (N/atomic weight)×m

Number of molecules in m gram = (N/molecular weight)×m

** (C) Size of an atom**

Volume of the atom, V = (4/3)πr3

Mass of the atom, m = A/N

Here, A is the atomic weight and N is the Avogadro’s number.

Radius, r =[3A/4πNρ]1/3\

Here ρ is the density.

This law states that the volume of a given amount of gas varies inversely as its pressure, provided its temperature is kept constant.

PV = Constant

It states that volume of a given mass of a gas varies directly as its absolute temperature, provided its pressure is kept constant.

V/T= Constant

V–V0/V0t = 1/273 = γp

Here γp (=1/273) is called the volume coefficient of gas at constant pressure.

It states that pressure of a given mass of a gas varies directly as its absolute temperature provided the volume of the gas is kept constant.

P/T = P0/T0 or P – P0/P0t = 1/273 = γp

Here γp (=1/273) is called the pressure coefficient of the gas at constant volume.

The pressure which it would exert if contained alone in the entire confined given space.

P= p1+p2+p3+……..

nRT/V = p1+p2+p3+……..

The rate of diffusion of gases varies inversely as the square root of the density of gases.

R∝1/√ρ or R1/R2 =√ρ2/ ρ1

So, a lighter gas gets diffused quickly.

It states that under similar conditions of pressure and temperature, equal volume of all gases contain equal numbers of molecules.

For m gram of gas, PV/T = nR = (m/M) R

Root mean square velocity of a gas is the square root of the mean of the squares of the velocities of individual molecules.

C= √[c12+ c22+ c32+…..+ cn2]/n = √3P/ ρ

The pressure of a gas is equal to two-third of kinetic energy per unit volume of the gas.

P= 2/3 E

Root mean square velocity of the molecules of a gas is proportional to the square root of its absolute temperature.

C= √3RT/M

Root mean square velocity of the molecules of a gas is proportional to the square root of its absolute temperature.

At, T=0, C=0

Thus, absolute zero is the temperature at which all molecular motion ceases.

K.E. per gram mol of gas = ½ MC2 = 3/2 RT

½ C2 = 3/2 rt

Here, ½ C2 = kinetic energy per gram of the gas and r = gas constant for one gram of gas.

Kinetic energy per molecule = ½ mC2 = 3/2 kT

Here, k (Boltzmann constant) = R/N

Thus, K.E per molecule is independent of the mass of the molecule. It only depends upon the absolute temperature of the gas.

R1/R2 = C1/C2 = √ρ2/ ρ1

It is the speed, possessed by the maximum number of molecules of a gas contained in an enclosure.

Vm= √[2kT/m]

Average speed of the molecules of a gas is the arithmetic mean so the speeds of all the molecules.

Vav= √[8kT/πm]

It is the square root of the mean of the squares of the individual speeds of the molecules of a gas.

Vrms = √[3kT/m]

Vrms > Vav > Vm

The number of possible independent ways, in which the position and configuration of the system may change.

In general, if N is the number of particles, not connected to each other, the degrees of freedom n of such a system will be,

n = 3N

If K is the number of constraints (restrictions), degree of freedom n of the system will be,

n = 3N –K

**(A) Mono-atomic gas: Degree of freedom of monoatomic molecule, n = 3**

**(B) Di-atomic gas:**

At very low temperature (0-250 K):- Degree of freedom, n = 3

At medium temperature (250 K – 750 K):- Degree of freedom, n = 5 (Translational = 3, Rotational = 2)

At high temperature (Beyond 750 K):- Degree of freedom, n = 6 (Translational = 3, Rotational = 2, Vibratory =1), For calculation purposes, n = 7

In any dynamical system, in thermal equilibrium, the total energy is divided equally among all the degrees of freedom and energy per molecule per degree of freedom is ½ kT.

E = ½ kT

**Points to remember:**

• In real gas, we assume that the molecules have finite size and intermolecular attraction acts between them.

• Real gases behave like perfect gas at high temperature and low pressure.

• Gaseous state of matter below critical temperature is called vapours. Below critical temperature gas is vapour and above critical temperature vapour is gas.

• The internal energy of a perfect gas consists only of kinetic energy of the molecules.

It is the branch of physics which deals with processes involving heat, work and internal energy. Thermodynamics is concerned with macroscopic behavior rather than microscopic behavior of the system. Everything external to the system is surrounding or environment. The actual or imaginary surface separating the system from the environment is called Boundary. It may be either movable or immovable.

**System:** Part of the universe under investigation.

**Open System:** A system which can exchange both energy and matter with its surroundings. A boundary which allows the exchange of matter is called permeable boundary. For example: Water heater, turbine, car radiator, the ocean, etc. To analyse an open system, energy of the system is always taken equal to the energy leaving the system.

**Closed System (m=constant):** A system which permits passage of energy but not mass, across its boundary. For example: a greenhouse is a closed system which exchanges heat but does not work with its surroundings. It depends on the properties of its boundary a system exchanges heat, work or both. An Adiabatic boundary does not allow the exchange of work between surroundings and system.

**Isolated system: **A system which can neither exchange energy nor matter with its surrounding. For example: an insulated gas cylinder, an insulated container.

**Surroundings:** Part of the universe other than the system, which can interact with it.

**Boundary:** Anything which separates the system from surrounding.

**State variables: **The variables which are required to be defined in order to define the state of any system i.e. pressure, volume, mass, temperature, surface area, etc.

**State Functions:** Property of a system which depends only on the state of the system and not on the path. Example: Pressure, volume, temperature, internal energy, enthalpy, entropy etc.

**Kinetic Energy:** Energy possessedby the atoms or molecules by virtue of their motion is called kinetic energy.

Sum total of kinetic and potential energies of atoms/molecules constituting a system is called the internal energy of the system.

• ΔU is taken positively, if the internal energy of the system increases.

• ΔU is taken negative, if the internal energy of the system decreases.

Work is said to be done when a force acting on a system displaces the body in its own direction.

dW = Fdx = PdV

W = P(Vf -Vi)

• If the gas expands, work is said to be done by the system. In this case Vf > Vi, therefore, W will be positive.

• If the gas is compressed, work is said to be done on the system. In this case Vf< Vi, therefore, work done is negative.

The thermodynamic state of a system can be determined by quantities like temperature (T), volume (V), pressure (P), internal energy (U) etc. These quantities are known as thermodynamic variables, or the parameters of the system.

**Equation of state**

A relation between the values of any of the three thermodynamic variables for the system, is called its equation of state.

Equation of state for an ideal gas is PV = RT

**Equilibrium of a system**

A system is said to be in equilibrium if its macroscopic quantities do not change with time.

Relation between joule and calorie: 1 joule = 4.186 cal

If the quantity of heat supplied to a system is capable of doing work, then the quantity ofheat absorbed by the system is equal to the sum of the increase in the internal energy of the system, and the external work done by it.

dQ = dU+dW

A process by which one or more parameters of the thermodynamic system undergoing a change is called a thermodynamic process or a thermodynamic change.

The process in which change in pressure and volume takes place at a constant temperature, is called a isothermal change.

The process in which change in volume and temperature of a gas takes place at a constant pressure is called an isobaric process.

The process in which changes in pressure and temperature take place in such a way that the volume of the system remains constant, is called an isochoric process.

The process in which change in pressure and volume and temperature takes place without any heat entering or leaving the system is called adiabatic change.

Such an expansion in which no external work is done and the total internal energy of the system remains constant and is called free expansion.

In a system in which the parameters acquire the original values, the process is called a cyclic process.

The process in which change in any of the parameters takes place at such a slow speed that the values of P,V, and T can be taken to be, practically, constant, is called a quasi-static process.

Cooling caused in adiabatic process - dT = PdV/Cv

Melting:- dU = mLf

Boiling:- dU = mLv – P(Vf -Vi)

Mayer’s formula:- Cp - Cv = R

The amount of heat required to raise the temperature of a unit mass of substance through 1ºC.

**Specific heat capacity at constant volume (cv):**The amount of heat required to raise the temperature of 1 g of the gas through 1ºC keeping volume of the gas constant.

Molar specific heat capacity, at constant volume (Cv), is defined as the amount of heat required to raise the temperature of 1 mole of gas through 1ºC keeping its volume constant.

Cv= Mcv

**Specific heat capacity at constant pressure (cp)**

Specific heat capacity, at constant pressure, is the amount of heat required to raise the temperature of 1 g of gas through 1ºC keeping its pressure constant.

Gram molecular specific heat capacity of a gas (Cp), at constant pressure, is defined as the amount of heat required to raise the temperature of 1 mole of the gas through 1ºC keeping its pressure constant.

Cp = Mcp

PV γ = Constant

Slope of isothermal:- dP/dV = -P/V

Slope of adiabatic:- dP/dV = -γP/V

Adiabatic gas constant:- γ = Cp/Cv

As,Cp>Cv,

So,γ>1

This signifies that, slope of the adiabatic curve is greater than that of isothermal.

• For isobaric process: zero

• For isochoric process: infinite

**Work done for isobaric process:- **W = P(V2-V1)

**Work done for isochoric process:- **W = 0

**Work done in isothermal expansion and compression**

W = 2.3026 RT log10Vf/Vi (isothermal expansion)

W = - 2.3026 RT log10Vf/Vi (isothermal compression)

**Work done during an adiabatic expansion**

= K/1-γ [Vf1-γ – Vi 1-γ] = 1/1-γ [P2V2-P1V1] = R/1- γ [T2-T1]

**Adiabatic constant (γ)**

γ = Cp/Cv = 1+2/f, Here f is the degrees of freedom.

It is a process which can be made to proceed in the reverse direction by a very slight change in its conditions so that the system passes through the same states as in direct process, and at the conclusion of which the system and its surroundings acquire the initial conditions.

**For Example**

• All isothermal and adiabatic process when allowed to proceed slowly,

are reversible, provided there is no loss of energy against any type of resistance.

• Friction, viscosity are other examples.

A process which cannot be made to be reversed in the opposite direction by reversing the controlling factor is called an irreversible process.

**For Example**

• Work done against friction

• Joule’s heating effect

• Diffusion of gases into one another

• Magnetic hysteresis

It is a device used to convert heat into mechanical energy

Work done, W = Q1-Q2

Efficiency η of an engine is defined as the fraction of total heat, supplied to the engine which is converted into work.

η= W/ Q1 = [Q1- Q2]/ Q1 = 1-[Q2/Q1]

**(A) Clausius statement:** The Heat cannot flow from a cold body to a hot body without the performance of work by some external agency.

**(B) Kelvin’s statement:** In Kelvin’s statement, it is impossible to obtain a continuous supply of energy by cooling a body below the coldest of its surroundings.

**(C) Planck’s statement:** In Planck’s statement, it is impossible to extract heat from a single body and convert the whole of it into work.

It is a device which is used to keep bodies at a temperature lower than that of surroundings.

Coefficient of performance of a refrigerator is defined as the amount of heat removed per unit work done on the machine.

β= Heat removed/work done = Q2/W = Q2/[Q1- Q2] = T2/[T1- T2]

Coefficient of performance of a refrigerator is not a constant quantity since it depends upon the temperature of body from which the heat is removed.

For a perfect refrigerator, W = 0 or Q1= Q2 or β =∞

(A) H = U+PV

(B) At constant pressure

dH = dU + pdV

(C) For system involving mechanical work only

dH =QP(At constant pressure)

(D) For exothermic reactions

dH is negative

(E) For endothermic reactions

dH is positive

dH = dU + dngRT

Here,dng= (Number of moles of gaseous products - Number of moles of gaseous reactants)

**Points to remember**

• In thermodynamics heat and work are not state variables, whereas internal energy is a state variable.

• Pressure, volume, temperature and mass are state variables. Heat and work are not state variables.

• A graphical representation of the state of a system with the help of two thermodynamic variables is called indicator diagram.

**Question: Type of ideal gas which will have the largest value for Cp– Cv**

**(A) Monoatomic**

**(B) Diatomic**

**(C) Polyatomic**

**(D) The value will be the same for all.**

Answer: (D) The value will be the same for all

**Question: A constant volume gas thermometer works on**

**(A) the principle of Archimedes**

**(B) Boyle’s law**

**(C) Pascal’s law**

**(D) Charle’s law**

Answer: (D) Charle’s law

**Question: Two moles of an ideal monatomic gas occupies a volume V at 27 degree C. The gas expands adiabatically to a volume 2V. Calculate (1) the final temperature and (2) change in its internal energy.**

**(1) (a) 189 K (b) 2.7 kJ**

**(2) (a) 195 K (b) 2.7 kJ**

**(3) (a) 189 K (b) -2.7 kJ**

**(4) (a) 195 K (b) -2.7 kJ**

Answer: (3) (a) 189 K (b) -2.7 kJ

Physics has a tough reputation. It’s less memorization-based than many other technical subjects. Here are a few points students can go through to learn the subject.

• Study with another student or in a small group.

• Pay attention in calss, take good notes, and focus on understanding.

• Work on problems that seem confusing, but do not spend too much time on one problem.

• Practice conceptual questions presented in class and at the end of each chapter.

• Do not memorize, understand the concept.

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