PV=N KB T$PV=N\phantom{\rule{0.22em}{0ex}}{K}_{B}\phantom{\rule{0.22em}{0ex}}T$P = PressureN = Total Number of moleculesKB$K}_{B$ = Boltznann constantT = TemperatureNA$N}_{A$=Avogadro Numbern = Number of molesPV=N

NA NAKB T$PV=\frac{N}{{N}_{A}}\phantom{\rule{0.22em}{0ex}}{N}_{A}{K}_{B}\phantom{\rule{0.22em}{0ex}}T$⟹ PV=n R T$PV=n\phantom{\rule{0.22em}{0ex}}R\phantom{\rule{0.22em}{0ex}}T$

Relation between Gas Constant and Boltznann constantR=NAKB$R={N}_{A}{K}_{B}$NA$N}_{A$=Avogadro NumberKB$K}_{B$ = Boltznann constant

Real gases approach ideal gas behaviour at low pressures and high temperatures.

Mean Free PathAverage distance molecule can travel without colliding is called mean free path.

The mean free path, in gases, is of the order of thousands of angstroms. The atoms are much freer in gases and can travel long distances without colliding. If they are not enclosed, gases disperse away. The force has a long range attraction and a short range repulsion. The atoms attract when they are at a few angstroms but repel when they come closer.

Internal energy of ideal gas depends only upon temperature. (It does not depend on pressure or volume).

Brownian motion It is the random motion of particles suspended in a fluid (a liquid or a gas) resulting from their collision with the fast-moving molecules in the fluid.

DiffusionThe spreading of a substance of its own accord is called diffusion . It is due to molecular motion.

Behaviour of GasesIn gas, molecules are far from each other and their mutual interactions are negligible except when two molecules collide.

The atoms attract when they are at a few angstroms but repel when they come closer.When a molecule approaches another molecule, there is a repulsive force between them, due to which the molecules behave as small hard spherical particles.

1. The mean free path is expected to vary inversely with the density of the gas.2. Higher the density, more will be the collisions and smaller will be the mean free path λ. 2. Mean free path is inversely proportional to the size of the molecule. Smaller the size of the molecule, less is the chance for collision and larger is the mean free path. 3. Mean free path is inversely proportional to d2$d}^{2$ ("d" is diameter of the molecule)

Mean Free Path ( 𝜆$\mathit{\lambda}$ )𝜆=1

2 𝜋 d2(N/V)$\mathit{\lambda}=\frac{1}{\sqrt{2}\phantom{\rule{0.22em}{0ex}}\mathit{\pi}\phantom{\rule{0.22em}{0ex}}{d}^{2}\phantom{\rule{0.22em}{0ex}}(N/V)}$d = diameter of moleculeN = Number of molecule enclosed in volume V

Avogadro’s hypothesisThe number of molecules per unit volume is the same for all gases at a fixed temperature and pressure. The number gas molecules in 22.4 litres of any gas is 6.02 × 1023. This is known as Avogadronumber and is denoted by NA$N}_{A$. The mass of 22.4 litres of any gas is equal to its molecular weight in grams at S.T.P (standard temperature 273 K and pressure 1 atm). This amount of substance is called a mole.

The perfect gas equation can be written asPV=𝜇RT$PV=\mathit{\mu}RT$

Average PressureP=1

3n m v2$P=\frac{1}{3}n\phantom{\rule{0.22em}{0ex}}m\phantom{\rule{0.22em}{0ex}}{v}^{2}$wheren = number of molecules per unit volume.v = root mean square velocitym = mass of one molecule

Kinetic Interpretation of TemperatureP=1

3n m v2$P=\frac{1}{3}n\phantom{\rule{0.22em}{0ex}}m\phantom{\rule{0.22em}{0ex}}{v}^{2}$⟹ PV=1

3n m V v2$PV=\frac{1}{3}n\phantom{\rule{0.22em}{0ex}}m\phantom{\rule{0.22em}{0ex}}V\phantom{\rule{0.22em}{0ex}}{v}^{2}$⟹ PV=1

3N m v2$PV=\frac{1}{3}N\phantom{\rule{0.22em}{0ex}}m\phantom{\rule{0.22em}{0ex}}{v}^{2}$⟹ PV=2

3×1

2 N m v2$PV=\frac{2}{3}\times \frac{1}{2}\phantom{\rule{0.22em}{0ex}}N\phantom{\rule{0.22em}{0ex}}m\phantom{\rule{0.22em}{0ex}}{v}^{2}$1) Internal energy E of an ideal gas is purely kinetic.the sample.2) E=1

2 N m v2$E=\frac{1}{2}\phantom{\rule{0.22em}{0ex}}N\phantom{\rule{0.22em}{0ex}}m\phantom{\rule{0.22em}{0ex}}{v}^{2}$ is the average translational kinetic energy of the molecules in the gas.⟹ PV=2

3 E$PV=\frac{2}{3}\phantom{\rule{0.22em}{0ex}}E$whereE = Internal Energyn = number of molecules per unit volume.v = root mean square velocitym = mass of one moleculeV = Volume

Law of Equipartition of EnergyKinetic energy of single moleculeKE=1

2mv2x+1

2mv2y+1

2mv2z$KE=\frac{1}{2}m{v}_{x}^{2}+\frac{1}{2}m{v}_{y}^{2}+\frac{1}{2}m{v}_{z}^{2}\phantom{\rule{0.22em}{0ex}}$Kinetic energy of single moleculeKE=3

2kBT $KE=\frac{3}{2}{k}_{B}T\phantom{\rule{0.22em}{0ex}}$So mean energy associated with every component =1

2kBT$=\frac{1}{2}{k}_{B}T$

Law of equipartition of energy is valid for high temperatures and not for extremely low temperatures where quantum effects become important.

Degree of freedom.Degrees of freedom of a system are defined as the total number of coordinates or independent quantities required to describe the position and configuration of the system completely.

Use of the law of equipartition of energy and calculate the specific heat of gasesMono-atomic GasDegree of fredom = 3So enerygy associated per molecule=3

2kBT$=\frac{3}{2}{k}_{B}T$So enerygy associated per mole E=3

2RT$E=\frac{3}{2}RT$Molar specific heat at constant volumeCv=dE

Use of the law of equipartition of energy and calculate the specific heat of gasesDiatomic Gas (rigid)Degree of fredom = 5 (3 translational and 2 rotational)So enerygy associated per molecule=5

2kBT$=\frac{5}{2}{k}_{B}T$So enerygy associated per mole E=5

2RT$E=\frac{5}{2}RT$Molar specific heat at constant volumeCv=dE

Use of the law of equipartition of energy and calculate the specific heat of gasesDiatomic Gas (non rigid vibrating molecule)Degree of fredom = 5 (3 translational , 2 rotational and 1 vibration)So enerygy associated per molecule=3

2kBT+2

2kBT+kBT$=\frac{3}{2}{k}_{B}T+\frac{2}{2}{k}_{B}T+{k}_{B}T$So enerygy associated per mole E=7

2RT$E=\frac{7}{2}RT$Molar specific heat at constant volumeCv=dE

Use of the law of equipartition of energy and calculate the specific heat of gasesPolyatomic Gas (for non linear molecule)Degree of fredom = 5 (3 translational , 3 rotational and f vibration)So enerygy associated per molecule=3

2kBT+3

2kBT+f kBT=(3+f)kBT$=\frac{3}{2}{k}_{B}T+\frac{3}{2}{k}_{B}T+f\phantom{\rule{0.22em}{0ex}}{k}_{B}T=(3+f){k}_{B}T$So enerygy associated per mole E=(3+f)RT$E=(3+f)RT$Molar specific heat at constant volumeCv=dE

Transmission of Heat Radiation1. Radiation is continuous emission of energy from the surface of any body because of its thermalenergy. 2. This emitted energy is termed as radiant and is in the form of electromagnetic waves. 3. Radiation is the fastest mode of transfer of heat. 4. The process of transfer of heat by radiation does not require any material medium since electromagnetic waves travel through vacuum. 5. Heat transfer by radiation is possible through vacuum as well as through a material medium transparent to this radiation. 6. Physical contact of the bodies that are exchanging heat is also not required.7. The electromagnetic radiation emitted by the bodies, which are at higher temperature with respect to the surroundings, is known as thermal radiation.

Interaction of Thermal Radiation and MatterQ $Q\phantom{\rule{0.22em}{0ex}}$= Total energy incident on a surface of an objectQa$Q}_{a}\phantom{\rule{0.22em}{0ex}$= Total energy absorbed by the surface of an objectQr$Q}_{r}\phantom{\rule{0.22em}{0ex}$= Total energy reflected by the surface of an objectQt$Q}_{t}\phantom{\rule{0.22em}{0ex}$= Total energy transmitted by the surface of an objectQ=Qa+Qr+Qt$Q={Q}_{a}+{Q}_{r}+{Q}_{t}$Coefficient of absorption : a=Qa

Q$a=\frac{{Q}_{a}}{Q}$Coefficient of reflection : r=Qr

Q$r=\frac{{Q}_{r}}{Q}$Coefficient of transmission : tr=Qt

Perfect transmitter ( tr$t}_{r$ = 1)If , r = 0 and a = 0, then tr$t}_{r$ = 1, then all the incident renergy is transmitted through the object. Then it is a perfect transmitter. The object is said to be completely transparent to the radiation.

Diathermanous ( tr≠0${t}_{r}\ne 0$ )A substance through which heat radiations can pass is known as a diathermanous substance.Coefficient of transmission , tr≠0${t}_{r}\ne 0$A diathermanous body is neither a good absorber nor a good reflector.Example : glass, quartz, sodium chloride, hydrogen, oxygen, dry air.

If tr$t}_{r$ = 1 and a + r = 1 then it is said to be opaque to the radiation.It do not transmit heat radiations incident on them. They are known as athermanous substances.

If tr$t}_{r$ = 1 and a = 0, then r = 1, then the incident energy is reflected by the object.These objects are perfect reflector. A good reflector is a poor absorber and a poor transmitter.

Perfect BlackbodyA body, which absorbs the entire radiant energy incident on it, is called an ideal or perfect blackbody. For a perfect blackbody, a = 1.

Consider two objects, which are opaque to thermal radiation, having the same temperature and same surface area. The surface of first object is well-polished. The surface of second object is painted black.The well-polished object reflects most of the energy falling on it and absorbs little. The black painted object absorbs most of the radiation falling on it and reflects little. But the rate of emission of thermal radiation must be equal to rate of absorption for both the objects, so that temperature is maintained. Black painted object absorbs more, hence it must radiate more to maintain the temperature. Good absorbers are always good emitters.Poor absorbers are poor emitters. Since each object must either absorb or reflect the radiation incident on it, a poor absorber should be a good reflector and vice versa. Hence, a good reflector is also a poor emitter. This is the reason for silvering he walls of vacuum bottles or thermos flasks.

Ferry’s BlackbodyA double walled hollow sphere having tiny hole or aperture, through which radiant heat can enter.Thespace between the walls is evacuated and outer surface of the sphere is silvered. The inner surface of sphere is coated with lamp-black. There is a conical projection on the inner surface of sphere opposite the aperture. The projection ensures that a ray travelling along the axis of the aperture is not incident normally on the surface and is therefore not reflected back along the same path. Radiation entering through the small hole has negligible chance of escaping back through the smallhole. A heat ray entering the sphere through the aperture suffers multiple reflections and is almost completely absorbed inside. So, the aperture behaves like a perfect blackbody.

A cavity radiator consists of a block of material with internal cavity. The inner and outer surfaces are connected by a small hole.The radiation falling on the block that enters through the hole, cannot escape back from it. Hence, the cavity acts as a blackbody. When the block is heated to high temperature, thermal radiation is emitted. This is called cavityradiation and resembles the radiation emitted by a blackbody. Its nature depends only on the temperature of the cavity walls and not on the shape and size of the cavity or the material ofthe cavity walls.

Emission of Heat Radiation1. All bodies at all temperatures above 0 K (absolute zero temperature) radiate thermal energy and at the same time, they absorb radiation received from the surroundings. 2. The amount of thermal radiation emitted per unit time depends on the nature of emitting surface, its area and its temperature. 3. Hotter bodies radiate at higher rate than the cooler bodies. 4. Light coloured bodies reflect most of the visible radiation whereas dark coloured bodies absorb most of the incident visible radiation.

All bodies radiate electromagnetic radiation when their temperature is above the absolute zero oftemperature.

Amount of heat radiated by a body depends on1. The absolute temperature of the body (T)2. The nature of the body – the material, nature of surface – polished or not, etc.3. Surface area of the body (A)4. Time duration of for which body emits radiation (t)

Emissive power (or Radiant Power)The quantity of heat radiated per unit area per unit time (or power emitted per unit area) is defined as emissive power or radiant power (R).R=Q

At$R=\frac{Q}{At}$

The absorptive power of a body is defined as the ratio of the energy absorbed in a given time to the radiant energy incident on it at the same instant of time.

At a given temperature, a perfect blackbody has maximum emissive power. Thus it is convenient to compare emissive power of a given surface with that of the perfect blackbody at the same temperature.

Coefficient of Emission or EmissivityThe coefficient of emission or emissivity (e) of a given surface is the ratio of the emissive power R of the surface to the emissive power R of a perfect black surface, at the same temperature.e=R

RB$e=\frac{R}{{R}_{B}}$

1. 0<e<1$0<e<1$2. Emissivity is larger for rough surfaces and smaller for smooth and polished surfaces.3. Emissivity also varies with temperature and wavelength of radiation to some extent.

Kirchhoff’s Law of Heat Radiation1. At a given temperature, the ratio of emissive power to coefficient of absorption of a body is equal to the emissive power of a perfect blackbody at the same temperature for all wavelengths.2. For a body emitting and absorbing thermal radiation in thermal equilibrium, the emissivity is equal to its absorptivity.So, a=e$a=e$

R$R$ = emissive power of body (Emissive power is the quantity of heat radiated (emitted) from unit area in unit time.)RB$R}_{B$ = emissive power of black bodyQ$Q$ = Heat incident on body in unit time in unit areaQa$Q}_{a$ = Heat absorbed by body in unit time in unit areaa$a$ = absorptivitye$e$ = emissivityQa=a Q${Q}_{a}=a\phantom{\rule{0.22em}{0ex}}Q$In thermal equilibriumQuantity of radiant heat absorbed by body A = Quantity of heat emitted by body A a Q=R $\phantom{\rule{0.22em}{0ex}}a\phantom{\rule{0.22em}{0ex}}Q=R\phantom{\rule{0.22em}{0ex}}$Q=RB$Q={R}_{B}$So, a=R

RB$a=\frac{R}{{R}_{B}}$Emissivity, e=R

RB$e=\frac{R}{{R}_{B}}$So, a=e$a=e$

Spectral Distribution of Blackbody Radiation1. The radiant energy emitted per unit area per unit time by a blackbody depends on its temperature. 2. Hot objects radiate electromagnetic radiation in a large range of frequencies. 3. The rate of emission per unit area or power per unit area of a surface is defined as a funtion of the wavelength λ$\lambda$ of the emitted radiation. 4. At low temperature, the power radiated is small and primarily lies in the long wavelength region. 5. As the temperature is increased, rate of emission increases fast. At each temperature, the radiant energy contains a mixture of different wavelengths. 6. At higher temperatures, the total energy radiated per unit time increases and the proportion ofenergy emitted at higher frequencies or shorter wavelengths also increases.7. The spectral distribution depended only on the absolute temperature T of a blackbody and was independent of the material.

Wien’s Displacement LawIt is observed that the wavelength, for which emissive power of a black body maximum, is inversely proportional to the absolute temperature of the blackbody.𝜆max∝1

T$\mathit{\lambda}}_{max}\propto \frac{1}{T$𝜆maxT=b${\mathit{\lambda}}_{max}T=b$b = Wien’s constant = 2.897×10-3 m K$2.897\times {10}^{-3}\phantom{\rule{0.22em}{0ex}}m\phantom{\rule{0.22em}{0ex}}K$

Stefan-Boltzmann Law of RadiationThe rate of emission of radiant energy per unit area or the power radiated per unit area of a perfect blackbody is directly proportional to the fourth power of its absolute temperature.R=𝜎T4$R=\mathit{\sigma}{T}^{4}$

If Q is the amount of radiant energy emitted in time t by a perfect blackbody of surface area AQ

At=𝜎T4$\frac{Q}{At}=\mathit{\sigma}{T}^{4}$

If a body is not perfectly black body then the energy radiated per unit area per unit time is still proportional to the fourth power of temperature but is less than that for the blackbody.R=e𝜎T4$R=e\mathit{\sigma}{T}^{4}$wheree : emissivity of the surface.

If the perfect blackbody having absolute temperature T is kept in a surrounding which is at a lower absolute temperature T0$T}_{0$ , then the energy radiated per unit area per unit time=𝜎T4$=\mathit{\sigma}{T}^{4}$Energy absorbed from surroundings per unit area per unit time =𝜎T40$=\mathit{\sigma}{T}_{0}^{4}$So, net loss of energy by perfect blackbody per unit area per unit time=𝜎(T4-T40)$=\mathit{\sigma}({T}^{4}-{T}_{0}^{4})$

For an ordinary body, net loss of energy per unit area per unit time=e𝜎(T4-T40)$=e\mathit{\sigma}({T}^{4}-{T}_{0}^{4})$