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CLASS XII PHYSICS Chapter 13 - Kinetic theory of gases and radiation
- 1. CHAPTER 3 Kinetic Theory of Gases and Radiation CLASSXII PHYSICS MAHARASHTRASTATEBOARD
- 2. Can you recall? 1. What are different states of matter? 2. How do you distinguish between solid, liquid and gaseous states? 3. What are gas laws? 4. What is absolute zero temperature? 5. What is Avogadro number? What is a mole? 6. How do you get ideal gas equation from the gas laws? 7. How is ideal gas different from real gases? 8. What is elastic collision of particles? 9. What is Dalton's law of partial pressures?
- 3. SOLID LIQUID GAS MOLECULES ATOMS Matter is made up of tiny particles called molecules. All types of matter are made up of extremely small particles. STATES OF MATTER
- 4. Scanning Tunnelling Microscope (STM) Molecules are always in a state of motion and even when inside matter, they never stop moving.
- 6. Gas law’s • Boyle’s Law: V ∝ 𝟏 𝑷 at constant T • Charles Law: V 𝜶 T at constant P • Avogadro’s Law: V 𝜶 𝐧 • Gay-Lussac’s Law: P 𝜶 T at constant V Combining these all law’s, we get V 𝜶 n T 𝟏 𝑷 ∴ 𝑷𝑽 𝜶 𝒏𝑻 PV = nRT Where, R is proportionality constant OR Universal Gas Constant R = 8.314 J 𝒎𝒐𝒍−𝟏 𝒌−𝟏
- 7. Mole’s Unit: mol Formula: Mole = 𝑴𝒂𝒔𝒔 𝑴𝒐𝒍𝒂𝒓 𝑴𝒂𝒔𝒔 Concept: 1 mole of substance → Contain 6.022 X 𝟏𝟎𝟐𝟑 No. of mole (n) = 𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒑𝒂𝒓𝒕𝒊𝒄𝒍𝒆𝒔 𝟔.𝟎𝟐𝟐 𝑿 𝟏𝟎𝟐𝟑
- 8. ATOM Weight of one atom of an element = 𝑨𝒕𝒐𝒎𝒊𝒄 𝑾𝒆𝒊𝒈𝒉𝒕 𝟔.𝟎𝟐𝟐 𝑿 𝟏𝟎𝟐𝟑 Weight of one molecule of a compound = 𝑴𝒐𝒍𝒆𝒄𝒖𝒍𝒂𝒓 𝑾𝒆𝒊𝒈𝒉𝒕 𝟔.𝟎𝟐𝟑 𝑿 𝟏𝟎𝟐𝟑 For Gas, n = 𝑵 𝑵𝑨 Where, n → mole N → Number of molecules in a gas 𝑵𝑨 → Avogadro's number OR Number of molecule in 1 mole of gas
- 9. IDEAL GAS Equation, PV = nRT PV = 𝑵 𝑵𝑨 RT …………. [n = 𝑵 𝑵𝑨 ] Here, R = 𝑵𝑨𝒌𝑩 Where, 𝒌𝑩 is the Boltzmann Constant So, PV = 𝑵 𝑵𝑨 𝑵𝑨𝒌𝑩𝑻 PV = N𝒌𝑩𝑻 Where, N is the Number of Particles NOTE: A gas obeying the equation of state i.e., PV = nRT at all pressure, and temperature is an ideal gas.
- 10. Behaviour of gas • Gas contained in container is characterized by its pressure, volume and temperature. • Particle of gas are always in constant motion. • It is very hard to evaluate the motion of single particle of gas. • We find average of physical quantities and then relate to the macroscopic view of gas.
- 11. Fig.(a): A gas with molecules dispersed in the container: A stop action photograph. Fig.(b): A typical molecule in a gas executing random motion.
- 12. IDEAL gaS • A gas obeying ideal gas equation is nothing but an ideal gas. • Intermolecular interaction is different or absent. REAL GAS • Compose of atom/molecule which do interact with each other. • If there is atom/molecule are so apart then we can say that the real gases are ideal. • This can be achieved by lowering pressure and increasing temperature.
- 13. MEAN FREE PATH • Represented by ‘𝝀’ • Defined as average distance travelled by a molecule with constant velocity between two successive collision. • Mean free path is inversely vary with density. • Mean free path also inversely proportional to size of the particle of gases. Conclusion: 𝝀 = 𝟏 𝒅𝟐𝝆 𝝀 = 𝟏 𝒅𝟐𝝆𝝅 𝟐 𝝀 = 𝟏 𝒅𝟐( 𝑵 𝑽 )𝝅 𝟐 ……….[𝝆 = 𝑵 𝑽 ]
- 14. EAXMPLE • Obtain the mean free path of nitrogen molecule at 0 °C and 1.0 atm pressure. The molecular diameter of nitrogen is 324 pm (assume that the gas is ideal). Solution: Given T = 0 °C = 273 K, P = 1.0 atm = 1.01×𝟏𝟎𝟓 Pa and d = 324 pm = 324 × 𝟏𝟎−𝟏𝟐 m. For ideal gas PV = N𝒌𝑩T, ∴ 𝑵 𝑽 = 𝑷 𝒌𝑩𝑻 Using Eq., mean free path 𝝀 = 𝟏 𝒅𝟐( 𝑵 𝑽 )𝝅 𝟐 = 𝒌𝑩𝑻 𝟐𝝅𝒅𝟐𝑷 = 𝟏.𝟑𝟖 𝑿 𝟏𝟎−𝟐𝟑 𝑱 𝑲 (𝟐𝟕𝟑 𝑲) 𝟐𝝅 𝟑𝟐𝟒 𝑿 𝟏𝟎−𝟏𝟐 𝒎 𝟐 (𝟏.𝟎𝟏 𝑿 𝟏𝟎𝟓 𝑷𝒂) = 0.8 X 𝟏𝟎−𝟕 𝒎 Note that this is about 247 times molecular diameter.
- 15. Pressure of ideal gas A cubical box of side L. It contains n moles of an ideal gas. Volume of that cube is, V = 𝑳𝟑 , Where L → 𝑳𝒆𝒏𝒈𝒕𝒉 𝒐𝒇 𝒔𝒊𝒅𝒆 We have to find relation between pressure P of gas with molecular speed Here, change in momentum in –component ∆ 𝑷𝒙 = 𝑭𝒊𝒏𝒂𝒍 𝒎𝒐𝒎𝒆𝒏𝒕𝒖𝒎 − 𝑰𝒏𝒊𝒕𝒊𝒂𝒍 𝑴𝒐𝒎𝒆𝒏𝒕𝒖𝒎 = - m𝒗𝒙 − 𝒎𝒗𝒙 = - 2 m𝒗𝒙 So, total momentum applied on the wall is 2 m𝒗𝒙 The total distance travelled by particle of gas during collision of molecule and wall is 2L −𝒗𝒙 m 𝒗𝒙
- 16. ∆𝒕 = 𝟐𝑳 𝒗𝒙 ……[Speed = 𝑫𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝑻𝒊𝒎𝒆 ] For 1-molecule, Average force = Average rate of change of momentum = 𝑴𝒐𝒎𝒆𝒏𝒕𝒖𝒎 𝑻𝒊𝒎𝒆 = 𝟐𝒎𝒗𝒙𝟏 ∆𝒕 = 𝟐𝒎𝒗𝒙𝟏 𝟐𝑳 𝒗𝒙𝟏 = 𝟐𝒎𝒗𝒙𝟏 𝟐𝑳 x 𝒗𝒙𝟏 Average Force (𝑭𝟏) = 𝒎𝒗𝒙𝟏 𝟐 𝟐𝑳 Similarly, for particle 2,3,4, ……with respect to x-component velocities are 𝒗𝒙𝟐 , 𝒗𝒙𝟑 , 𝒗𝒙𝟒 , … . . Average Force (𝑭) = 𝒎(𝒗𝒙𝟏 𝟐 +𝒗𝒙𝟐 𝟐 + 𝒗𝒙𝟑 𝟐 + ……..) 𝑳
- 17. Pressure(P) = 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝐹𝑜𝑟𝑐𝑒 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠ℎ𝑎𝑑𝑒𝑑 𝑝𝑎𝑟𝑡 = 𝒎(𝒗𝒙𝟏 𝟐 +𝒗𝒙𝟐 𝟐 + 𝒗𝒙𝟑 𝟐 + ……..) 𝑳 𝐿2 = 𝒎(𝒗𝒙𝟏 𝟐 +𝒗𝒙𝟐 𝟐 + 𝒗𝒙𝟑 𝟐 + ……..) 𝑳𝟑 ………….(i) Average Velocity (𝑣𝑥 2 ) = 𝒗𝒙𝟏 𝟐 +𝒗𝒙𝟐 𝟐 + 𝒗𝒙𝟑 𝟐 + …….𝒗𝒙𝑵 𝟐 𝑵 N𝑣𝑥 2 = 𝒗𝒙𝟏 𝟐 + 𝒗𝒙𝟐 𝟐 + 𝒗𝒙𝟑 𝟐 + … … . 𝒗𝒙𝑵 𝟐 Put it in equation (i) Pressure(P) = 𝒎N𝑣𝑥 2 𝑳𝟑 Pressure(P) = 𝒎N𝑣𝑥 2 𝑽 ……..[V=𝐿3 ] ……………(ii)
- 18. Now, as there is perfectly collision in molecule 𝟏 𝟐 𝒎𝒗𝒙 𝟐 = 𝟏 𝟐 𝒎𝒗𝒚 𝟐 = 𝟏 𝟐 𝒎𝒗𝒛 𝟐 𝒗𝒙 𝟐 = 𝒗𝒚 𝟐 = 𝒗𝒛 𝟐 …………………….. (a) For average velocity of all component (x, y, z) 𝒗𝟐 = 𝒗𝒙 𝟐 + 𝒗𝒚 𝟐 + 𝒗𝒛 𝟐 𝒗𝟐 = 𝒗𝒙 𝟐 + 𝒗𝒙 𝟐 + 𝒗𝒙 𝟐 ……… From equation (a) 𝒗𝟐 = 𝟑 𝒗𝒙 𝟐 𝒗𝒙 𝟐 = 𝟏 𝟑 𝒗𝟐 Put in equation (ii) P = 𝒎𝑵 ( 𝟏 𝟑 𝒗𝟐) 𝑽 = 𝒎𝑵𝒗𝟐 𝟑𝑽 = 𝟏 𝟑 𝑵 𝑽 𝒎 𝒗𝟐
- 19. ROOTMEAN SQUARE(rms) SPEED P = 𝟏 𝟑 𝑵 𝑽 𝒎 𝒗𝟐 ∴ 𝟑 𝑷 𝑽 = 𝑵 𝒎 𝒗𝟐 𝒗𝟐 = 𝟑 𝑷 𝑽 𝑵 𝒎 We know that, P V = n R T 𝒗𝟐 = 𝟑 𝒏 𝑹 𝑻 𝑵 𝒎 = 𝟑 ( 𝑵 𝑵𝑨 ) 𝑹 𝑻 𝑵 𝒎 …………[n = 𝑵 𝑵𝑨 ] 𝒗𝟐 = 𝟑 𝑵 𝑹 𝑻 𝑵𝑵𝑨𝒎 = 𝟑 𝑹 𝑻 𝑵𝑨𝒎 𝒗𝟐 = 𝟑 𝑹 𝑻 𝑵𝑨𝒎 𝒗𝒓𝒎𝒔 = 𝟑 𝑹 𝑻 𝑴𝟎 …….[𝑴𝟎= 𝑵𝑨𝐦]
- 20. INTERPRETATIONOF TEMPERATURE IN KTG We know that, P = 𝟏 𝟑 𝑵 𝑽 𝒎 𝒗𝟐 P V = 𝟏 𝟑 𝑵 𝒎 𝒗𝟐 = 𝟐 𝟑 𝑵 ( 𝒎 𝒗𝟐 𝟐 ) ……………(i) In ideal gas there is no interaction molecule so P.E. = 0. Here, Total Energy (E) = K.E. + P.E. = N ( 𝒎 𝒗𝟐 𝟐 ) + 0 = N ( 𝒎 𝒗𝟐 𝟐 ) …………….(ii) Put in equation (i) P V = 𝟐 𝟑 E ……….(iii) Using Ideal Gas, P V = N 𝒌𝑩 𝑻 = 𝟐 𝟑 E ……….(iv) N 𝒌𝑩 𝑻 = 𝟐 𝟑 E E = 𝟑 𝟐 N 𝒌𝑩 𝑻 𝑬 𝑵 = 𝟑 𝟐 𝒌𝑩 𝑻
- 21. Example At 300 K, what is the rms speed of Helium atom? [mass of He atom is 4u, 1u = 1.66 × 𝟏𝟎−𝟐𝟕 kg; 𝒌𝑩= 1.38 × 𝟏𝟎−𝟐𝟑 J/K] Solution: Given T = 300 K, m = 4 × 1.66 × 𝟏𝟎−𝟐𝟕 kg Average K. E = 𝟏 𝟐 𝒎 𝒗𝟐 = 𝟑 𝟐 𝒌𝑩 𝑻 𝒗𝟐 = 𝟑𝒌𝑩 𝑻 𝒎 = 𝟑 𝑿 𝟏.𝟑𝟖 𝑿𝟏𝟎−𝟐𝟑 𝑿 𝟑𝟎𝟎 𝟒 𝑿1.66 × 𝟏𝟎−𝟐𝟕 = 187.05 X 𝟏𝟎𝟒 𝒗𝒓𝒎𝒔 = 𝒗𝟐 = 13.68 X 𝟏𝟎𝟐 = 1368 m/s
- 22. 𝑳𝑨𝑾 𝑶𝑭 𝑬𝑸𝑼𝑰𝑷𝑨𝑹𝑻𝑰𝑻𝑰𝑶𝑵 Energy of single molecule is K.E = 𝟏 𝟐 𝒎𝒗𝒙 𝟐 + 𝟏 𝟐 𝒎𝒗𝒚 𝟐 + 𝟏 𝟐 𝒎𝒗𝒛 𝟐 For gas at temperature T, The average K.E per molecule is <K.E.> <K.E> = < 𝟏 𝟐 𝒎𝒗𝒙 𝟐 > + < 𝟏 𝟐 𝒎𝒗𝒚 𝟐 > + < 𝟏 𝟐 𝒎𝒗𝒛 𝟐> But the mean distance is 𝟑 𝟐 𝒌𝑩 𝑻 𝟏 𝟐 𝒌𝑩 𝑻 is the molecular energy in single x or y or z component.
- 23. DEGREEOF FREEDOM The molecule is free to move in whole space. Molecule can move in 3-D space. “Degree of freedom of a system are defined as the total number of co-ordinate required to describe the position and configuration of the system completely.” DIATOMIC MOLECULE Molecules having 3-translational degrees (x, y, z) Diatomic molecule rotate around axis Suppose, a molecule moving along axis z with 𝑰𝒛 with 𝒘𝒛 K.E = 𝟏 𝟐 𝑰𝒛 𝒘𝒛 𝟐 For axis y , K.E = 𝟏 𝟐 𝑰𝒚 𝒘𝒚 𝟐 So, Total energy = E (Translational) + E (Rotational) = [ 𝟏 𝟐 𝒎𝒗𝒙 𝟐 + 𝟏 𝟐 𝒎𝒗𝒚 𝟐 + 𝟏 𝟐 𝒎𝒗𝒛 𝟐] + [ 𝟏 𝟐 𝑰𝒛 𝒘𝒛 𝟐 + 𝟏 𝟐 𝑰𝒚 𝒘𝒚 𝟐 ] Fig.: The two independent axes z and y of rotation of a diatomic molecule lying along the x-axis.
- 24. From equipartition of energy: Each and every molecule separated with 𝟏 𝟐 𝒌𝑩 𝑻 molecule. For, Translational motion → 3 (dof) → [ 3 X 𝟏 𝟐 𝒌𝑩 𝑻 ] Rotational motion → 2 (dof) → [ 2 X 𝟏 𝟐 𝒌𝑩 𝑻 ] Vibrational motion → 2 (dof) → [ 2 X 𝟏 𝟐 𝒌𝑩 𝑻 ] → [ 𝟏 𝟐 m 𝒖𝟐 + 𝟏 𝟐 k 𝒓𝟐 ] • The energy of molecule [ 𝟏 𝟐 𝒌𝑩 𝑻 ] is valid for high temperature and not valid for extremely low temperature where quantum effects become important.
- 25. SPECIFIC HEAT CAPACITY If temperature of gas increase (Cause considerable change in volume and pressure) 𝑪𝑽 → Specific heat at constant volume 𝑪𝑷 → Specific heat at constant pressure MAYER’S RELATON Consider, 1 mole of gas (Ideal) enclosed in cylinder Let P, V, T be the pressure, volume and temperature dT = Change in temperature (Volume is constant → 𝑵𝒐 𝒘𝒐𝒓𝒌 𝒊𝒔 𝒅𝒐𝒏𝒆) d𝑸𝟏 → Increase in the internal energy dE → 𝑰𝒏𝒕𝒆𝒓𝒏𝒂𝒍 𝒆𝒏𝒆𝒓𝒈𝒚 ∴ 𝒅𝑸𝟏 = 𝒅𝑬 = 𝑪𝑽𝒅𝑻 … … … … 𝒊 Where, 𝑪𝑽 → 𝑴𝒐𝒍𝒂𝒓 𝒔𝒑𝒆𝒄𝒊𝒇𝒊𝒄 𝒉𝒆𝒂𝒕 𝒂𝒕 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝒗𝒐𝒍𝒖𝒎𝒆
- 26. If gas is heated with same given temperature at constant pressure then, dV → Change (increase) in volume Here, work is done to pull backward the piston dw = p dv ∴ 𝒅𝑸𝟐 = dE + dw = 𝑪𝑷𝒅𝑻 ………(ii) Where, 𝑪𝑷 → Molar specific heat at constant pressure From equation (i) ∴ 𝑪𝑽𝒅𝑻 + 𝒅𝒘 = 𝑪𝑷𝒅𝑻 𝑪𝑽𝒅𝑻 + 𝑷 𝒅𝑽 = 𝑪𝑷𝒅𝑻 𝑷 𝒅𝑽 = 𝑪𝑷𝒅𝑻 - 𝑪𝑽𝒅𝑻 𝑷 𝒅𝑽 = (𝑪𝑷 - 𝑪𝑽 ) 𝒅𝑻 ……………… (iii)
- 27. For, 1 mole of gas, PV = RT …………(n = 1) But, when pressure is constant P dV = R dT (𝑪𝑷 - 𝑪𝑽 ) 𝒅𝑻 = R dT 𝑪𝑷 - 𝑪𝑽 = 𝑹 ………(Mayer’s relation) Now, 𝑪𝑷 - 𝑪𝑽 = 𝑹 𝑱 Where, J – mechanical equivalent of heat Here, 𝑪𝑷 = 𝑴𝟎 𝑺𝑷 𝑪𝑽 = 𝑴𝟎 𝑺𝑽 𝑴𝟎 = Molar mass of gas 𝑺𝑽, 𝑺𝑷 = Respective principle of heat 𝑴𝟎 𝑺𝑷 - 𝑴𝟎 𝑺𝑽 = 𝑹 𝑱 (𝑺𝑷 - 𝑺𝑽) = 𝑹 𝑴𝟎𝑱
- 28. a) MONOATOMICGAS For translational motion → 3 X 𝟏 𝟐 𝒌𝑩 𝑻 → 𝟑 𝟐 𝒌𝑩 𝑻 𝑪𝑽 = 𝒅𝑬 𝒅𝒕 = 𝟑 𝟐 𝒌𝑩𝑻𝑵𝑨 𝑻 ……….(𝒌𝑩𝑵𝑨 = 𝑹) 𝑪𝑽 = 𝟑 𝟐 𝒌𝑩N = 𝟑 𝟐 𝑹 𝑪𝑷 = 𝑹 + 𝑪𝑽 = R + 𝟑 𝟐 𝑹 𝑪𝑷 = 𝟓 𝟐 𝑹 ∴ 𝜸 = 𝑪𝑷 𝑪𝑽 = 𝟓 𝟐 𝑹 𝟑 𝟐 𝑹 𝜸 = 𝑪𝑷 𝑪𝑽 = 𝟓 𝟑
- 29. b) diaTOMICGAS Gases like 𝑶𝟐, 𝑵𝟐, 𝑪𝑶, 𝑯𝑪𝒍 𝒉𝒆𝒍𝒅 𝒂𝒕 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝒕𝒆𝒎𝒑𝒆𝒓𝒂𝒕𝒖𝒓𝒆 𝑻. Each molecule have 3-Translational and 2-Roational dof. E = [3 X 𝟏 𝟐 𝒌𝑩 𝑻] + [ 2 X 𝟏 𝟐 𝒌𝑩 𝑻] = [ 𝟑 𝟐 𝑵𝑨𝒌𝑩 𝑻] + [ 𝟐 𝟐 𝑵𝑨𝒌𝑩 𝑻] = 𝟓 𝟐 𝑵𝑨𝒌𝑩 𝑻 𝑪𝑽 = 𝒅𝑬 𝒅𝒕 = 𝟓 𝟐 𝒌𝑩𝑻𝑵𝑨 𝑻 ……….(𝒌𝑩𝑵𝑨 = 𝑹) 𝑪𝑽 = 𝟓 𝟐 𝒌𝑩𝑵𝑨 = 𝟓 𝟐 𝑹 𝑪𝑷 = 𝑹 + 𝑪𝑽 = R + 𝟓 𝟐 𝑹 = 𝟕 𝟐 𝑹 ∴ 𝜸 = 𝑪𝑷 𝑪𝑽 = 𝟕 𝟐 𝑹 𝟓 𝟐 𝑹 = 𝟕 𝟓 ……….(For rigid rotator body)
- 30. For non rigid vibrating molecule E = Translational Energy + Rotational Energy + Vibrational Energy E = [3 X 𝟏 𝟐 𝑵𝑨𝒌𝑩 𝑻] + [ 2 X 𝟏 𝟐 𝑵𝑨𝒌𝑩 𝑻] + [ 2 X 𝟏 𝟐 𝑵𝑨𝒌𝑩 𝑻] = 𝟕 𝟐 𝑵𝑨𝒌𝑩 𝑻 𝑪𝑽 = 𝒅𝑬 𝒅𝒕 = 𝟕 𝟐 𝒌𝑩𝑻𝑵𝑨 𝑻 ……….(𝒌𝑩𝑵𝑨 = 𝑹) 𝑪𝑽 = 𝟕 𝟐 𝒌𝑩𝑵𝑨 = 𝟕 𝟐 𝑹 𝑪𝑷 = 𝑹 + 𝑪𝑽 = R + 𝟕 𝟐 𝑹 = 𝟗 𝟐 𝑹 ∴ 𝜸 = 𝑪𝑷 𝑪𝑽 = 𝟗 𝟐 𝑹 𝟕 𝟐 𝑹 = 𝟗 𝟕
- 31. b) POLYaTOMICGASES E = Translational Energy + Rotational Energy + (f x Vibrational Energy) Where, f → Number of vibrational dof E = [3 X 𝟏 𝟐 𝑵𝑨𝒌𝑩 𝑻] + [ 3 X 𝟏 𝟐 𝑵𝑨𝒌𝑩 𝑻] + [ f X 2 X 𝟏 𝟐 𝑵𝑨𝒌𝑩 𝑻] = 𝟏 𝟐 𝑵𝑨𝒌𝑩 𝑻 (3+3+2f) = (3 + f) 𝑵𝑨𝒌𝑩 𝑻 𝑪𝑽 = 𝒅𝑬 𝒅𝒕 = (𝟑+𝒇)𝒌𝑩𝑻𝑵𝑨 𝑻 ……….(𝒌𝑩𝑵𝑨 = 𝑹) 𝑪𝑽 = (3+f)𝒌𝑩𝑵𝑨 = (3+f) 𝑹 𝑪𝑷 = 𝑹 + 𝑪𝑽 = R + (3+f) 𝑹 = 𝟑 + 𝒇 𝑹 ∴ 𝜸 = 𝑪𝑷 𝑪𝑽 = (𝟒+𝒇) (𝟑+𝒇)
- 32. HEAT Heat is the form of energy which can be formed from one object to another object. MODES OF CONDUCTION 1)Conduction 2)Convection 3)Radiation • Does not required material medium for transfer • Emission of heat → radiant energy → Electromagnetic wave • Can travel through both material and vacuum medium. • Fastest way of heat transfer. Requires material medium for transfer.
- 33. INTERACTION OF THERMAL RADIATIONANDMATTER When any thermal radiation falls on object then some part is reflected, some part is absorbed and some part transmitted. Consider, Q → Total amount of heat energy incident 𝑸𝒂 → 𝑨𝒎𝒐𝒖𝒏𝒕 𝒐𝒇 𝒉𝒆𝒂𝒕 𝒂𝒃𝒔𝒐𝒓𝒃𝒆𝒅 𝑸𝒓 → 𝑨𝒎𝒐𝒖𝒏𝒕 𝒐𝒇 𝒉𝒆𝒂𝒕 𝒓𝒆𝒇𝒍𝒆𝒄𝒕𝒆𝒅 𝑸𝒕 → 𝑨𝒎𝒐𝒖𝒏𝒕 𝒐𝒇 𝒉𝒆𝒂𝒕 𝒕𝒓𝒂𝒏𝒔𝒎𝒊𝒕𝒕𝒆𝒅 Total energy (Q) = 𝑸𝒂 + 𝑸𝒓 + 𝑸𝒕 ∴ 𝟏 = 𝒂 + 𝒓 + 𝒕𝒓 Where, a = 𝑸𝒂 𝑸 , r = 𝑸𝒓 𝑸 , 𝒕𝒓 = 𝑸𝒕 𝑸 Coefficient of absorption, reflection, transmission
- 34. COEFFICIENT OF HEAT READIATION • Coefficient of absorption/absorptive power/absorptivity “The ratio of amount of heat absorbed to total amount of heat incident.” a = 𝑸𝒂 𝑸 • Coefficient of reflection/reflectance “The ratio of amount of heat reflected to total amount of heat incident.” r = 𝑸𝒓 𝑸 • Coefficient of transmission or transmittance “The ratio of amount of heat is transmitted to total amount of heat incident.” 𝒕𝒓 = 𝑸𝒕 𝑸
- 35. Cases a)Perfect transmitter: r = 0, a = 0, 𝒕𝒓 ≠ 1 Object is said to be completely transparent. b) Di-athermanous substance: 𝒕𝒓 ≠ 0 “Substance through which heat radiation can pass is called di- athermanous” Neither a good absorber nor good reflector E.G., Quartz, Sodium Chloride, Hydrogen, Oxygen c) Athermanous substance: 𝒕𝒓 = 0, a + r = 1 “Substance which are largely opaque to thermal radiation i.e. do not transmit heat.”
- 36. d) Good reflector/poor absorber/poor transmitter: 𝒕𝒓 = 0, a = 0, r = 1 e) Perfect blackbody: 𝒕𝒓 = 0, r = 0 and a = 1 “All the incident energy is absorbed by the object such an object is called a perfect blackbody.” NOTE: All the a (absorption), r (reflection) and t (transmission) depends upon wavelength of incident radiation. “A body which absorb entire radiant energy incident on it, is called an ideal or perfect blackbody.” Platinum black absorb nearly 97 % of incident radiant heat on it. Good absorber are good emitter and poor absorber are poor emitter. Fig.: Ferry’s blackbody.
- 37. EMISSION OF HEAT RADIATION Pierre prevost published a theory of radiation known as theory of exchange of heat. At temperature 0 K or above it. • Body radiate thermal energy and at same time they absorb radiation received from surrounding. • Emission per unit time depends upon surface of emission, its area and size. • Hotter body radiate at higher rate than the cooler body
- 38. AMOUNT OF HEAT RADIATED BY THE BODY DEPENDS ON – • The absolute temperature of body • The nature of body (Polished or not, size, colour) • Surface area of body (A) • Time duration Amount of heat radiated 𝜶 Surface area of body and time duration Q 𝜶 At Q = Rat Where, R → Emissive power/Radiant power R = 𝑸 𝑨𝒕 SI Unit – J 𝒎−𝟐 𝒔−𝟏 Dimension - [𝑳𝟎 𝑴𝟏 𝑻−𝟑 ]
- 39. EMISSIVITY “The coefficient of emission or emissivity (e) of a given surface is the ratio of the emissive power R of the surface to emissive power 𝑹𝑩 of perfect black body at same temperature.” e = 𝑹 𝑹𝑩 For perfect blackbody, e = 1, Perfect reflector, e = 0 KIRCHOFF’S LAW OF HEATRADIATION “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 wavelength.” 𝑹 𝒂 = 𝑹𝑩 ∴ 𝑹 𝑹𝑩 = 𝒂 ∴ 𝒂 = 𝒆 At thermal equilibrium, emissivity is equal to absorptivity.
- 40. WEIN’S DISPLACEMENTLAW The wavelength for which emissive power of a blackbody is maximum is inversely proportional to the absolute temperature of the blackbody. 𝝀𝒎𝒂𝒙 𝜶 𝟏 𝑻 𝝀𝒎𝒂𝒙 = 𝒃 𝑿 𝟏 𝑻 Where, b → Wein’s constant b = 2.897 X 𝟏𝟎−𝟑 mk Law is useful to determine distant length.
- 41. SREFAN-BOLTZMANN LAWOF RADIATION “The rate of emission of radiant energy per unit area or the power radiated per unit area of a perfect blackbody is directly proportional to fourth power of its temperature.” R 𝜶 𝑻𝟒 R = 𝝈 𝑻𝟒 Where, 𝝈 → 𝑺𝒕𝒆𝒇𝒂𝒏′ 𝒔 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝝈 = 𝟓. 𝟔𝟕 𝑿 𝟏𝟎−𝟖 Unit → J 𝒎−𝟐 𝒔−𝟏 𝒌−𝟒 Dimension → [𝑳𝟎 𝑴𝟏 𝑺−𝟏 𝑲−𝟒 ] We know that, R = 𝑸 𝑨𝒕 So, R = 𝝈 𝑻𝟒 → 𝑸 𝑨𝒕 = 𝝈 𝑻𝟒 …………(For black body) For ordinary body, R = 𝐞 𝝈 𝑻𝟒
- 42. Let, T → 𝑨𝒃𝒔𝒐𝒍𝒖𝒕𝒆 𝒕𝒆𝒎𝒑𝒆𝒓𝒂𝒕𝒖𝒓𝒆 𝒐𝒇 𝒃𝒍𝒂𝒄𝒌𝒃𝒐𝒅𝒚 𝐤𝐞𝐞𝐩 𝐚𝐭 𝐥𝐨𝐰𝐞𝐫 𝐚𝐛𝐬𝐨𝐥𝐮𝐭𝐞 𝐭𝐞𝐦𝐩𝐞𝐫𝐚𝐭𝐮𝐫𝐞 𝑻𝑶 Energy radiated per unit time = 𝝈 𝑻𝟒 Energy radiated per unit time in surrounding = 𝝈 𝑻𝟎 𝟒 Net loss of energy by perfect blackbody per unit time = 𝝈 𝑻𝟒 - 𝝈 𝑻𝟎 𝟒 = 𝝈 ( 𝑻𝟒 - 𝑻𝟎 𝟒 ) For ordinary body = 𝐞𝝈 ( 𝑻𝟒 - 𝑻𝟎 𝟒 )