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JEE Advanced 2023 Revision Notes for Thermal Physics

Last updated date: 18th May 2024
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JEE Advanced 2023 Revision Notes for Thermal Physics - Free PDF Download

Thermal physics is an excellent segment of the higher secondary Physics syllabus. This chapter explains the concepts related to heat, heat transfer, temperature, statistical mechanics, thermodynamics, kinetic theory, etc. It deals with the transfer of heat from one body to the other. This chapter will also explain what thermal energy is and how it is transmitted in different energy forms. The laws and derivations will also be explained in this chapter. To understand the concepts, Thermal Physics notes for JEE Advanced will be the best study material to refer to.


JEE Advanced Revision Notes


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JEE Advanced

Chapter Name:

Thermal Physics

Academic Session:



English Medium



Available Material:

Chapter-wise Revision Notes with PDF

These revision notes are compiled by the experts to provide a better understanding of the fundamental concepts of thermal physics. Utilising these notes will help students to study and recall theories, laws, definitions, derivations, etc in no time.

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Access JEE Advanced Revision Notes Physics Thermal Physics


  • Heat is a type of thermal energy that flows from a higher temperature body to a lower temperature body when they come into contact.

  • A body's heat, or thermal energy, is the sum of all of its constituent particles' kinetic energies due to translational, vibrational, and rotational motion.

  • We know that, Joule (J) is the SI unit of heat energy.

  • The calorie is the most common unit of heat energy. We know that,  $1{\text{cal}} = 4.18{\text{J}}$

  • A calorie is the amount of heat required to raise the temperature of 1g of water by 1 0C.

  • We know that, Mechanical energy or work (W) can be converted into heat (Q) by $1{\text{W}} = {\text{JQ}}$

Here $J$= Joule's mechanical equivalent of heat. 

  • ${\text{J}}$ is a conversion factor (not a physical quantity) and its value is $4.186{\text{J}}/{\text{cal}}$.


  • The degree of hotness or coldness of a body is referred to as its temperature. A thermometer is a device that is used to measure the temperature.

  • We know that, highest possible temperature achieved in laboratory is about 108 K while lowest possible temperature attained is 10-8 K.

  • Cryogenics is a branch of physics concerned with the production and measurement of temperatures below 0 degrees Celsius, whereas pyrometer is concerned with the measurement of extremely high temperatures. The sun's core temperature is 107 degrees Fahrenheit, while its surface temperature is 600 degrees Fahrenheit.

  • NTP or STP implies $273.15{\text{K}}\left( {{0^\circ }{\text{C}} = {{32}^\circ }{\text{F}}} \right)$.

Different Scale of Temperature

  • Celsius Scale: In Celsius scale of temperature, the melting point of ice is taken as ${0^\circ }{\text{C}}$ & ${100^\circ }{\text{C}}$ as the boiling point of water & space between these two points is divided into 100 equal parts.

  • Fahrenheit Scale: In Fahrenheit scale of temperature, the melting point of ice is taken as ${32^\circ }{\text{F}}$ & 211 $^\circ {\text{F}}$ as the boiling point of water & the space between these two points is divided into 180 equal parts.

  • Kelvin scale: In Kelvin scale of temperature, the melting point of ice is taken as $273{\text{K}}$ & $373{\text{K}}$ as the boiling point of water & the space between these two points is divided into 100 equal parts.

Relation Between Different Scales of Temperatures

$\dfrac{C}{{100}} = \dfrac{{F - 32}}{{180}} = \dfrac{{K - 273}}{{100}} = \dfrac{R}{{80}}$


Thermometric Property

A thermometric property is a property of an object that changes with temperature. The following are examples of thermometers with various thermometric properties:

  1. Pressure of a Gas at Constant Volume:

$\dfrac{{{p_1}}}{{{T_1}}}{\text{ }} = \dfrac{{{p_2}}}{{{T_2}}}$

${p_t}{\text{ }} = {p_0}\left( {1 + \dfrac{t}{{273}}} \right)$

$t{\text{ }} = {\left( {\dfrac{{{p_t} - {p_0}}}{{{p_{100}} - {p_0}}} \times 100} \right)^\circ }{\text{C}}$

Here,$p,{p_{100}}$ & ${p_t}$ are pressure of a gas at constant volume ${0^\circ }{\text{C}},{100^\circ }{\text{C}}$ and${t^\circ }{\text{C}}$.

A constant volume gas thermometer can measure temperature from $ - {200^\circ }{\text{C}}$ to${500^\circ }{\text{C}}$.

  1. Electrical Resistance of Metals

    1. ${R_t} = {R_0}\left( {1 + at + \beta {t^2}} \right)$

Here  $\alpha $ and $\beta $ are constants for a metal.

Then $\beta $ is too small therefore we can take

  1. ${R_t} = {R_0}(1 + at)$

Here, $\alpha  =$  temperature coefficient of resistance and ${R_0}$ & ${R_{{t^\prime }}}$ are electrical resistances at ${0^\circ }{\text{C}}$ & ${t^\circ }{\text{C}}$.

  1. $\alpha  = \dfrac{{{R_2} - {R_1}}}{{{R_1}{t_2} - {R_2}{t_1}}}$

Here ${R_1}$ & ${R_2}$ are electrical resistances at temperatures ${t_1}$ & ${t_2}$.

  1. $t = \dfrac{{{R_t} - {R_0}}}{{{R_{100}} - {R_0}}} \times {100^\circ }{\text{C}}$

Here ${R_{100}}$ is the resistance at${100^\circ }{\text{C}}$.

We know that, Platinum resistance thermometer can measure temperature from $ - {200^\circ }{\text{C}}$ to ${1200^\circ }{\text{C}}$.

  1. Length of Mercury Column in a Capillary Tube

${I_t} = {I_0}(1 + \alpha t)$

Here $\alpha  =$  coefficient of linear expansion and ${I_0},{I_t}$ are lengths of mercury column at ${0^\circ }{\text{C}}$ & ${t^\circ }{\text{C}}$.

  1. Thermo Electro Motive Force

    1. When two thermocouple junctions are kept at different temperatures, a thermo-emf is created between them, which varies with the temperature difference between them.

    2. Thermo-emf

$E = at + b{t^2}$

Here $a$ and $b$ are constants for the pair of metals.

  1. We know that, Unknown temperature of hot junction when cold junction is at ${0^\circ }{\text{C}}$.

$t = {\left( {\dfrac{{{E_t}}}{{{E_{100}}}} \times 100} \right)^\circ }{\text{C}}$

Here ${E_{100}}$ is the thermo-emf when hot junction is at${100^\circ }{\text{C}}$.

We know that, A thermo-couple thermometer can measure temperature from $ - {200^\circ }{\text{C}}$ to${1600^\circ }{\text{C}}$.

  1. Thermal Equilibrium

The bodies are said to be in thermal equilibrium when there is no heat transfer between them when they are in contact.

Specific Heat:

  • The specific heat of a substance is the amount of heat required to raise the temperature of a unit mass of the substance.

  • Specific heat can be denoted by c or s.

  • Specific heat SI unit is joule/kilogram- $^\circ {{\text{C}}^\prime }\left( {J/{\text{kg}}{ - ^\circ }{\text{C}}} \right)$. Specific heat dimensions are$\left[ {{{\text{L}}^2}{{\text{T}}^{ - 2}}{\theta ^{- 1}}} \right]$.

  • We know that, the specific heat of water is $4200{\text{J}}{\text{k}}{{\text{g}}^{ - 10}}{{\text{C}}^{ - 1}}$ or$1{\text{cal}}{{\text{g}}^{ - 1}}{{\text{C}}^{ - 1}}$, which high compared with most other substances.

  • Gases Have Two Types of Specific Heat

    • At constant volume, the specific heat capacity$\left( {{C_v}} \right)$.

    • At constant pressure, the specific heat capacity (C).

    • We know that, Specific heat at constant pressure $\left( {{C_p}} \right)$ is greater than specific heat constant volume$\left( {{C_v}} \right)$, i.e., ${C_p} > {C_v}$.

    • For molar specific heats we can write  ${C_p} - {C_v} = R$

Here $R =$  gas constant and this relation are called Mayer's formula.

  • The ratio of a gas's two main specific heats is denoted by$\gamma $

$\gamma  = \dfrac{{{C_p}}}{{{C_V}}}$

  • We know that, the value of $\gamma $ depends on atomicity of the gas.

  • The ratio of a gas's two primary specific heats is written as $Q = mc\Delta t$, Here, m= mass of the substance, $c =$ specific heat of the substance and $\Delta {\text{t}} =$  change in temperature.

Thermal (Heat) Capacity

  • A body's heat capacity is equal to the amount of heat energy required to raise its temperature through${1^\circ }{\text{C}}$.

  • We know that, Heat capacity =$ mc \Delta {T}$

Here $c =$  specific heat of the substance of the body and $m =$  mass of the body.

  • Heat capacity SI unit is joule/kelvin $({\text{J}}/{\text{K}})$.

Water Equivalent

  • It is the amount of water whose thermal capacity is equal to the body's heat capacity. It is represented by the letter W.

  • Then we know that, ${\text{W}} = {\text{ms}} =$  Heat capacity of the body.

  • Specific heat of water is 1cal/g-$^\circ C$  or 4200 J/kg-K.

  • Specific heat of ice is 0.5 cal/g-$^\circ C$  or 2100 J/kg-K.

Change in Phase:

A substance can exist in three possible phases viz., solid, liquid and gas. Transition from one phase to another are accompanied by the absorption or liberation of heat and usually by change in volume, even when the transition occurs at constant temperature.

Latent Heat:

  • Latent heat is the heat energy absorbed or released at constant temperature per unit mass for state change.

  • The heat energy absorbed or released during a state change is given by ${\text{Q}} = {\text{mL}}$

  • Here ${\text{m}} =$  mass of the substance and ${\text{L}} =$  latent heat.

  • Latent heat unit is \[cal/g\]or $J/kg$ and its dimension is$\left[ {{M^0}{L^2}{T^{ - 2}}} \right]$.

  • For water at its normal boiling point or condensation temperature $\left( {{{100}^\circ }{\text{C}}} \right)$, the latent heat of vaporization is $L = 540{\text{cal}}/{\text{g}}$$= 40.8{\text{kJ}}/{\text{mol}}$ $= 2260{\text{kJ}}/{\text{kg}}$

  • For water at its normal freezing or melting point $\left( {{0^\circ }{\text{C}}} \right)$, the latent heat of fusion is ${\text{L}} = 80{\text{cal}}/{\text{g}} = 60{\text{kJ}}/{\text{mol}}$$= 336{\text{kJ}}/{\text{kg}}$

  • It is more painful to get burnt by steam rather than by boiling was ${100^\circ }{\text{C}}$ gets converted to water at ${100^\circ }{\text{C}}$, then it gives out 536 ${\text{cal}}/{\text{gm}}$ heat. So, it is clear that steam at ${100^\circ }{\text{C}}$ has more heat than wat ${100^\circ }{\text{C}}$ (i.e., boiling of water).

  • When snow falls, the temperature of the atmosphere drops dramatically because the snow absorbs heat from the atmosphere. So, when snow falls in the mountains, one does not feel too cold, but when ice melts, he feels too cold.

  • Ice cream has a stronger shivering effect on teeth than water (obtained from ice). This is because ice cream absorbs a lot of heat from your teeth when you eat it.


  • Melting is the process of converting a solid into a liquid at a constant temperature.


  • Evaporation is the conversion of liquid into vapour at any temperature (even below the boiling point).


  • When a liquid is gradually heated, at a certain temperature, the saturated vapour pressure of the liquid equals atmospheric pressure, and bubbles of vapour rise to the liquid's surface. This is referred to as liquid boiling.

  • The boiling point of a liquid is the temperature at which it begins to boil. The boiling point of water increases as pressure increases and decreases as pressure decreases.


  • Sublimation is the process of converting a solid into a vapour.

Hoar Frost:

  • Hoar frost refers to the transformation of vapours into solids.


  • This is the branch of heat transfer concerned with measure heat. Calories or kilocalories are commonly used to measure heat.

Principle of Calorimetry

  • When a hot body and a cold body are mixed, the heat lost by the hot body equals the heat gained by the cold body.

  • Heat lost = Heat gain

Thermal Expansion

Thermal expansion refers to the increase in size caused by heating. Thermal expansion can be classified into three types.

  1. Expansion of solids

  2. Expansion of liquids

  3. Expansion of gases

Expansion of Solids

  • There are three types of expansion that occur in solids.

  • Linear expansion is the term used to describe the expansion of length as a result of heating.

  • We know that, Increase in length ${I_2} = {I_1}(1 + a\Delta t)$

  • Here, ${I_1}$ and ${I_2}$ are initial and final lengths, $\Delta {\text{t}} =$  change in temperature and $\alpha  =$  coefficient of linear expansion.

  • We know that, Coefficient of linear expansion $\alpha  = \left( {\dfrac{{\Delta l}}{l} \times \Delta t} \right)$

  • Here $l =$  real length and $\Delta l =$  change in length and $\Delta {\text{t}} =$  change in temperature.

Superficial Expansion: 

  • The term "superficial expansion" refers to the expansion of an area as a result of heating.

  • We know that, Increase in area ${{\text{A}}_2} = {{\text{A}}_1}(1 + \beta \Delta {\text{t}})$

  • Here, ${A_1}$ and ${A_2}$ are initial and final areas and $\beta $ is a coefficient of superficial expansion.

  • We know that, Coefficient of superficial expansion $\beta  = \left( {\dfrac{{\Delta A}}{A} \times \Delta t} \right)$

  • Here, ${\text{A}} =$  Area, $\Delta {\text{A}} =$  change in area and $\Delta {\text{t}} =$  change in temperature.

Cubical Expansion:

  • Cubical expansion refers to the expansion of volume as a result of heating.

  • We know that, Increase in volume ${V_2} = {V_1}(1 + \gamma \Delta t)$

  • Here ${V_1}$ and ${V_2}$ are initial and final volumes and $\gamma $ is a coefficient of cubical expansion.

  • We know that, Coefficient of cubical expansion $\gamma  = \dfrac{{\Delta V}}{{V \times \Delta t}}$

  • Here $V =$  real volume, $\Delta V =$  change in volume and $\Delta t =$  change in temperature. Relation between coefficients of linear, superficial and cubical expansions $\beta  = 2\alpha $ and $\gamma  = 3\alpha $ Or $\alpha :\beta :\gamma  = 1:2:3$

Expansion of Liquids

Only volume expansion occurs in liquids when heated.

  1. Apparent Expansion of Liquids: When the expansion of a container    containing liquid on heating is not taken into account, the observed expansion is referred to as apparent expansion of liquids.

    1. We know that, Coefficient of apparent expansion of a liquid $\left( {{\gamma _a}} \right) = \dfrac{{{\text{ apparent increase in volume }}}}{{{\text{ original volume }} \times {\text{ rise in temperature }}}}$

  2. Actual Expansion of Liquids When the expansion of the container containing the liquid on heating is considered, the observed expansion is referred to as real expansion of liquids.

    1. We know that, Coefficient of real expansion of a liquid

$\left( {{\gamma _r}} \right) = \dfrac{{{\text{ real increase in volume }}}}{{{\text{ original volume }} \times {\text{ rise in temperature }}}}$

Here, Both, ${\gamma _r}$, and ${\gamma _{\text{a}}}$ are measured in $^\circ {{\text{C}}^{ - 1}}$.

  1. We can show that ${\gamma _r} = {\gamma _{\text{a}}} + {\gamma _g}$

Here, ${\gamma _r}$ and ${\gamma _{\text{a}}}$ are coefficient of real and apparent expansion of liquids and ${\gamma _{\text{g}}}$ is coefficient of cubical expansion of the container.

Anomalous Expansion of Water

  • If temperature of water is increased from ${0^\circ }{\text{C}}$, then its volume decreases up to ${4^\circ }{\text{C}}$, becomes minimum at ${4^\circ }{\text{C}}$ and then increases. Behavior of water around ${4^\circ }{\text{C}}$ is called, anomalous expansion water.

Expansion of Gases

  • In gases, there are two types of coefficients of expansion.

    • Volume Coefficient (${\gamma_V}$): 

The change in volume per unit volume per degree Celsius at constant pressure is known as the volume coefficient.

${\gamma _V} = \dfrac{{{V_2} - {V_1}}}{{{V_0}\left( {{t_2} - {t_1}} \right)}}$.

Here${{\text{V}}_0},{{\text{V}}_1}$, and ${{\text{V}}_2}$ are volumes of the gas at ${0^\circ }{\text{C}},{{\text{t}}_1}^\circ {\text{C}}$ & ${{\text{t}}_2}^\circ {\text{C}}$.

  • Pressure Coefficient${\gamma_p}$: 

The change in pressure per unit pressure per degree Celsius at constant volume is referred to as the pressure coefficient.

${\gamma _p} = \dfrac{{{p_2} - {p_1}}}{{{p_0}\left( {{t_2} - {t_1}} \right)}}$

Here  ${p_0},{p_1}$ and ${p_2}$ are pressure of the gas at ${0^\circ }{\text{C}},{t_1}^\circ {\text{C}}$ and ${t_2}^\circ {\text{C}}$.

Practical Applications of Expansion

  1. Space is left between the ends of two rails when rails are laid down on the ground.

  2. The transmission cables are loosely attached to the poles.

  3. The iron rim that goes on a cart wheel is always slightly smaller in diameter than the wheel itself.

  4. A glass stopper stuck in the neck of a glass bottle can be removed by warming the bottle's neck.

Important Points

  • A pendulum clock becomes slow and loses time in the summer due to an increase in its time period.

  • We know that, Loss of time in a time period $\Delta T = \dfrac {1}{2} a\Delta \theta T$.  The loss of time in any given time interval t can be expressed as 

$\Delta T = \dfrac {1}{2} \alpha \Delta \theta t$

  • At higher temperatures, a scale will expand and scale readings will be lower than true values, resulting in

True value = scale reading $(1 + \alpha \Delta t)$

Here, $\Delta $t is the temperature difference.

However, at lower temperatures, the reading will be higher or the true value will be lower.

Latent Heat of Fusion: 

The total amount of heat required to convert unit mass of solid into liquid at its melting point is called latent heat of fusion. 

Latent Heat of Vapourisation: 

The total amount of heat required to convert unit mass of liquid into vapour at its boiling point is called latent heat of vapourisation.

If Q is the amount of heat absorbed or liberated by m amount of substance, then L is given by Q = mL.

Latent heat of fusion of ice is 80 cal/g or 336 kJ/kg. 

Latent heat of vapourisation of water is 540 cal/g or 2259 kJ/kg at one atmospheric pressure.

Kinetic Theory of Gases:

According the Kinetic Theory of Gases:

  • The gas is composed of small indivisible large number of particles called molecules. The properties of the individual molecules are same as that of the gas as a whole.

  • The size of molecule is negligibly small in comparison to the average distance between the molecules.

  • Intermolecular forces between the molecules are negligibly small. Thus, all internal energy of the gas is kinetic.

  • Effect of gravity on them is negligibly small and Newton laws are valid for their motion.

  • The molecules of the gas are in a state of continuous random motion, moving in all directions with all possible velocities.

  • The collision between the molecules and with the walls of the container are perfectly elastic. Between two collision a molecule moves in a straight path with a uniform velocity. The average distance moved by any molecule between two successive collisions is called mean free path.

  • The time of collision is negligibly small in comparison to the time of the free path between the molecules.

Boyle’s Law: 

According to the Boyle’s law the volume of a given mass of a gas is inversely proportional to its pressure, the temperature remaining constant.

⇒ PV = constant

Charle’s Law: 

According to the Charle’s law at constant pressure, the volume of a given amount (mass) of a gas increase or decreases by $\dfrac{1}{{273}}$  of volume at 0$^\circ C$  for each 1$^\circ C$  rise or fall of temperature.

$\Rightarrow \dfrac{V}{T}$  = constant

Gay-Lussac’s Law or Pressure Law: 

According to the pressure law at constant pressure, the pressure of a given amount (mass) of a gas increase or decreases by $\dfrac{1}{{273}}$  of volume at 0$^\circ C$  for each 1$^\circ C$  rise or fall of temperature.

$\Rightarrow \dfrac{P}{T}$  = constant

Ideal Gas Equation:

If we relate P, V and T all together by gas laws, we will get PV = nRT, where n is the number of moles of the gas and R is the universal gas constant.

R = 1.98 cal/mol-K = 8.314 J/mol-K = 0.082 lit-atm/mol-K

Vander Waal's Equation:

The ideal gas equation PV = nRT, can be used for real gases at high temperature and low pressure at which intermolecular forces are negligibly small. Vander Waal modified this equation so that it can be used for real gases at a wide range of temperature and pressure.

  1. Volume Correction: Due to finite size of molecule, some part of volume of gas is occupied by the molecules. So, the space available for the motion of molecules of gas will be slightly less than the volume of gas. Hence effective volume becomes (V – b).

  2. Pressure Correction: Due to intermolecular forces, molecules do not exert that force on the walls of the container as they would have exerted in absence of intermolecular forces.

The effective pressure as calculated is $\left( {P + \dfrac{a}{{{V^2}}}} \right)$

For n moles of gas, we have:

$\left( {P + \dfrac{a}{{{V^2}}}} \right)(V - nb) = nRT$

Dalton’s Law of Partial Pressure:

The total amount of pressure exerted by a number of non-reacting gases when taken together in a container is equal to the sum of their partial pressures.

Partial pressure of a gas = (Mole fraction of that gas) × (Total pressure)

Root Mean Square Speed:

It is defined as the square root of mean of squares of the speed of different molecules.

According to kinetic theory of gases, ${v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $

Average Speed:

It is defined as the arithmetic mean of the speeds of the molecules of a gas at a given temperature.

According to kinetic theory of gases, ${v_{avg}} = \sqrt {\dfrac{{8RT}}{{\pi M}}} $

Most Probable Speed:

It is the speed with which the largest number of molecules in a gas are moving.

According to kinetic theory of gases, 

${v_{mp}} = \sqrt {\dfrac{{2RT}}{M}} $

Note: ${v_{rms}} > {v_{avg}} > {v_{mp}}$

Degrees of Freedom:

The degrees of freedom are the minimum number of information required to know about the system.

The independent possible motions are translational, rotational and vibrational, so there are three types of degrees of freedom.

  1. Translational Degrees of Freedom: The maximum number of translational degrees of freedom can be three.

  2. Rotational Degrees of Freedom: The maximum number of rotational degrees of freedom can be three.

  3. Vibrational Degrees of Freedom: Their numbers depend on atoms in the molecule and their arrangement. These degrees of freedom are considered at a very high temperature.

  4. Monoatomic Gas: 

    1. The degrees of freedom of monoatomic gas molecules are due to three independent translational motions along x, y, and z-axis. 

    2. So, the number of degrees of freedom of a monoatomic gas is 3.

  5. Diatomic or Triatomic Linear Gas: 

    1. The degrees of freedom of diatomic gas molecules are due to translation and rotation both. 

    2. If the line joining the two atoms (particles) is taken as the z-axis, then moment of inertia and hence rotational kinetic energy about z-axis becomes zero. 

    3. The molecule has three degrees of freedom of translation and two degrees of freedom of rotation. 

    4. So, the number of degrees of freedom of a Diatomic or triatomic linear gas is 5.

  6. Triatomic non-linear or Polyatomic Gas: 

    1. A non-linear molecule has non-zero moment of inertia about any axis, so there are three translational and three rotational degrees of freedom. 

    2. Total number of degrees of freedom are 6.

  7. Law of Equipartition of Energy: 

    1. For a system in thermal equilibrium, the total energy of a dynamic system is equally distributed among its various degrees of freedom. 

    2. The energy associated with each degree of freedom is $\dfrac{1}{2}RT$  per mole.

    3. Energy of a gas =$\dfrac{1}{2} fRT$, where f is the number of degrees of freedom.

Zeroth Law of Thermodynamics

Statement: If two bodies A and B are in thermal equilibrium with a third body C separately, then bodies A and B will be in thermal equilibrium with one another.

Triple Point of Water

  • The triple point of water is defined as the pressure and temperature at which water coexists in equilibrium in all three states of matter, namely ice, water, and vapour.

  • The triple point of pure water is at 0.01°C (273.16K, 32.01°F) and 4.58 mm (611.2Pa) of mercury and is used to calibrate thermometers.

First Law of Thermodynamics:

The first law of thermodynamics is basically based on the law of conservation of energy. According to this law the heat Q supplied to a system is equal to the sum of the change in internal energy of that system (∆U) and work done by the system (W).

Q = W + ∆U

Reversible and Irreversible Process:

  • Reversible Process: 

    • Any process which returns to its initial state of thermodynamic variables at each stage of variation as it proceeds in a direct process is called a reversible process. 

    • A complete reversible process is an ideal concept as it can never be realised in practice because dissipative forces cannot be completely eliminated.

  • Irreversible Process: 

    • A process is called irreversible if it cannot be retracted back exactly in the opposite direction. 

    • During an irreversible process, work is always done to overcome friction or other non-conservative forces. Energy is also dissipated. 

    • Most of the processes occurring in nature are irreversible.

Second Law of Thermodynamics:

Kelvin-Plank Statement: It is impossible to construct an engine that can convert heat totally into work without producing any other effect. According to this statement, the efficiency of any heat engine always be less than 100%.

Clausius Statement: For a self-acting machine or engine, it is impossible to transfer heat from a cold body to a relatively hotter body without the help of an external agency.

Newton's Law of Cooling:

  • If the temperature difference is minimal, Newton's rule of cooling states that the rate at which an exposed body changes temperature by radiation is roughly equal to the difference between the object's temperature and its surroundings.

  • Newton's law of cooling states that the rate of heat loss from a body is directly proportional to the temperature differential between the body and its surroundings.

Newton's Law of Cooling Formula

  • More quickly heat is transmitted, or more quickly the body's temperature changes, the greater the temperature difference between the system and its surroundings. The formula for Newton's law of cooling is as follows:

$\mathrm{T}(\mathrm{t})=\mathrm{T}_{\mathrm{s}}+\left(\mathrm{T}_{0}-\mathrm{T}_{\mathrm{s}}\right) \mathrm{e}^{-k t}$

Here,$t=$ time,, $T(t)=$ temperature of the given body at time $t$,, $\mathrm{T}_{\mathrm{s}}=$ surrounding temperature,$T_{0}=$ initial temperature of the body,$k=$ constant.

List of Important Formulae

  1. Specific Heat Capacity

${\text{s}} = \dfrac{{\text{Q}}}{{{\text{m}}\Delta \theta }}$

unit $ \to {\text{J}}/{\text{kg}}{ - ^\circ }{\text{C}}($ S. I. system)

$ \to {\text{cal}}/{\text{g}}{{\text{m}}^ - }^\circ {\text{C}}$

  1. Heat Capacity (C)

${\text{C}} = {\text{ms}} = \dfrac{{\text{Q}}}{{\Delta \theta }}$

  1. Water Equivalent 

${\text{W}} = {\text{ms}}$

  1. Latent Heat

    1. The heat required to change the unit mass of a solid to its liquid state at its melting point is called Latent Heat of Fusion. For changing from ice to water 

${L_f} = 80{\text{cal}}/{\text{gm}}$.

  1. The heat required to change the unit mass of the liquid to its vapour state at its boiling points is called Latent Heat of Vaporization. For changing from water to steam 

${L_v} = 540{\text{cal}}/{\text{gm}}$.

  1. $Q = mL$

  1. Principle of Calorimetry

Heat released by I substance $\Rightarrow $ Heat absorbed by II substance

${{\text{m}}_1}{{\text{s}}_1}\left( {{{\text{T}}_1} - {\text{T}}} \right) = {{\text{m}}_2}{{\text{s}}_2}\left( {T - {{\text{T}}_1}} \right)$

Final temp $T = \dfrac{{{m_1}{s_1}{T_1} + {m_2}{s_2}{T_2}}}{{{m_1}{s_1} + {m_2}{s_2}}}$

  1. Solution Technique to Find Specific Heat of a Solid 

${S^\prime } = \dfrac{{\left( {{m_1} + W} \right)s\left( {\theta  - {\theta _1}} \right)}}{{{m_2}\left( {{\theta _2} - \theta } \right)}}$

  1. To find the Latent Heat of Vaporization

$L = \dfrac{{\left( {{m_1} + W} \right)s\left( {\theta  - {\theta _1}} \right)}}{{{m_2}}} - s\left( {{\theta _2} - \theta } \right)$

  1. Mechanical Equivalent of Heat (J)

${\text{W}} = {\text{JH}}$ Unit of ${\text{J}} =$  Joule/cal.

$ \uparrow  \uparrow J = 4.2{\text{J}}/{\text{cal}}.$

  1. Conversion of Temp. Scale

$\dfrac{{{{\text{T}}_{\text{C}}} - 0}}{{100}} = \dfrac{{{{\text{T}}_{\text{F}}} - 32}}{{180}} = \dfrac{{{{\text{T}}_{\text{R}}} - 0}}{{80}} = \dfrac{{{{\text{T}}_{\text{K}}} - 273}}{{100}}$

$\Rightarrow \dfrac{{{T_C}}}{5} = \dfrac{{{T_{{F^ - }}} - 32}}{9} = \dfrac{{{T_R}}}{4} = \dfrac{{{T_{{K^K}}} - 273}}{5}$

$\dfrac{{\Delta {{\text{T}}_C}}}{5} = \dfrac{{\Delta {{\text{T}}_{\text{F}}}}}{9} = \dfrac{{\Delta {{\text{T}}_{\text{R}}}}}{4} = \dfrac{{\Delta {{\text{T}}_{\text{K}}}}}{5}$

$\Delta  \to $ change

${{\text{T}}_{\text{C}}} \to $ Celsius

${{\text{T}}_{\text{F}}} \to $ Fahrenheit

${{\text{T}}_{\text{R}}} \to $ Reaumer

${{\text{T}}_{\text{K}}} \to $ Kelvin

Numerical Examples

  1. The 20 -gram water at ${80^\circ }{\text{C}}$ is mixed with 20 -gram honey at ${60^\circ }{\text{C}}$. Find the common temperature of the mixture (specific heat capacity of water is $1{\text{cal}}/{{\text{g}}^\circ }{\text{C}}$ and specific heat of honey is $\left. {4{\text{cal}}/{{\text{g}}^\circ }{\text{C}}} \right)$.

(a). ${60^\circ }{\text{C}}$

(b). ${64^\circ }{\text{C}}$

(c). ${70^\circ }{\text{C}}$

(d). ${80^\circ }{\text{C}}$

Answer: ‘b’.

Complete solution:

According to question given that,

Mass of water $\left( {{m_w}} \right) = 20{\text{gm}}$

The temperature of water $\left( {{T_W}} \right) = {80^\circ }{\text{C}}$

Specific heat capacity of water $\left( {{{\text{S}}_{\text{W}}}} \right) = 1{\text{cal}}/{{\text{g}}^\circ }{\text{C}}$

Mass of honey $\left( {{{\text{m}}_H}} \right) = 20{\text{gm}}$

The temperature of honey $\left( {{{\text{T}}_{\text{H}}}} \right) = {60^\circ }{\text{C}}$

Specific heat capacity of honey $\left( {{{\text{S}}_{\text{H}}}} \right) = 4{\text{cal}}/{{\text{g}}^\circ }{\text{C}}$

Let ${\text{T}}$ is final temperature:

The heat is released as the water cools down from ${80^\circ }{\text{C}}$ to ${{\text{T}}^\circ }{\text{C}}$.

${\text{Q}} = {\text{m}}s\Delta \theta [{\text{Q}} =$  heat, ${\text{m}} =$  mass of a body, ${\text{s}} =$  specific heat of a body and $\Delta \theta  =$  change in temperature]

$\Rightarrow Q = (20{\text{gm}}) \times \left( {1{\text{cal}}/{\text{g}}{{\text{m}}^\circ }{\text{C}}} \right) \times {(80 - {\text{T}})^\circ }{\text{C}}$

$\Rightarrow Q = (1600 - 20{\text{T}}){\text{cal}}$

Heat gained by the honey:

$\Rightarrow \Delta {\theta ^\prime } = {(T - 20)^\circ }{\text{C}}$

$\Rightarrow {Q^\prime } = {\text{m}}s\Delta \theta  = 20 \times 4 \times (T - 60) = (80T - 4800){\text{cal}}$${\text{ Heat gained }}\left( {{Q^\prime }} \right) = {\text{ heat loss }}(Q)$

$\Rightarrow (1600 - 20T) = (80T - 4800)$

$\Rightarrow T = {64^\circ }{\text{C}}$

$\therefore {\text{ Final temperature }}(T) = {64^\circ }{\text{C}}$

  1. The pressure of an ideal gas undergoing isothermal change is increased by $10\% $. The volume of the gas must decrease by about

(a) $0.1\% $

(b) $9\% $

(c) $10\% $

(d) $0.9\% $

Answer: ‘b’.

Complete solution:

An ideal gas is a gas whose molecules take up very little space and have no molecular interactions between them. It always follows a law known as the ideal gas equation.

$PV = nRT$

Here, ${\mathbf{P}}$ is pressure, ${\mathbf{V}}$ is the volume occupied by the gas, ${\mathbf{n}}$ is the number of moles of the gas and $T$ is the temperature of the gas. $R$ is the universal gas constant. 

An isothermal process is one in which the temperature remains constant. Here isothermal process is given

Temperature $(T) =$  constant

${\text{PV}} = {\text{nRT}} =$  constant

${{\text{P}}_1}{{\text{V}}_1} = {{\text{P}}_2}{{\text{V}}_2}\quad  \ldots $ (1)

${{\text{P}}_1}$ is initial pressure, ${{\text{P}}_2}$ is final pressure, ${{\text{V}}_1}$ is initial volume and ${{\text{V}}_2}$ is final pressure

Let's say initial pressure $\left( {{P_1}} \right) = P$ and initial volume $\left( {{V_1}} \right) = V$

Since pressure is increased by $10\% $

Final pressure $\left( {{{\text{P}}_1}} \right) = {\text{P}} + {\text{P}} \times 10\%  = 1.1{\text{P}}$

Use equation (1);

${P_1}{V_1} = {P_2}{V_2}$

$P \times {V_1} = 1.1P \times {V_2}$

${{\text{V}}_2} = {{\text{V}}_1}/1.1 = 10{\text{V}}/11$

Decrease in volume $(\Delta V) = {V_1} - {V_2} = V - 10V/11 = V/11$ $\% $ Decrease in volume $= \dfrac{{\Delta V}}{V} \times 100 = \dfrac{V}{{11V}} \times 100 = 9\% $

Common Mistakes Students Make When Studying For Exams

  1. While studying thermal physics for JEE preparation we should always start from basics, most of the students start with advanced thermal physics where they will find it difficult to solve due to lack of basics.

  2. While preparing for the Thermal physics chapter of JEE exam we should focus on very minute things such as value of calorie, conversion of temperatures, etc.. 

  3. After studying the entire chapter we should dedicate a good amount of time for the revision. Lack of revision will lead to confusions, it is always better to revise the parts we studied and make a note of it.

Importance of JEE Advanced Physics Revision Notes Thermal Physics

As mentioned earlier, this chapter will cover the key areas of heat energy and how it is transferred from one object to the other. It will explain how heat energy behaves in a system and how it travels. The laws of thermodynamics will be introduced with proper explanation and derivation of formulae.

This chapter will explain what entropy is and how it can be measured. The study of entropy will also enable the students to understand how a system performs or does work. Studying entropy will also help us to understand how it represents the disorder degree. Similarly, it will also explain how the presence of heat is the measure of temperature in heated bodies.

The chapter on Thermal Physics will also explain what enthalpy is and how it can be calculated. The establishment of the formula is done after a proper explanation of the scientific terms involved. Students will learn how to calculate enthalpy by understanding all the scientific terms involved.

This is just a glimpse of this unique chapter that explains how important it is in studying physics. This chapter is a significant part of all engineering courses studied these days. The JEE Advanced exam also gives it a higher priority and sets fundamental questions from this chapter to check the aptitude level of the aspirants. This is why it is smarter to use the Thermal Physics revision notes for JEE Advanced to comprehend the concepts better.

Benefits of Vedantu’s Thermal Physics Revision Notes

  • These notes will deliver a simpler version of the fundamental concepts of all scientific terms, laws, derivation of formulae, etc. for your understanding. You will find an easier medium to comprehend these concepts and complete studying this chapter faster.

  • The revision notes can be downloaded in PDF format. These notes can be used as excellent revision material as all the concepts are explained in a concise format. Rest assured that these notes will offer accurate knowledge to you to gain more confidence.

  • Use these notes to study and recall the principles of thermal physics before an exam. There is no need to make notes when you have this file with you.

  • Use the sample questions to check your preparation level and find out the sections you need to work on more.

Download Thermal Physics Revision Notes for JEE Advanced PDF

The revision notes are available in the form of a free PDF version. Add these notes to your study material and use them to revise this chapter better. Learn how the experts have easily explained the concepts. Find accurate answers to the fundamental questions by using the easier format of these notes to remember everything you have studied.

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FAQs on JEE Advanced 2023 Revision Notes for Thermal Physics

1. What is thermodynamics?

It is a crucial branch of Physics where we study the relation between heat and other forms of energy.

2. What determines the flow of heat?

The difference in temperature determines the flow of heat from one point to the other.

3. What is temperature?

Temperature is the measurement of the coldness or hotness of a body on different scales. It is measured in Kelvin, Celsius, or Fahrenheit.

4. When do the molecules of a body reach a perfect order?

According to the definition of entropy, when a body is at absolute zero temperature, all the molecules are arranged to maintain a perfect order.