Notes for School Exams Target IIT-JEE 2015 Thermodynamicsvidyadrishti.com/Thermodynamics II.pdf · Target IIT-JEE 2015 . Physics . Thermodynamics . Author: Pranjal K. Bharti (B. Tech.,

THERMODYNAMICS Author: Pranjal K. Bharti (B. Tech., IIT Kharagpur) www.vidyadrishti.org

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2013-2015

Notes for School Exams Target IIT-JEE 2015 Physics

Thermodynamics Author: Pranjal K. Bharti (B. Tech., IIT Kharagpur)

H. O. D. Physics, Concept Bokaro Centre

Mb: 7488044834

© 2007 P. K. Bharti All rights reserved.

www.vidyadrishti.org



Thermodynamics: Thermodynamics is the science of energy conversion involving heat and other forms of energy, most notably mechanical work. It studies and interrelates the macroscopic variables, such as temperature, volume and pressure, which describe physical, thermodynamic systems.

Heat: Heat always refers to the transfer of energy between systems (or bodies), not to energy contained within the systems (or bodies). • Heat is energy in transit. • Heat of a body or heat contained in a body is a

meaningless statement. • S.I. unit of heat: J (joules)

Temperature: Loosely speaking, temperature is the degree of hotness or coldness of a body or environment.

• A body that feels hot usually has a higher temperature than a similar body that feels cold.

• On the microscopic scale, temperature is defined as the average kinetic energy of microscopic motions of a single particle in the system per degree of freedom. We will learn about temperature at microscopic level in the chapter ‘Kinetic Theory of Gases’.

• On the macroscopic scale, temperature is the unique physical property that determines the direction of heat flow between two objects placed in thermal contact. If no heat flow occurs, the two objects have the same temperature; otherwise heat flows from the hotter object to the colder object.

NOTE: • Heat & temperature are two different physical

quantities. Heat is a form of energy which flows from a body at higher temperature to the body at lower temperature. Temperature is a measure of the internal energy of the system, while heat is a measure of how energy is transferred from one system (or body) to another.

THERMAL EQUILIBRIUM: Two bodies are said to be in thermal equilibrium if no transfer of takes place when they are placed in contact. Thus, two objects are in thermal equilibrium with each other when they are at the same temperature.

Answer the following question: Q1. All objects sitting in a room should reach the same temperature. Yet if you pick up a cup made of glass, it feels cooler than a cup made of plastic. Many people unconsciously recognize plastic by its relative "warm feel." How can two objects be the same temperature and yet one feels cooler? What mistaken assumption are we making?

Zeroth law of Thermodynamics • If objects A and B are separately in thermal

equilibrium with a third object C, then objects A and B are in thermal equilibrium with each other.

• In other words, states that if two bodies A and C have same temperature and B and C have also same temperature then A and B have also same temperature.

• We can think of temperature as the property that determines whether an object is in thermal equilibrium with other objects.

TEMPERATURE SCALE & THERMOMETERS There are three temperature scales in use today:

• Celsius Scale • Fahrenheit Scale • Kelvin Scale

Celsius Scale

• Some important temperatures on Fahrenheit Scale:

• Freezing point of water: 0o C

• Boiling point of water: 100 oC

• Human body temperature: 32 oC

• Room Temperature: 25 oC

• Absolute Zero : – 273.15 oC

Absolute Zero

Absolute Zero is the lowest possible temperature, at which

point the atoms of a substance transmit no thermal energy –

they are completely at rest. It is 0 degrees on the Kelvin scale,

which translates to -273.15 oC (or -459.67 degrees

Fahrenheit).

Fahrenheit Scale

• Some important temperatures on Fahrenheit Scale:

• Freezing point of water: 32 oF

• Boiling point of water: 212 oF

• Human body temperature: 98.6 oF

• Room Temperature: 77 oF

• Absolute Zero: – 459 oF

• NOTE: Fahrenheit Scale is used by doctors.

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Kelvin Scale

An absolute temperature scale is based on the definition that the volume of a gas at constant (low) pressure is directly proportional to temperature and that 100 degrees separates the freezing and boiling points of water. Some important temperatures on Fahrenheit Scale: • Freezing point of water: 273.15 K • Boiling point of water: 373.15 K • Human body temperature: 402.15 K • Room Temperature: 298.15 K • Absolute Zero: 0 K

NOTE: • We don’t use degree K (

oK) to denote temperature on

Kelvin scale. • There are no negative Kelvin temperatures. • Kelvin is used as S.I. unit of temperature. Relation between Kelvin, Celsius and Fahrenheit scales:

We can write

Kelvin Scale

Celcius Scale

Point of interest Freezing pointBoiling Point Freezing point

Point of interest Freezing pointBoiling Point Freezing point

Point of interest Freezing pointBoili

− −

−= −

−=

Fahrenheit Scaleng Point Freezing point −

273.15 0 32

373.15 273.15 100 0 212 32 K C F− − −

⇒ = =− − −

273.15 32

100 100 180 K C F −

=⇒−

=

Hence, we have

9 32 & 273.155

F C K C= + = +

Answer the following questions: Q2. What is the S.I. unit of temperature?

Q3. Convert 32 oF into Celsius and Kelvin scale.

Q4. Convert 100 oC into Fahrenheit and Kelvin scale.

Q5. Is it correct to say that temperature of a body is – 459 K ? Q6. Consider the following pairs of materials. Which pair represents two materials, one of which is twice as hot as the other? (a) boiling water at 100°C, a glass of water at 50°C (b) boiling water at 100°C, frozen methane at -50°C (c) an ice cube at -20°C, flames from a circus fire-eater at 233°C (d) No pair represents materials one of which is twice as hot as the other

Answers: 2. K

3. 32 oF = 0

oC = 273.15 K (Use formulae)

4. 100 oC = 212

oF = 373.15 K (Use formulae)

5. No, because there are no negative Kelvin temperatures. 6. (c) The phrase “twice as hot” refers to a ratio of temperatures in Kelvin scale. When the given temperatures are converted to Kelvins, only those in part (c) are in the correct ratio. Thermal expansion • Most materials expand when their temperatures increase.

When an object’s temperature is raised, every linear dimension increases in size. This phenomenon is known as thermal expansion.

• The term thermal expansion refers to the increase in size of an object as that object is heated. With relatively few exceptions, all objects expand when they are heated and contract when they are cooled.

• Perhaps the most important exception to this rule is water. Water contracts as it cools from its boiling point to about 4°C (39.2°F). At that point, it begins to expand as it cools further to its freezing point. This unusual effect explains the fact that ice is less dense than water.

• When an object is heated or cooled, it expands or contracts in all dimensions. However, for practical reasons, scientists and engineers often focus on two different kinds of expansion: linear expansion (expansion in one direction only) and volume expansion (expansion in all three dimensions). There is also third kind of expansion known as area expansion (expansion in two dimensions only).

Kelvin Scale

Celsius Scale

Fahrenheit Scale

Freezing point

Boiling point

Point of interest

212 o

F

32 o

F

F

100 o

C

0 o

C

C

273.15 K

373.15 K

K



Q7. Do holes become larger or smaller when heated? Answer: When an object’s temperature is raised, every linear dimension increases in size. This includes any holes in the material, which expand in the same way as if the hole were filled with the material. Keep in mind the notion of thermal expansion as being similar to a photographic enlargement. Linear expansion • A good example of linear expansion (expansion in one

direction only) railroad tracks distorted because of thermal expansion on a very hot July day.

• Most materials expand as they are heated and contract if they are cooled. Thus their length is a function of temperature. If the length of an object is L and the temperature changes by a differential amount dT, then the differential change in the length dL is given by:

dL LdTα= (linear expansion)

where α is a constant known as the coefficient of linear expansion.

Clearly,

dL

LdTα = (Coefficient of linear expansion)

• Thus we can define coefficient of linear expansion as

change in length per unit length of a solid when temperature changes by 1 K or 1° C at constant pressure.

• Sometimes we are interested in average linear expansion. • Suppose a rod of material has a length L at some initial

temperature T. When the temperature changes by ∆T, the length changes by ∆L. Experiments show that if ∆T is not too large, ∆L is directly proportional to ∆T.

• If two rods made of the same material have the same temperature change, but one is twice as long as the other, then the change in its length is also twice as great. Therefore ∆L must also be proportional to L.

• Thus average thermal expansion is given by:

L L Tα∆ = ∆ (Average linear thermal expansion)

• Hence,

L

L Tα ∆

=∆

(Average coefficient of linear expansion)

• Let the new length of rod after expansion or contraction becomes L’. Clearly, L’ = L + ∆L. Hence, new length L’ of rod becomes

( )' 1L L Tα= + ∆ Area expansion • If the area of an object is A and the temperature changes

by a differential amount dT, then the differential change in the area dA is given by:

dA AdTβ= (Volume expansion)

where β is a constant known as the coefficient of area expansion.

Clearly,

dA

AdTβ = (Coefficient of area expansion)

• Thus we can define coefficient of area expansion as

change in area per unit area of a solid when temperature changes by 1 K or 1° C at constant pressure.

• Sometimes we are interested in average area expansion. • Suppose area of material is A at some initial temperature

T. When the temperature changes by ∆T, the area changes by ∆Α. Experiments show that if ∆T is not too large, ∆A is directly proportional to ∆T.


A

A Tβ ∆

=∆

(Average coefficient of area expansion)

• Let the new area of the material after expansion or

contraction becomes A’. Clearly, A’ = A + ∆A. Hence, new area A’ becomes

( )' 1A A Tβ= + ∆

L+dL

L T

T+dT

L’=L+∆L

L T

T+∆T

T

A

A+ dA

T+dT



Volume expansion • If the volume of an object is V and the temperature

changes by a differential amount dT, then the differential change in the volume dV is given by:

dV VdTγ= (Volume expansion)

where γ is a constant known as the coefficient of volume expansion.

Clearly,

dV

VdTγ = (Coefficient of volume expansion)

• Thus we can define coefficient of volume expansion as

change in volume per unit volume of a solid when temperature changes by 1 K or 1° C at constant pressure.

• Sometimes we are interested in average volume expansion.

• Suppose volume of material is V at some initial

temperature T. When the temperature changes by ∆T, the volume changes by ∆V. Experiments show that if ∆T is not too large, ∆V is directly proportional to ∆T.


V

V Tγ ∆

=∆

(Average coefficient of volume expansion)

• Let the new volume of the material after expansion or

contraction becomes V’. Clearly, V’ = V + ∆V. Hence, new area V’ becomes

( )' 1V V Tγ= + ∆

Relation between α and β Let us consider a square sheet of side L at temperature T.

Therefore, area of the square at temperature

A = L2 …(i)

When the temperature changes by ∆T, the edge changes by

∆L. Therefore, new area of square becomes

( )2 2 2 2A A L L L L L L+ ∆ = + ∆ = + ∆ + ⋅∆ …(ii)

As change in length ∆L is very small, its square ∆L2 must be

negligible. Therefore, neglecting ∆L2 from (ii) we get, 2 2A A L L L+ ∆ = + ⋅∆ …(iii)

Subtracting (i) from (iii), we get change in area ∆A of the

square plate, which is

2A L L∆ = ⋅∆ …(iv).

Now, using definition of coefficient of area expansion we get,

2

2 (using (i) and (iv))

2 ...(v)

AA T

L LL T

LL T

β

β

β

∆=

∆⋅∆

⇒ =∆∆

⇒ =∆

From definition of coefficient of linear expansion we get,

LL T

α ∆=

∆ …(vi)

Using (v) and (vi), we get

2β α= (Relation between α and β)

T T+dT

V V+dV

T

A

A+ dA

T+dT

L+∆L L



Relation between α and γ Let us consider a cubical body of edge L at temperature T.

Therefore, volume of the cube at temperature

V = L3 …(i)

When the temperature changes by ∆T, the edge changes by

∆L. Therefore, new volume of cube becomes

( )3V V L L+ ∆ = + ∆

3 2 2 33 3V V L L L L L L⇒ + ∆ = + ⋅∆ + ⋅∆ + ∆ …(ii)

As change in length ∆L is very small, its square ∆L2 and cube

∆L3 must be negligible. Therefore, neglecting ∆L2 and ∆L3

from (ii) we get, 3 3V V L L L+ ∆ = + ⋅∆ …(iii)

Subtracting (i) from (iii), we get change in volume ∆V of the

cube, which is

3V L L∆ = ⋅∆ …(iv).

Using definition of coefficient of volume expansion we get,

3

3 (using (i) and (iv))

3 ...(v)

VV T

L LL T

LL T

γ

γ

γ

∆=

∆⋅∆

⇒ =∆∆

⇒ =∆

From definition of coefficient of linear expansion we get,

LL T

α ∆=

∆ …(vi)

Using (v) and (vi), we get

3γ α= (Relation between α and γ)

Relation between α, β and γ

We have seen from last page that

β = 2α and γ = 3α Combining these two expressions, we get

2 3β γα = = (Relation between α, β and γ )

Thermal stress and strain • Do you know we can use concept of linear expansion to

find out strain developed in a rod? Using concept of Young’s modulus we can also find out stress. How?

• Suppose a uniform rod of natural length L is held fixed by clamping its two ends at temperature T. Clearly the tension is zero at this temperature T.

• Suppose we temperature is increased by ∆T. Thus, rod will also try to expand. But, as the rod is fixed at the clamps, its length remains the same as the length at T. Hence, strain develops in the rod.

• Suppose the rod length is increased by ∆L when

temperature is increased by temperature ∆T in the case when rod is not clamped at the ends.

• Assuming coefficient of linear expansion a to be constant over ∆T, we have

∆L = α L∆T …(i)

• Hence strain developed in the rod is

Change in lengthStrainInitial length

LL

∆= =

Hence from (i), we have

Strain L L T TL L

α α∆ ∆= = = ∆

• Hence, thermal strain is given by

Thermal strain Tα= ∆

• If Y is the Young’s modulus of the material, we can find

thermal stress as

Stress = (Young’s Modulus) x Strain

Thermal stress Y Tα⇒ = ∆

Note: Here stress & strain are longitudinal stress & longitudinal strain respectively.

T T+dT

L+∆L L



Specific heat capacity

The specific heat capacity (or specific heat) of a substance may be defined as the amount of heat required to raise the temperature of unit mass of the substance through one degree Celsius. It depends on the nature of the substance and its temperature. If an amount of heat ΔQ is needed to raise the temperature of m mass of a substance through ΔT, then specific heat is given by

Qcm T

∆=

× ∆ (Specific heat capacity)

The SI unit of specific heat is J kg–1 K–1 which is same as J kg–1 0C–1 and CGS unit is cal g–1 oC–1. Clearly, the amount of heat required to raise the temperature of m mass of a substance through ΔT is ΔQ = mc ΔT

Molar specific heat

The molar specific heat of a substance is defined as the amount of heat required to raise the temperature of one mole of the substance through one degree. It depends on the nature of the substance and its temperature. If an amount of heat ΔQ is required to raise the temperature of n moles of a substance through ΔT, then molar specific heat is given by

QCn T∆

=∆

(Molar Specific heat)

The SI unit of molar specific heat is J mol–1 K–1 which is same as J mol–1 0C–1 and CGS unit is cal mol–1 oC–1. Therefore, the amount of heat required to raise the temperature of n moles of a substance through ΔT is ΔQ = nC ΔT.

Heat capacity or thermal capacity

The heat capacity of a body is defined as the amount of heat required to raise its temperature through one degree. If an amount of heat ΔQ is required to raise the temperature of a substance through ΔT, then molar specific heat is given by

QST

∆=

∆ (Heat capacity)

By definition, heat capacity = mass × specific heat

S = mc

The SI unit of heat capacity is JK–1 and the CGS unit is JoC–1. Therefore, the amount of heat required to raise the temperature of a substance through ΔT is ΔQ = S ΔT

Water equivalent

The water equivalent of a body is defined as the mass of water which requires the same amount heat as is required by the given body for the same rise of temperature. Water equivalent = Mass × Specific heat

w = mc The SI unit of water equivalent is kg and the CGS unit is g.

Calorimetry

The branch of physics that deals with the measurement of heat is called calorimetry.

Principle of calorimetry or the law of mixtures

The principle of calorimetry states that the heat gained by the cold body must be equal to the heat lost by the hot body, provided there is no exchange of heat with the surroundings.

Heat gained = Heat lost

This principle is a consequence of the law of conservation of energy and useful for solving problems relating to calorimetry.

Few Definitions

Melting point: The temperature at which the solid and the liquid states of a substance coexist in thermal equilibrium with each other is called its melting point.

Boiling point: The temperature at which the liquid and vapour states of a substance coexist in thermal equilibrium with each other is called its boiling point. The boiling point of a substance at standard atmospheric pressure is called its normal boiling point.

Sublimation: The process of transition of a substance from the solid state to the vapour state without passing through the liquid state is called sublimation, and the substance is said to sublime. Substances like dry ice (solid CO2), iodine, naphthalene and camphor undergo sublimation when heated. During the sublimation process, the solid and vapour states of a substance coexist in thermal equilibrium with each other.

Note: The melting point of those substances which expand on melting (e.g., paraffin wax, phosphorus, sulphur, etc.) increases with the increase in pressure while the melting point of those substances which contract on melting (e.g., ice, cast iron, bismuth etc.) decreases with increase in pressure.



Regelation: The phenomenon in which ice melts when pressure is increased and again freezes when pressure is removed is called regelation (re = again; gelare = freeze):

This phenomenon of freezing is called regelation. Practical applications of regelation: (a) By pressing snow in our hand, we can transform it into a

snow-ball. When snow is pressed, its crystals melt. As the pressure is released, water refreezes forming a snow-ball.

(b) Skating is possible due to the formation of water layer below the skates. Water is formed due to the increase of pressure and it acts as a lubricant.

Latent Heat

The amount of heat required to change the state of unit mass of a substance at constant temperature and pressure is called latent heat of the substance. If m mass of a substance undergoes a change from one state to another, then the amount of heat required for the process is

ΔQ = mL

where L is the latent heat of the substance and is a characteristic of the substance. Its value also depends on the pressure. Clearly,

QLm

∆=

(Latent heat)

SI unit of latent heat is J kg–1 and CGS unit of latent heat is cal g–1.

Latent heat of fusion: The amount of heat required to change the state of unit mass of a substance from solid to liquid at its melting point is called latent heat of fusion or latent heat of melting. It is denoted by Lf.

Latent heat of vaporization: The amount of heat required to change the state of unit mass of a substance from liquid to vapour at its boiling point is called latent heat of vaporization or latent heat of boiling. It is denoted by Lv.

KINETIC THEORY OF GASES

Assumptions of Kinetic Theory of gases

1. All gases consists of molecules. The molecules of a gas are all alike and differ from those of other gases.

2. The molecules of a gas are very small in size as compared to the distance between them.

3. The molecules of a gas behave as perfectly elastic spheres.

4. The molecules are always in random motion. 5. During their random motion, the molecules collide against

one another and the walls of the containing vessel. The collisions of the molecules with one another and with the walls of the vessel are perfectly elastic. This implies that total kinetic energy is conserved. The total momentum is conserved as usual.

6. Between two collisions, a molecule moves along a straight line and the distance covered between two successive collisions is called the free path of the molecule.

7. The collisions are almost instantaneous i.e. the time during which a collision takes place is negligible as compared to the time taken by the molecule to cover the free path.

8. The molecules do not exert any force on each other except during collisions.

Notations used in this chapter: Avogadro number 23 16.02 10 molAN −= × Universal gas constant R = 8.314 J/K-mol Boltzmann constant 231.38 10 J/Kk −= × No. of moles n Total number of molecules N Mass of each molecule m Total mass of molecules M Molecular mass M0 Density ρ Average speed v Root mean square speed rmsv Most probable speed pv

No. of moles (n)

0

total massmolecular mass

Mn nM

= ⇒ =

total no. of moleculesAvogadro number A

Nn nN

= ⇒ =

Some useful relations

0 (total mass)M nM Nm= =

( )0 molecular massAM mN=

(density)A

M NV N

ρ = =

AR N k= Root mean square speed

22 2 2 & rms rms

vv v v v

N= = =∑



Equation of state of an ideal gas

PV = nRT

Pressure of an ideal gas

2 21 13 3rmsp v vρ ρ= =

Kinetic energy of an ideal gas

Total KE of the gas 21 32 2rmsK Mv nRT= =

Average KE of each molecule = 21 32 2rmsmv kT=

In thermal equilibrium KE of all molecules are equal

2 21 ,1 2 ,2

1 12 2rms rmsm v m v=

Important proportionality relations 2

rmsK v∝ (KE) 2

rmsT v∝ (Absolute temperature)

Relation between KE and pressure of an ideal gas 23

pV KE=

Speeds in terms of temperatures

0

3 3rms

kT RTvm M

= = (rms speed)

0

8 8kT RTvm Mπ π

= = (average or mean speed)

0

2 2p

kT RTvm M

= = (most probable speed)

Graham’s law of diffusion

Rate of diffusion, 1rmsr v

ρ∝ ∝

1 2

2 1

rr

ρρ

=

Dalton’s law of partial pressures

1 2 ... np p p p= + + +

Equation of state of a real gas (van der Waals equation)

( )2

ap V b nRTV

+ − =

Mean free path:

22kTd P

λπ

=

STP (standard temperature and pressure)

T = 273 K and P = 1 atm

Pressure exerted by a gas

Derivation of 21 3

P ρ υ= :

Consider a gas contained in a vessel. Let n be the number of gas molecules per unit volume inside the vessel and m be the mass of each molecule. Let v be the velocity of any gas molecule at any instant. Let v x, v y and v z are components of the velocity v along X, Y and Z-axis respectively. The molecule moves with momentum mvx along X-axis and strikes against the right face. Since the collision is perfectly elastic in nature, the molecule rebounds back with same speed vx. Therefore, change in momentum of the molecule along X-axis = −2mvx. Now, in time ∆ t, all those molecules will hit area A, therefore, such molecules lie in volume A × vx ∆ t. Since the number of molecules per unit volume is n, the number of such molecules is nAvx ∆ t. In fact, on the average, half of this number is expected to move along negative X-axis and the other half along positive X-axis. Therefore, The number of molecules hitting area A in time ∆ t along

positive X-axis = 12 xnAv t∆

And the total change in momentum of molecules in time ∆ t along X-axis,

( ) 2122x x x xp mv nAv t mnAv t∆ = − × ∆ = − ∆

Since different molecules or even the same molecule at different times may possess different speeds, in the above expression for ∆ px; it is reasonable to replace 2

xv by 2 ,xv where 2 ,xv is mean square component of speeds of the molecules

along X-axis. Therefore, 2

x xp mnAv t∆ = − ∆ Therefore, force exerted by walls of the vessel on the gas molecules along X-axis is given by

2xx x

pF mnAv

t∆

= = −∆

If px is pressure exerted by gas molecules along X-axis, then 2

2x xx x

F mnAvP mnv

A A= = =

If Py and Pz are pressures along Y-and Z-axis respectively, then 2

y yP mnv= and 2z zP mnv=

Pressure exerted by gas molecules must be same in all direction i.e. Px = Py = Pz = P (say)

In other words, ( )13 x y zP P P P= + +

( ) ( )2 2 2 2 2 21 13 3x y z x y zP mnv mnv mnv mn v v v⇒ = + + = + +

213

P mnv⇒ =



where 2 2 2 2

x y zv v v v= + + is mean square velocity of the gas molecules. Now, m n = ρ, density of the gas. Therefore,

213

P vρ=

Mean free path

The mean free path ( λ )of a gas may be defined as the average distance travelled by the gas molecule in two successive collisions.

22kTd P

λπ

= (mean free path)

where, λ = mean free path k = Boltzmann constant T = absolute temperature of the gas d = molecular diameter P = pressure of the gas

LAWS OF THERMODYNAMICS

Thermodynamics: Thermodynamics is the branch of science that deals with the concepts of heat and temperature and the inter-conversion of heat and other forms of energy.

Heat

Heat always refers to the transfer of energy between systems (or bodies), not to energy contained within the systems (or bodies). • Heat is energy in transit. • Heat of a body or heat contained in a body is a

meaningless statement. • S.I. unit of heat: J (joules) • It is represented by Q, ∆Q or dQ

Sign conventions used: • Heat absorbed by a system is positive. Heat given out by a

system is negative.

Equation of state of an ideal gas

The mathematical relation between the pressure, volume and temperature of a thermodynamics system is called its equation of state. For example, the equation of state for n moles of an ideal gas can be written as

PV = nRT (Equation of state of an ideal gas)

where, P = pressure V = volume n = number of moles R = universal gas constant T = temperature

Thermodynamics equilibrium: A system is said to be in the state of thermodynamic equilibrium if the macroscopic variables describing the thermodynamics state of the system do not change with time.

Thermal equilibrium: Two systems are said to be in thermal equilibrium with each other if they have the same temperature.

Zeroth law of thermodynamics. It states that if two systems A and B are separately in thermal equilibrium with a third system C, then A and B are also in thermal equilibrium with each other.

Concept of temperature: Zeroth law of thermodynamics implies that temperature is a physical quantity which has the same value for all systems which are in thermal equilibrium with each other.



Work done by a gas

• Work done for small change in volume: dW = pdV

• Total work done

2

1

area of the region between curve and axis

V

V

W pdV

PV V

=

= −

∫

Sign Convention for work

• Compression (volume decreases): – ve work

• Expansion (volume increases): + ve work

Expansion: + ve work Compression: – ve work

Work done for a cyclic process

Any process in which the system returns to its initial state after undergoing a series of changes in known as a cyclic process. + ve work (how?) – ve work (how?)

• Work done = area of shaded region (how?)

• Note:

Change in internal energy in a cyclic process = zero (We will study about internal energy later)

To prove 2

1

V

V

W pdV= ∫ :

Consider a gas contained in a cylinder of cross-sectional area A and provided with a frictionless movable piston. Let P be the pressure of the gas. Force exerted by the gas on the piston, F = PA Suppose the gas expands a little and pushes out the piston through a small distance dx. The work done by the gas is

dW = Fdx = PA dx

dW = P dV …(i)

where dV = Adx, is the change in volume of the gas. The total work done by the gas when its volume increase from V1 to V2 will be

2

1

.V

V

W dW PdV= =∫ ∫

Internal energy

The internal energy of a system is the sum of molecular kinetic and potential energies in the frame of reference relative to which the centre of mass of the system is at rest. Important points about internal energy: 1. Internal energy of an ideal gas is purely kinetic in nature.

In an ideal gas, there are no molecular forces of attraction. 2. Internal energy of an ideal gas is a function of

temperature. U = f(T)

3. Change in internal energy ∆U is zero in two cases a) Isothermal process: ∆U = 0 b) Cyclic process: ∆U = 0

First law of thermodynamics

According to the First Law of Thermodynamics, if some heat is supplied to a system which is capable of doing work, the heat energy (∆Q) supplied to the system goes in partly to increase the internal energy of the system (∆U) and the rest in work done on the environment (∆W).

∆U = Q – W (1st law of thermodynamics)

where, ∆U = change in internal energy Q = heat given to the system W = work done by the system

Gas

dx

PA

V

p

V1 V2

A C

B D

V

p

V

p

V

p

V

p

Q

W ∆U



First Law of Thermodynamics is simply the general law of conservation of energy applied to any system in which the energy transfer from or to the surroundings is taken into account.

Sign convention:

Heat given to the system = positive Heat lost by the system = negative

Work done by the system = positive Work done on the system = negative

Different kind of processes

1. Quasistatic process • Very slow process • Always in thermodynamic equilibrium

2. Reversible process • Any process which can be made to proceed in the

reserve direction by variation in its conditions such that any change occurring in any part of the direct process is exactly reversed in the corresponding part of reverse process is called a reversible process.

• Necessary conditions for a reversible process: i. Quasistatic process

ii. No dissipation of energy due to dissipative forces (viscosity, friction, inelasticity, etc.)

• Note: Unless otherwise stated all processes at this level is assumed to be reversible process.

3. Cyclic process • Any process in which the system returns to its initial

state after undergoing a series of changes in known as a cyclic process. • dU = 0 for a cyclic process

4. Isochoric process (volume constant)

• Equation of state for ideal gas: 1 2

1 2

p pT T

=

• Work done: W = 0

• 1st law of thermodynamics: ∆U = Q – W

∆U = Q

5. Isobaric process (pressure constant)

• Equation of state for ideal gas: 1 2

1 2

V VT T

=

• Work done:

( )2

1

2 1

V

V

W pdV p V V= = −∫

W = p∆V

6. Isothermal process (temperature constant)

• Equation of state for ideal gas:

1 21 1 2 2

2 1

p Vp V PVp V

= ⇒ =

• Change in internal energy: ∆U = 0


0 = Q – W Q = W

• Work done by ideal gas in isothermal process:

2 1

1 2

ln lnV pW nRT nRTV p

= =

(Work done by ideal gas in isothermal process)

Proof:

For an ideal gas undergoing isothermal process, we have

nRTpV nRT pV

= ⇒ =

Therefore, work done by gas is

1

nRT nRTW pdV dV pV V

W nRT dVV

= = =

=

∫ ∫

∫

Taking proper limits, we have

[ ] ( )2

2

11

2 1

2 2

1 1

1 ln ln ln

ln 2.303log

VV

VV

W nRT dV nRT V nRT V VV

V VW nRTV V

= = = −

⇒ = =

∫

Also for isotheral process

1 2

2 1

p Vp V

= .

Therefore

2 1

1 2

ln lnV pW nRT nRTV p

= =

7. Adiabatic process (No heat transfer)

• Q = 0


∆U = – W

• Work done by an ideal gas (proof is given later)

( )1 21 1 2 2

1 1nR T TPV PVW

γ γ−−

= =− −

where p

v

CC

γ =



SPECIFIC HEAT CAPACITIES OF GASES

Molar specific heat

• The molar specific heat of a substance is defined as the amount of heat required to raise the temperature of one mole of the substance through one degree. It depends on the nature of the substance and its temperature.

• If an amount of heat ΔQ is required to raise the temperature of n moles of a substance through ΔT, then molar specific heat is given by

QCn T∆

=∆

(Molar Specific heat)

• S.I. Unit: J/mol-K • Specific heat capacity depends on the process of the

system. There can be as many specific heat capacities as many the number of processes. We are generally interested in two types of specific heat of gases.

1. Specific heat capacity at constant volume (CV): The process is obtained at constant volume.

2. Specific heat capacity at constant pressure (CP): The process is obtained at constant pressure.

Specific heat capacity at constant volume, CCV

• Definition: Specific heat capacity at constant volume CV is defined as the amount of heat energy required to raise the temperature of 1 mol of that substance by 1°C when the process is at constant volume.

vv

QC

n T∆

=∆

(specific heat capacity at constant volume)

where CV

= specific heat capacity at constant volume ∆QV = heat energy transferred at constant volume n = number of moles ∆T = change in temperature

Specific heat capacity at constant pressure, CCpp

• Definition: Specific heat capacity at constant pressure Cp is

defined as the amount of heat energy required to raise the temperature of 1 mol of that substance by 1°C when the process is at constant pressure.

PP

QCn T∆

=∆

(specific heat capacity at constant pressure)

where CP

= specific heat capacity at constant pressure

∆QP = heat energy transferred at constant pressure n = number of moles ∆T = change in temperature

Relation between change in internal energy ∆U and Cv

• Let us consider a thermodynamic process at constant volume such that ∆QV amount of heat energy transferred. Clearly work done by gas at constant volume,

∆W = 0. • From 1st Law of thermodynamics we know that,

∆U = ∆Q – ∆W ∆U = ∆Qv

• Therefore, from definition of Cv , we have

vv

Q UCn T n T∆ ∆

= =∆ ∆

• At differential level we can write this relation as

dU = n Cv dT (Relation between dU and CV)

Important points about internal energy

• The internal energy of an ideal gas depends only on its temperature, not on its pressure or volume.

• For an ideal gas the change in internal energy in any process is given by dU = nCvdT , whether the volume is constant or not.

• The internal energy of a non-ideal gas cannot depend only on temperature; it must depend on pressure as well. Specific heat capacity at constant temperature, CCT

• Clearly at change in temperature ∆T = 0 at constant temperature. Therefore,

10T

QCn T∆

= = = ∞∆

Specific heat capacity at for adiabatic process, CCadia

• Clearly ∆Q = 0 for adiabatic process. Therefore,

0adiaQC

n T∆

= =∆

Relation between Cp

and Cv of an ideal gas

• There is an important relationship between Cp and Cv of

an ideal gas :

Cp – Cv

= R

(Relation between Cp and Cv of an ideal gas)

where R = 8.3I4J/mol K. (universal gas constant ) • It is obvious from this relation that C

p > C

v for an ideal

gas. Difference between Cp and C

v is equal to R.



Proof:

• Let us consider an ideal gas undergoing constant-pressure process. Clearly, the amount of heat dQ given to the gas is

dQ = n Cp dT …(i)

• The work dW done by the gas in this constant-pressure process is dW = pdV

• We can also express dW in terms of the temperature change dT by using the ideal gas equation, pV = nRT. Because pressure p is constant, the change in volume V is proportional to the change in temperature T:

dW = p dV = n R dT …(ii) • Now we substitute eqns. (i) and (ii) into the first law of

thermodynamics,

dU = dQ – dW dU = n Cp dT – nRdT …(iii)

• The internal energy depends only on temperature, therefore, we can use dU = n Cv

dT. Thus putting dU = n

Cv dT in equation (iii), we get

n Cv dT = n Cp

dT – nRdT

Cv = Cp

– R

(Dividing each term by the common factor n dT) Cp

– Cv

= R

The ratio of specific heats, γ

• The ratio of specific heats is denoted by γ and is equal to Cp/Cv

. Hence,

= p

v

cc

γ

(ratio of specific heats)

• Since Cp > C

v , therefore γ > 1

• This quantity γ plays an important role in adiabatic processes for an ideal gas.

• Value of γ for different ideal gases: (Remember these; we will explain these in a later section)

• Monatomic ideal gas: 5 1.673

γ = =

Very close examples : He, Ne, Ar, Kr, Xe

• Diatomic ideal gas with no vibration: 7 1.45

γ = =

Very close examples : H2, O

2, N

2, CO, HCl

• Diatomic ideal gas with vibration: 9 1.297

γ = =

Adiabatic Processes for an Ideal Gas

Proof of constant PV γ = (Important)

• Let us consider n mole of an ideal gas undergoing an adiabatic, reversible process. When a gas expands adiabatically, heat transferred to gas dQ = 0

• The change in the internal energy is given by dU = n C

v dT

• Hence, from the first law of thermodynamics, we have, dU = dQ – dW

n Cv dT = 0 – pdV

n Cv dT = – pdV …(i)

• Taking the total differential of the equation of state of an ideal gas, PV = nRT, we get P dV + V dP = nR dT

PdV VdPdTnR+

= ...(ii)

• Substituting this value of dT in (i), we get,

( )( ) ( )

( )

( )

[ ]

2 2

1 1

rearranging

–

rearranging

ln

v

v

v v

p v p v p v

p

v

p

v

V P

V P

V

PdV VdPnC PdVnR

C PdV VdP RPdV

C R PdV C VdP

C PdV C VdP C C R C C R

CPdV VdP

CC

PdV VdPC

dV dPV P

V

γ γ

γ

γ

+ = −

⇒ + = −

⇒ + = −

⇒ = − = ⇒ = +

⇒ = −

⇒ = − =

⇒ = −

⇒

∫ ∫

[ ] ( )( ) ( )

( )

2 2

1 1

2 1 2 1 1 2

2 1

1 2

2 1

1 2

2 1

1 2

1 1 2 2

ln Integrating

ln ln ln ln = ln ln

ln ln ln ln ln

ln ln

con

ln ln

V P

P

A

P

V V P P P P

V P AA BV P B

V P A B BV P

V PV P

PVP

PVV γ

γ

γ

γ γ

γ

γ

= −

⇒ − = − − −

⇒ = − =

⇒ = =

⇒ =

⇒ =

⇒ =

stant

• Thus for an ideal gas undergoing adiabatic reversible process, we have

constant PV γ = • This equation can also be written in terms of temperature

T Vγ-1

= constant

1 constantTP

γ

γ − =



• Relation between V and T for an ideal gas undergoing

adiabatic process: Since we have to find out relation between V & T,

therefore we have to eliminate P from PV= nRT and PVγ

= constant. Since PV = nRT P = nRT/V . Therefore, putting this

value of P in PVγ = constant, we have

(nRT/V ) Vγ = constant

nRT Vγ-1

= constant

TVγ-1

= constant/nR = another constant (because n & R are constant)

Hence, T Vγ-1

= constant

Similarly, it can be proved that

1 constantTP

γ

γ − =

P-V diagram of an ideal gas undergoing an adiabatic process and isothermal process

• It has been found that adiabatic curve is steeper than isothermal curve.

• For an adiabatic process,

pVγ= constant

ln P + ln V = γ ln k (taking logarithm)

0dP dVP V

γ⇔ + = (differentiating )

• Therefore, slope of adiabatic curve,

( ) ...adiabatic

dP p idV V

γ = −

• For an isothermal process, • T = constant

Therefore, PV = nRT = constant PV = constant PdV + VdP = 0 (differentiating) • Therefore, slope of isothermal curve,

• ( ) ...

isothermal

dP P iidV V

⇔ =

• Comparing (i) & (ii)

• Slope of adiabatic curve = – γ(slope of isothermal curve) • Therefore, adiabatic curve is steeper than isothermal

curve.

Example

Q. Consider the process A and B in figure. It is possible that (a) Both the process are isothermal (b) Both the processes are adiabatic (c) A is isothermal and B is adiabatic (d) A is adiabatic and B is isothermal

Answer:

1st figure: A is isothermal and B is adiabatic 2nd figure: A is adiabatic and B is isothermal

Work done by an Ideal Gas undergoing adiabatic

process

• Proof of ( )1 21 1 2 2

1 1nR T TPV PVW

γ γ−−

= =− −

• Let us consider an ideal gas undergoing an adiabatic process. Suppose the gas has initial pressure P

1 and initial

volume V1. Pressure and volume finally changes to P

2 and

V2 adiabatically.

• As the process is adiabatic, we have PVγ = constant = K.

• We have arbitrarily assumed constant to be K. Hence,

PVγ = K

P1V

1

𝛄𝛄 = P

2V

2

𝛄𝛄 = K …(1)

• Hence, net work done by the gas when volume changes from V1 to V2 is

2 2

1 1

22

1 1

11 1

1

1 12 1

2 12 1

2 2 1 1

(integrating)1

1

1 (rearranging)1

1 [ ] 1

V V

V V

VVy

V V

K KW PdV dV PV K PV V

VW K V dV K

KW V V

K KW V VV V

W PV PV

γ

γ γ

γ γ

γ

γ

γ

γ

− +−

− + − +

= = = ⇔ =

⇒ = = − +

⇒ = − − +

⇒ = − − +

⇒ = −− +

∫ ∫

∫

( )

1 1 2 2 1 21 2

1 21 1 2 2

( =K and )

1 1

K KPV PV P PV V

nR T TPV PVW

γ γγ γ

γ γ

= ⇔ = =

−−⇒ = =

− −



Example

Q. Air is kept in a container having walls which are slightly

conducting. The initial temperature and volume are 27 0C

(equal to the temperature of surrounding) and 800 cm3

respectively. Find the rise in the temperature if the gas is

compressed to 200 cm3 (a) in a short time (b) in a long

time. Assume air to be an ideal gas and take 𝛄𝛄 = 1.4. • SOLUTION:

(a) When the gas is compressed in a short time, the process is assumed as adiabatic (because there is not enough time so that heat transfer can take place between system and surrounding). Thus,

• T Vγ-1

= constant

1 1 2 2

12 1

1.40

2 1

02

800(300) ( 27 300 )200

522 249

TV T V

VT TV

T T C K

T K C

γ γ

γ

⇒ =

⇒ =

⇒ = = =

⇒ = =

• Therefore, rise in temperature = 249 – 27 = 222 0C .

(b) When the gas is compressed in a long time the process is

very slow. Therefore, the temperature of the system is always equal to the temperature of the surrounding. Hence the process is isothermal, i.e. T = constant = temperature of surrounding = 270C

• Thus, rise in temperature = 0.

• DISCUSSION:

This example is very important from competition point of view. From this problem, we have seen that a process can be assumed

1. Adiabatic: If the gas is compressed or expanded in a short time (i.e., change in volume is very rapid)

2. Isothermal: If the gas is compressed or expanded in a very long time (i.e., change in volume is very slow) and gas can exchange heat with the surrounding.

HEAT ENGINE

Heat engine. It is a device which converts continuously heat energy into mechanical energy in a cyclic process. As shown in figure, a heat engine has the following essential parts: (a) Source: It is a heat reservoir at higher temperature T1. It

is supposed to have infinite thermal capacity so that any amount of heat can be drawn from it without changing its temperature.

(b) Sink. It is a heat reservoir at a lower temperature T2. It has also infinite thermal capacity so that any amount of heat can be added to it without changing its temperature.

(c) Working substance. Working substance is any material (solid, liquid or gas) which performs mechanical work when heat is supplied to it.

Working. In every cycle of operation, the working substance absorbs a definite amount of heat Q1 from the source at higher temperature T1, converts a part of this heat energy into mechanical work W and rejects the remaining heat Q2 to the sink at lower temperature environment. As the working substance returns to its initial state after completing one cycle, there is no change in its internal energy. Hence by first law of thermodynamics, net heat absorbed in a cycle = Work done, i.e., Q1 – Q2 = W

Efficiency of a heat engine. The efficiency of a heat engine is defined as the ratio of the net work done by the engine in one cycle to the amount of heat absorbed by the working substance from the source. The efficiency of heat engine is given by

1 2 2

1 1 1

Work done by engine (output)Heat absorbed from the source (input)

or 1Q Q QW

Q Q Q

η

η

=

−= = = −

Efficiency of a heat engine is always less than unity. The efficiency of a steam engine varies from 12 to 16%. The maximum efficiency of a petrol engine is 26% and that of a diesel engine is 40%.

Q2

W

Q1

Cold reservoir or Sink (T2)

Hot reservoir or Source (T1)

Heat engine



CARNOT ENGINE

Carnot engine. It is an ideal reversible heat engine which operates between two temperatures T1 (source) and T2 (sink). It is a device which converts continuously heat energy into mechanical energy in a cyclic process. Construction. As Carnot engine has the following main parts: (a) Cylinder: It has conducting base and insulating

walls. It is fitted with an insulting and frictionless piston.

(b) Source: It is a heat reservoir at a higher temperature T1 from which the engine draws heat.

(c) Sink: It is heat reservoir at a lower temperature T2 to which any amount of heat can be rejected by the engine.

(d) Working substance. The working substance is an ideal gas contained in the cylinder.

Carnot cycle. The working substance is carried through a reversible cycle of the following four steps: Step 1 → 2: Isothermal expansion (at temperature T1) of the gas taking its state from (P1, V1, T1) to (P2, V2, T1). The heat absorbed by the gas (Q1) from the reservoir at temperature T1. Work done by ideal gas in this step is given by

212 1

1

lnVW nRTV

=

As the process is isothermal, ∆U = 0, hence from first law of thermodynamics, we have

21 12 1

1

ln ...(i)VQ W nRTV

= =

Step 2 → 3: Adiabatic expansion of the gas taking its state from (P2, V2, T1) to (P3, V3, T2). Work done by ideal gas in this step is given by

( )1 223 ...(ii)

1nR T T

Wγ

−=

−

Step 3 → 4: Isothermal compression (at temperature T2) of the gas taking its state from (P3, V3, T2) to (P4, V4, T2). The heat released by the gas (Q2) to the reservoir at temperature T2.

As in step 1 → 2 we have,

342 34 2 2 34 2

3 4

ln ln ...(iii)VVQ W nRT Q W nRT

V V

− = = ⇒ = − =

Step 4 → 1: Adiabatic compression of the gas taking its state from (P4, V4, T2) to (P1, V1, T1). Work done by ideal gas in this step is given by

( )2 141 ...(iv)

1nR T T

Wγ

−=

− Total work done by the gas in the complete cycle is

( ) ( )12 23 34 41

1 2 2 12 41 2

1 3

2 41 2

1 3

ln ln +1 1

ln ln ...(v)

W W W W W

nR T T nR T TV VW nRT nRTV V

V VW nRT nRTV V

γ γ

= + + +

− − ⇒ = + + − −

⇒ = +

The efficiency η of the Carnot engine is

32

41 2 2

1 1 1 21

1

ln1 1 (from (i) & (iii))

ln

VnRTVQ Q QW

Q Q Q VnRTV

η

− = = = − = −

3

42

1 2

1

ln1 ...(vi)

ln

VVT

T VV

η

⇒ = −

Now since step 2 → 3 is an adiabatic process, 1 1 1

1 2 2 3 constant TV TV T Vγ γ γ− − −= ⇒ = 1

31

2 2

...(vii)VT

T V

γ −

⇒ =

Similarly for step 4 → 1, we have, 1

1 4

2 1

...(viii)T VT V

γ −

=

From (vii) and (viii), we have, 32

1 4

...(ix)VV

V V=

From (vi) and (ix), we have,

4

32 2

1 12

1

ln1 1

ln

VVT T

T TVV

η η

= − ⇒ = −

Hence, for a Carnot Engine, we have,

2 2

1 1

1 1Q TQ T

η = − = − (efficiency of the Carnot engine)



• Carnot’s Theorem: All reversible engines operating

between the same two temperatures have equal efficiency and no engine operating between the same two temperatures can have an efficiency greater than this.

• It is a consequence of the second law of thermodynamics.

REFRIGERATOR • Refrigerator. A refrigerator is a Carnot’s heat engine

working in the reverse direction. • Thus, when a Carnot engine works in opposite direction

as a refrigerator, it will absorb an amount of heat Q2 from the sink at lower temperature T2. As heat is to be removed from the sink at lower temperature, an amount of work equal to W = Q1 – Q2 (using 1st law of Thermodynamics) is performed by the compressor of the refrigerator to remove heat from sink and then to reject the total heat Q1 to the source (atmosphere) through the radiator fixed at its back.

• Coefficient of performance: It may be defined as the

ratio of the amount of heat removed (Q2) per cycle to the mechanical work (W) required to be done on it. It is denoted by β.

Thus, 2 2

1 2

Q QW Q Q

β = =−

• The expression for coefficient of performance can be put into another form. We have,

2

11 2

2

1

1

QQQ QQ

β = =− −

• Now, for a Carnot’s cycle, 1 1

2 2

Q TQ T

=

2

1 1 2

2

1

1

TT T TT

β∴ = =−−

2 2

1 2 1 2

Q TQ Q T T

β∴ = =− −

(coefficient of performance of a refrigerator)

• It may be pointed out that higher the value β, more is the efficiency of a

refrigerator and as the refrigerator works, T2 goes on decreasing

(without causing any practical change in the value of T1) and the value of coefficient of performance goes on decreasing.

SECOND LAW OF THERMODYNAMICS

• Kelvin-Planck statement: It is not possible to design a heat engine which works in a cyclic process and whose only result is to take heat from a body at a single temperature and convert it completely into mechanical work.

• Claussius statement: It is not possible to design a

refrigerator which works in a cyclic process and whose only result is to transfer heat from a body to a hotter body.

Degrees of freedom

• The number of independent terms in the expression of energy is called degree of freedom.

• Monatomic Gas: A monatomic gas (mono means one) can be assumed of containing only one type of atoms. Thus every molecule of a monatomic gas behaves as a particle in space. Hence, molecules of a monatomic gas has three degrees of freedom.

• Diatomic Gas: Every molecule of a diatomic gas behaves as a two-particle dumbbell-shaped system.

i. Molecules of a diatomic gas has five degrees of freedom without considering vibration and potential energy.

ii. Molecules of a diatomic gas has seven degrees of freedom considering vibration and potential energy.

Equipartition of Energy

• The principle of equipartition of energy states that each degree of freedom has, on average, an associated kinetic energy per molecule of ½ kT, where k is the Boltzmann constant and T is the absolute temperature.

• Thus, average energy associated with each degree of freedom = ½ k T

• Hence average energy per molecule of a monatomic gas = 32

kT

• Hence average energy per molecule of a diatomic gas

(without vibration) = 52

kT

• Hence average energy per molecule of a diatomic gas

(with vibration) = 72

kT

• Note: We can find out Cp

& Cv and hence, γ using the

concept of equipartition of energy. • Monatomic ideal gas:

52pC R= , 3

2vC R= &

5 1.673

p

v

CC

γ = = =

Diatomic ideal gas with no vibration: 72pC R= , 5

2vC R= &

7 1.45

p

v

CC

γ = = =

• Diatomic ideal gas with vibration: 92pC R= , 7

2vC R= &

9 1.297

p

v

CC

γ = = =



Step by step procedure to find C

p & C

v

STEP I: • Let us consider a sample of amount n moles of an ideal

gas. • Therefore, the total number of molecules = nN

A where

NA is the Avogadro number.

STEP II: • Using equipartition of energy, we can find out the

average internal energy per molecule. STEP III:

• Total internal energy = (total number of molecules) ×(average energy per molecule) U = nN

A × (average energy per molecule)

STEP IV:

• Use 1 v v

dUdU nC dT Cn dT

= ⇒ =

STEP V: • C

p – C

v = R to get C

p .

Example

• To find out Cp

& Cv of a monatomic gas. Hence, find out

𝛄𝛄. • Solution:

STEP I: • Let us consider a sample of amount n moles of an ideal

gas. • Therefore, the total number of molecules = nN

A where

NA is the Avogadro number.

STEP II: • From equipartition of energy we know that average

energy per molecule of a monatomic gas = 32

kT

STEP III: U = nN

A (average energy per molecule)

U = nNA

× 32

kT

U = 32

nRT (using NAk = R )

STEP IV:

• Using 1 3

2v vdUdU nC dT C R

n dT= ⇒ = =

STEP V:

3 52 2p v p vC C R C C R R R− = ⇒ = + = =

Therefore, specific heat ratio

53

p

v

CC

γ = =

In a similar manner we can find out γ for diatomic gases.

HEAT TRANSFER

• Heat can be transferred from one place to the other by any

of three possible ways : conduction, convection and radiation. In the first two processes, a medium is necessary for the heat transfer. Radiation, however, does no have this restriction. This is also the fastest mode of heat transfer, in which heat is transferred from one place to the other in the form of electromagnetic radiation.

Conduction

⋅ Transfer of heat through a substance in which heat is transported without direct mass transport is called conduction. This mode of heat transfer is generally seen in solids.

⋅ Consider a metal rod whose ends are in thermal contact with a hot reservoir at temperature T

1 and a

cold reservoir temperature T

2. The

sides of the rod are covered with insulating medium, so the transport of heat is along the rod, not through the sides.

⋅ After sometime, a steady state is reached; the temperature of the bar decreases uniformly with distance from T1 to T2; (T1 > T2). It is found experimentally that in this steady state, the rate of flow of heat (or heat current) H is proportional to the temperature difference (T1 – T2) and the area of cross section A and is inversely proportional to the length L :

2 1

1 2

= ...(i)

T TdQ kAdt L

T TdQ kAdt L

− = −

− ⇒

⋅ The constant of proportionality K is called the thermal conductivity of the material. The greater the value of K for a material, the more rapidly will it conduct heat.

Thermal Resistance (R)

⋅ Eq. (i) in differential form can be written as

...(ii)/

dQ T THdt L kA R

∆ ∆= = =

⋅ Here, ΔT = temperature difference (TD) and

thermal resistance of the rod.LRkA

= =

⋅ Just like we solve electric circuit we can solve conduction problem using the same concept.

General expression for conduction

dQ dTkAdt dx

= −

⋅ dTdx

is also known as temperature gradient.

T1 T2

L



Radiation

⋅ This mode of heat transfer doesn’t require any medium, and is fastest mode of heat transfer. Example − the fastest mode of heat transfer is the radiation received by earth, coming from sun.

⋅ All bodies emit radiant energy at all temperatures in all direction, which increases with increasing temperature and is independent of the presence of other bodies.

⋅ This radiant energy emitted by bodies is termed as thermal radiation or simply radiation. When this radiation falls on a material surface, a part of it can be absorbed and thermal energy of the receiving surface is increased.

⋅ Recent studies show that the thermal radiation is nothing but the energy of electromagnetic radiation. Every body at any temperature emits electromagnetic waves.

⋅ The rate at which a body radiates energy is dependent on temperature, its area and nature of surface.

⋅ A body is not only radiating the energy but is also absorbing from the surroundings.

⋅ If radiation emitted > radiation absorbed, then body gets cooled down, while if reverse is the case, the temperature of the body increases.

⋅ When the body is in thermal equilibrium with the surroundings, then also it radiates energy, but the same rate as it absorbs.

⋅ Rough and dark surface are good absorbers while smooth, shiny and bright surface are good reflectors.

Black body radiations

⋅ A body which absorbs all the radiations incident upon it is termed as a perfect black body. Such a body will emit radiation at the fastest rate. the radiations emitted by a black body are termed as black body radiations.

⋅ A perfect black body maintained at a suitable temperature absorbs all radiations incident on it. A perfect black body can’t be realized in practice.

Stefan-Boltzmann’s Law

⋅ For a perfect black body, the thermal energy radiated per unit time is given by,

4dQ H ATdt

σ= =

where A = surface area of body, T = its absolute temperature σ = an universal constant.

⋅ For a non-black body, 4dQ H eATdt

σ= = where e is the

‘emissivity’ of surface its value depends on the nature of the material of the black body ( Stefan-Boltzmann’s Law)

⋅ For a non-black body, the value of e is less than 1. For a perfect black body, e = 1.

Kirchhoff’s Law

⋅ Kirchhoff explained that a good absorber of thermal radiation is also a good emitter of it.

⋅ He showed that for a body a = e i.e., absorptivity = emissivity.

⋅ As a perfect black body is a perfect absorber, so it is a perfect emitter too.

Cooling of a Body Through Radiation

⋅ When a body having temperature T is placed in a surrounding of temperature ( )0 ,T T< then the temperature

of the body starts falling down. ⋅ Rate at which heat is emitted,

41dQ aATdt

σ=

⋅ Rate at which heat is absorbed,

420

dQ aATdt

σ=

⋅ Net rate of heat loss

[ ]4 40 H eA T T e aσ = − =

⋅ Rate of cooling

4 40

dT eA T Tdt ms

σ = − −

Wien’s Displacement Law

⋅ The figure above shows the experimental curves for radiation emitted by a black body versus wavelength for different temperatures.

⋅ According to Wien’s law,

mTλ = constant

where mλ is the wavelength corresponding to maximum intensity (energy content) of radiation emitted by body at temperature T.

⋅ The most significant feature of the curves obtained is that they are universal i.e., black-body radiation curves obtained depend only on the temperature, and not on the shape.

T1

T2

T3

Eλ

λ

T1 > T2 >T3

3mλ 1mλ 1mλ



Newton’s Law of Cooling

⋅ Newton’s law of cooling states that the rate of cooling (or rate of loss of heat) of a body is directly proportional to the temperature difference between the body and its surroundings, provided the temperature difference is small.

⋅ Consider a hot body at temperature T. Let 0T be the temperature of its surroundings. According to Newton’s law of cooling,

⋅ Rate of loss of heat ∝ Temperature difference between the body and its surroundings

or ( )0dQ T Tdt

− ∝ −

or ( )0 ...(1)dQ k T Tdt

− = −

where k is proportionality constant depending upon the area and nature of the surface of the body.

⋅ Let m be the mass and c the specific heat of the body at temperature T. If the temperature of the body falls by small amount dT in time dt, then the amount of heat lost is dQ = mc dT

∴ Rate of loss of heat is given by dQ dTmcdT dt

=

⋅ Combining the above equations, we get

( )0dTmc k T Tdt

− = −

or ( ) ( )0 0 ...(2)dT k T T K T Tdt mc

= − − = − −

where K = k/mc is another constant. The negative sign indicates that as the time passes, the temperature of the body decreases. The above equation can be written as

0

dtdT KT T

= −−

⋅ On integrating both sides, we get

0

1 dT K dtT T

= −−∫ ∫

or ( )0log ...(3)e T T Kt c− = − +

or 0kt cT T e− +− =

or 0c ktT T e e−− +

or 0 ....(4)ktT T Ce−− +

where c is a constant of integration and C = ec . Equation (1), (2), (3) and (4) are the different mathematical representations for Newton’s law of cooling. Using equation (4), one can calculate the time of cooling of a body through a particular range of temperature.

⋅ Approximate solution for Newton’s law of cooling is,

1 2 1 202

T T T TK Tt− + = −

where t is the time in which

temperature of body changes from T1 to T2.



Physics Classes by Pranjal Sir

(Admission Notice for XI & XII - 2014-15)

Batches for Std XIIth

Batch 1 (Board + JEE Main + Advanced): (Rs. 16000) Batch 2 (Board + JEE Main): (Rs. 13000) Batch 3 (Board): (Rs. 10000) Batch 4 (Doubt Clearing batch): Rs. 8000

About P. K. Bharti Sir (Pranjal Sir)

• B. Tech., IIT Kharagpur (2009 Batch) • H.O.D. Physics, Concept Bokaro Centre • Visiting faculty at D. P. S. Bokaro • Produced AIR 113, AIR 475, AIR 1013 in JEE -

Advanced • Produced AIR 07 in AIEEE (JEE Main)

Address: Concept, JB 20, Near Jitendra Cinema, Sec 4, Bokaro Steel City Ph: 9798007577, 7488044834 Email: [email protected] Website: www.vidyadrishti.org

Physics Class Schedule for Std XIIth (Session 2014-15) by Pranjal Sir

Sl. No.

Main Chapter Topics Board level JEE Main Level JEE Adv Level

Basics from XIth Vectors, FBD, Work, Energy, Rotation, SHM

3rd Mar to 4th Apr 14

1. Electric Charges and Fields

Coulomb’s Law 5th & 6th Apr 5th & 6th Apr 5th & 6th Apr Electric Field 10th & 12th Apr 10th & 12th Apr 10th & 12th Apr Gauss’s Law 13th & 15th Apr 13th & 15th Apr 13th & 15th Apr Competition Level NA 17th & 19th Apr 17th & 19th Apr

2. Electrostatic Potential and Capacitance

Electric Potential 20th & 22nd Apr 20th & 22nd Apr 20th & 22nd Apr Capacitors 24th & 26th Apr 24th & 26th Apr 24th & 26th Apr Competition Level NA 27th & 29th Apr 27th & 29th Apr, 1st, 3rd &

4th May PART TEST 1 Unit 1 & 2 4th May NA NA

NA 11th May 11th May 3.

Current Electricity Basic Concepts, Drift speed, Ohm’s Law, Cells, Kirchhoff’s Laws, Wheatstone bridge, Ammeter, Voltmeter, Meter Bridge, Potentiometer etc.

6th, 8th, 10th, 13th May



Competition Level NA 15th & 16th May 15th, 16th, 17th, 18th & 19th May

PART TEST 2 Unit 3 18th May NA NA NA 20th May 20th May

SUMMER BREAK 21st May 2013 to 30th May 2013 4. Moving charges and

Magnetism Force on a charged particle (Lorentz force), Force on a current carrying wire, Cyclotron, Torque on a current carrying loop in magnetic field, magnetic moment

31st May, 1st & 3rd Jun



Biot Savart Law, Magnetic field due to a circular wire, Ampere circuital law, Solenoid, Toroid

5th, 7th & 8th Jun 5th, 7th & 8th Jun 5th, 7th & 8th Jun

Competition Level NA 10th & 12th Jun 10th, 12th, 14th & 15th Jun PART TEST 3 Unit 4 15th Jun NA NA

NA 22nd Jun 22nd Jun 5. Magnetism and Matter 17th, 19th & 21st

Jun 17th, 19th & 21st Jun

Not in JEE Advanced Syllabus

mailto:[email protected]�

http://www.vidyadrishti.org/�



6.

Electromagnetic Induction

Faraday’s Laws, Lenz’s Laws, A.C. Generator, Motional Emf, Induced Emf, Eddy Currents, Self Induction, Mutual Induction

24th, 26th & 28th Jun



Competition Level NA 29th Jun & 1st Jul 29th Jun, 1st, 3rd & 5th Jul PART TEST 4 Unit 5 & 6 6th Jul NA NA

NA 13th Jul 13th Jul 7. Alternating current AC, AC circuit, Phasor,

transformer, resonance, 8th, 10th & 12th Jul 8th, 10th & 12th Jul 8th, 10th & 12th Jul

Competition Level NA 15th July 15th & 17th July 8. Electromagnetic Waves 19th & 20th July 19th & 20th July Not in JEE Advanced

Syllabus PART TEST 5 Unit 7 & 8 27th Jul 27th Jul 27th Jul

Revision Week Upto unit 8 31st Jul & 2nd Aug 31st Jul & 2nd Aug 31st Jul & 2nd Aug

Grand Test 1 Upto Unit 8 3rd Aug 3rd Aug 3rd Aug

9.

Ray Optics

Reflection 5th & 7th Aug 5th & 7th Aug 5th & 7th Aug Refraction 9th & 12th Aug 9th & 12th Aug 9th & 12th Aug Prism 14th Aug 14th Aug 14th Aug Optical Instruments 16th Aug 16th Aug Not in JEE Adv

Syllabus Competition Level NA 19th & 21st Aug 19th, 21st, 23rd, 24th Aug

10.

Wave Optics

Huygens Principle 26th Aug 26th Aug 26th Aug Interference 28th & 30th Aug 28th & 30th Aug 28th & 30th Aug Diffraction 31st Aug 31st Aug 31st Aug Polarization 2nd Sep 2nd Sep 2nd Sep Competition Level NA 4th & 6th Sep 4th, 6th, 7th, 9th, 11th Sep

PART TEST 6 Unit 9 & 10 14th Sep 14th Sep 14th Sep REVISION ROUND 1 (For JEE Main & JEE Advanced Level): 13th Sep to 27th Sep

Grand Test 2 Upto Unit 10 28th Sep 28th Sep 28th Sep

DUSSEHRA & d-ul-Zuha Holidays: 29th Sep to 8th Oct

11.

Dual Nature of Radiation and Matter

Photoelectric effect etc 9th & 11th Oct 9th & 11th Oct 9th & 11th Oct

Grand Test 3 Upto Unit 10 12th Oct 12th Oct 12th Oct

12.

Atoms 14th & 16th Oct 14th & 16th Oct 14th & 16th Oct

13. Nuclei 18th & 19th Oct 18th & 19th Oct 18th & 19th Oct X-Rays NA 21st Oct 21st & 25th Oct

PART TEST 7 Unit 11, 12 & 13 26th Oct NA NA 14. Semiconductors Basic Concepts and Diodes,

transistors, logic gates 26th, 28th, 30th Oct & 1st Nov

26th, 28th, 30th Oct & 1st Nov

Not in JEE Adv Syllabus

15.

Communication System 2nd & 4th Nov 2nd & 4th Nov Not in JEE Adv Syllabus

PART TEST 8 Unit 14 & 15 9th Nov 9th Nov NA Unit 11, 12 & 13 Competition Level NA 8th, 9th & 11th Nov 8th, 9th, 11th, 13th & 15th

Nov PART TEST 9 Unit 11, 12, 13, X-Rays NA 16th Nov 16th Nov

Revision Round 2

(Board Level)

Mind Maps & Back up classes for late registered students

18th Nov to Board Exams



Revision Round 3

(XIth portion for JEE)

18th Nov to JEE 18th Nov to JEE 18th Nov to JEE

30 Full Test Series Complete Syllabus Date will be published after Oct 2014





Documents

Notes for School Exams Target IIT-JEE 2015 Thermodynamicsvidyadrishti.com/Thermodynamics II.pdf · Target IIT-JEE 2015 . Physics . Thermodynamics . Author: Pranjal K. Bharti (B. Tech.,