HEAT AND THERMODYNAMICS - Lecture Notes

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) CHAPTER - 00 HEAT AND THERMODYNAMICS Thermometry Heat Heat is the form of energy which gives the sensation of hotness or coldness of a body. It is the physical cause of sensation of hotness or coldness of a body. S.I. unit : Joule Temperature Degree of hotness or coldness of a body is temperature. It determines the direction of heat flow. S.I. unit : Kelvin Thermometry is the technique for the quantitative determination of thermodynamic temperature Different temperature Scale Scale LFP UFP n Representation Celsius 00C 1000C 100 10C Fahrenheit 320F 2120F 180 10F Reaumer 00R 800R 180 10R Kelvin 273 K 373 K 100 1 K  Let X and Y are two temperature scales then X  XLFP  Y  YLFP  a cons tan t XUFP  XLFP YUFP  YLFP 1

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) Conversion of temperature Scales Let x be a faulty / Reference Scale x  xLFP  C  0  F  32  R  0 K  273 xUFP  xLFP 100  0 212  32 80  0 373  273 x  xLFP  C  F  32  R K  273 F1 9 C1  32 xUFP  xLFP 100 180 80 180  Relation between temperature scales Celsius and Fahrenheit 5 C  F  32 100 180 F2 9 C2  32 C  F  32 5 59 F  9 C  32 [F2  F1]  9 [C2  C1] 5 5 0F  F  9 C Slope = 9 5 32 5  C  5 F 9 0C Common Re ading in celsius and Celsius and Kelvin Fahrenheit is  400 C K  273 Fahrenheit and Kelvin  F  32 K  273 100 100  180 100 C  K  273 F  32  K  273 K  C  273 95 No common Reading in Celsius and Common Reading in Fahrenheit and kelvin Kelvin is 574.25 Since the no. of divisions in both the scales are same 2 Change in temperature in Celsius is equivalent to change in temperature Fahrenheit

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) Absolute zero When we extrapolate the pressure versus temperature graph of low density ideal gases it will reach the absolute minimum temperature known as “The absolute zero”. Its value is –273.15oC. Absolute zero is the foundation of kelvin scale. It is taken as 0 kelvin. To find unknown temperature using triple point Let X be a thermometric property Xt  Thermometric property at an unknown temperature t XTr  Thermometric property at triple point unknown temperature  t  Xt  273.16 X Tr for eg. t  Pt  273.16 PTr Note :- As the no. of divisions in a temperature scale increases “size of the degree” decreases. The zero value in Kelvin scale is known as “the absolute zero”. It is 0 Kelvin or –273.150C. In modern temperature scales absolute zero and triple point of water are taken as fixed reference points for calibration. Triple point of water 0.010C or 273.16 K @ 4.58 mm Hg pressure or at pressure 6.11 × 102 Pa or 6 × 10–3 atm Thermometers Devices used for the measurement of temperature Principle : Linear variation of thermometric property with temperature Eg : Pressure of a gas Volume of a liquid 3

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) Resistance of metal Light Intensity Thermo emf, magnetic properties Liquid Thermometers Mercury and alcohol are thermometric liquids Advantages of Hg High conductivity, high boiling point, low specific heat, high visibility, high sensitivity to heat, high angle of contact Gas Thermometers (Most Sensitive) Constant Volume Gas Thermometers P  T Based on Gaylussac’s law or Pressure Law or Charles law of pressure Constant Pressure Gas Thermometers V  T Based on Charles law Resistance Thermometers Eg : Platinum Resistance Thermometer Germanium Resistance Thermometer To find unknown temp Let x be a thermometric property x0 = thermometric property at 00C x100 = thermometric property at 1000C xt = thermometric property at t0C then unknown temp t  xt  x0 100 x100  x0 for pressure t  Pt  P0 100 P100  P0 for resistance t  Rt  R0 100 R100  R0 4

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) Thermal Expansion Generally, solids are found to expand on heating. This is called thermal expansion. Solids are made up of atoms and molecules. At a given temperature atoms and molecules in the solid are located at some equilibrium distance. When the solid is heated, the amplitude of vibration of atoms and molecules increases. Therefore effective interatomic separation increases. This causes thermal expansion. Thermal expansion of solids is of three types 1) Linear expansion 2) Areal expansion 3) Volume expansion Linear Expansion Increases in length of the solid on heating. Suppose L is the original length of a rod. Let L be a small increase in length of the rod when its temperature is increased by small amount t It is found that L  L L  t L  Lt L  Lt   L  change in length L1  L  L L1  L  Lt Lt original length  Rise in temp. final length L1  L1 t  is the constant of proportionality and is called co-efficient of linear expansion.  is defined as small change in length per unit original length per oC change in temperature. Value of  depends on material of the solid. 2. Area expansion / Superficial expansion Increase in surface are of solid with on heating. Let A be the change in area with temperature change A  A A  t A  At A  At   A   change in area At original area  rise in temp. 5

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE)   co-efficient of areal expansion unit : o C1 K 1 final area A1  A  A A1  A  At A1  A1 t 3. Volume expansion / cubical expansion Increase in volume of the solid on heating Let V be the original volume of the solid. V is the change in volume with rise in temperature t V  V V  t V  Vt V  Vt   V   change in volume Vt original volume  rise in temp. final volume V1  V  V unit : o C1 K 1 V1  V  Vt V1  V1 t Note : , ,  are characteristics of substances but it is not strictly a constant. It usually depends on temperature. Variation of  of copper with respect to temperature is given by From the graph  become constant at very high temperature 6

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) Expansion of Solids Cubical Three types of Expansion is possible in solids Expansion 1) Linear Expansion (in length)   cubical 2) Areal Expansion (in area) Expansivity  V 3) Volume Expansion (in volume) Areal Linear Expansion Vt Expansion  V  V t   linear   superficial Expansivity Expansivity Co-efficient of   L  A expansion Lt At Change in value L L t  A A t Fractional change L   t  A  t V   t L A V % change L 100   t 100  A 100  t 100  V 100   t 100 Final Value L A V L '  L[1 t] A '  A [1 t] V '  V [1 t] Relation between , and  We have      A   V t AT VT When we consider a cube of side  It has a face area A  2 and volume V  3 Since A  2 7

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) A  2  A   A  2  At t   2 also V  3 V  3  V  V  3  Vt t   3 for isotropic solids   1   2   3  ::  1: 2:3 for Anisotropic solids [Materials which show different properties in different dimensions]   x  y  z Application of Linear Expansion Time loss/gain of a pendulum clock Since invar (iron-nickel alloy) has low values of expansion co-efficients it is used to make pendulum. But when temperature increases there will be slight increase in length of pendulum and will finally results in change in time period. we have time period T  2  / g T  2 1/2 T  k 1/2 g Fractional change T 1      T 2   Fractional change in Time period for unit time T  1   T2 8

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) Time loss/gain in a day in seconds 1 day = 24 hrs = 86400 s t  1    86400 2 In summer Temperature  length  time period  Time loss, clock will become slow In winter Temperature  length  time period  Time gain, clock will become fast Thermal Stress When a metal rod is rigidly fixed at its both ends so that it is prevented from expansion or contraction. On heating stress will be developed due to thermal strain it is called thermal stress Thermal strain = l   t l Y  Thermal stress Thermal strain Thermal stress  Y  Thermal strain  Y t Tension or force  Thermal stress  Area  YA  t Note : Thermal stress developed in a rod is depending on the material of rod and rise in temperature and is independent of length of the rod.  If the rod is free to expand, then thermal stress developed on the rod will be zero. Bimetallic Strip Two metal rods of same length and different  joined together to form a bimetallic strip. On heating bimetallic strip will bend in such a way that metal having larger value of  will come on the convex side and metal having smaller value of  will come on the concave side. On cooling the reverse happens. A > B A B Radius of bimetallic arc R d if  A  B [A  B ]t 9

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) Expression for Radius of bimetallic strip We have 1  1 t Let  be the initial length of both the rods. But after heating final lengths of A and B will be different due to different     1  At Angle  arc A R    1  B t    LA B Rd   LB R   R d 1 d ........... (1) A RR  B also   1 At .............(2) A 1 Bt  B from (1) and (2) 1 d  1 At R 1 Bt 1 d  1 At1 Bt1 R 1 d  1  A t 1  Bt R d  A  B t R R d A  B  t 10

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) Expansion of Liquids For heating a liquid, it has to be kept in a container. On heating the container will also expand. Real Expansion of = Apparent Expansion of + Expansion of Liquid Liquid the container     Real apparent container   Real apparent  3vessel Anomalous Expansion Water Volume of given amount of water decreases with increase in temperature from 0 to 40C. But beyond 40C water will normally expand. Water has the least volume and maximum density at 40C Volume Pmax 00C 40C Temp It plays an important role in the survival of aquaticf life in cold winter season in polar regions. There is expansion of water above and below 40C Condition for two metal rods of different length and  to have same difference in length at all temperature For difference in length independent of temperature  = constant 1  2 11 t   2 t 2 11  2 2 1  2 2 1 The correct the reading of a metallic scale Case 1 : When scale is expanding True reading  Scale reading(1  t)    of the scale 11

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) t  temp diff (temp at which measurement is taken and temperature at which scale is calibrated Case 2 : When object (only) is expanding Measurement value = True value (1 0t) 0   of object material Case 3 : When both scale and object are expanding The value = scale value [1 (s  0 )] t 0   of the object s   of the scale Variation of density with temperature density  mass volume m v as temp increases volume increases and density decreases ' m v' ' m '  V [1 t] 1 t  '   (1  t) expanding binomially Variation of upthrust with temperature Apparent weight = Actual weight – Upthrust At normal temperature Upthrust FB  VLg When temperature increases by t Upthrust FB'  V ' L' g FB'  V [1 st]  L g [1 Lt] 12

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) FB'  VLg 1 st  1  L t   FB'  FB 1  s t  1  L t   FB1  1 st FB 1 L t In other way FB1 FB[1 (s  L ) t Special Cases Metal disc with hole   D2  D1 t o2C  t1oC D2  D1 D1 t2  t1 R2  R1   R2  R1 D  diameter R  radius R1 t2  t1 13

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) Broken Ring t 2oC  t1oC   x2  x1 x  Gap between ends x1 t2  t1 1. When a metallic disc with an inner hole is heated the diameter of the hole will also increases. 2. When a metallic sphere with a cavity is heated size and volume of cavity increases. 3. When a broken ring with gap between ends is heated the gap will also increases. Calorimetry Calorimetry means measurement of heat. It deals with specific measurement of specific heat and latent heat of different substances. Joules mechanical equivalent of heat (J) Joule found that when mechanical work is converted into heat (Q) the ratio of W and Q is always a constant W  J or W  JQ Q J  joules mechanical equivalent of heat (It is a conversion factor) J = 4.186 Joule/ Calorie Calorie It is the amount of heat energy required to rise the temperature of 1 g water by 10C (14.50C to 15.50C) 1 Calorie = 4.2 J Principle of Calorimetry Law of mixtures Heat lost by hot body = heat gained by cod body [when no heat is lost to the surroundings] Specific heat capacity (C) It is the amount of heat energy required to rise the temperature of unit mass (1 g on 1 kg) of any substance by 10C or 1 K for a given mass m heat required is H  m  C 14

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) H is called heat capacity or thermal capacity for a given mass m and t rise in temperature Amount of heat required Q mct dQ = mcdt If C is varying with temperature Unit of specific heat capacity t2 C  Q  Joule mt kg K Q   mcdt t SI unit : J Kg–1K–1 C Q  Calorie CGS unit : Calg–1 0C–1 mt g  0C for water C = 4200 J Kg–1K–1 C = 1 Calg–1 0C–1 for ice C = 2100 J Kg–1K–1 C = 0.5 Calg–1 0C–1 Latent Heat It is the amount of heat energy exclusively utilized for phase transition at certain fixed temperatures like melting point or boiling point etc. It can be Latent heat of fusion Lf  and Latent heat of vaporisation LV  Lf   heat required to convert unit mass of solid into liquid state at melting point For a given mass m for fusion Q  mLf LV   amount of heat enegy required to convert unit mass of liquid into its gaseous state at boiling point. For a given mass m Q  mLv for vaporisation Latent heat of fusion (Lf) For solid  Liquid Transition @ melting point Eg : 1 g ice @ 00 C Lf  1 g water @ 00C Lf of ice = 80 cal/g Lf of ice = 3.35105 J / Kg Latent heat of Vapourisation (Lv) For liquid  Gas transition @ boiling point Eg : 1 g water @ 1000C LV  1 g steam @ 1000C LV of water = 540 Cal/g LV of water = 2.26 106 J / Kg 15

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) Unit of Latent Heat Capacity Q = mL S.I. Unit : Joule/kilogram L = Q/m CGS Unit : Calorie/gram Water Equivalent It is the amount of water which has got the same heat capacity as that of a given substance. Let H = mC for a given substance for water H  mC mC  mC m  mC C Water equivalent is numerically equal to heat capacity of the substance in CGS system Temperature - Time Graph Graph plotted with time along x axis and temperature along y axis Phase Change Increase in temperature dQ  mL dQ  mcdT p  dQ  mL p  dQ  mcdT dt dt dt dt L   P  dt dT  P   1  m  dt  m  C L = a constant × dt dT  a constant  1 dt C L  dt 1 L  time interval Slope  L  change in x value C C 1 S lo p e 16

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) P-T diagram A graph between the temperature T and pressure P of the substance is called phase diagram or P-T diagram. Phase Diagram Note : for water melting point decreases with increase in pressure 17

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) Heating Curve Regelation Melting of ice at lower temperature due to increase in pressure and refreezing when pressure is with drawn is regelation Heat Transfer Transport of heat energy from one point to another can be done in 3 ways Conduction Convection Radiation Conduction Particle to particle heat transfer without actual transport of matter. Its common to solids and mercury Gravity has no effect in conduction Metals are good conductors of heat Heat Current (H) Rate of flow of heat energy through a conductor H Q unit : Watt t Thermal Conductivity The ability of a conductor to allow the passage of heat energy through it 18

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) Expression for Thermal conductivity Consider a metallic rod with length  , uniform cross sectional area A. Its both ends are maintained at two different temperature T1 and T2 (T1 > T2). Steady State is a condition in which heat current through every cross section of the conduction become a constant. There is no further absorption of heat energy by molecules. Only effective method of heat transfer is conduction is steady state. At Steady State Rate of heat flow H Q  area of cross section t Q t  temperature gradient Q A Q  T1  T2 t t  Temperature gradient  Temp diff length  T1  T2  Q  KA (T1  T2 ) Q  K A T t  t  Unit of thermal conductivity H  K A T  K H  Watt  m A T m2  Kelvin Unit of K  Watt m–1K–1 Thermal Resistance (RT) It is the ability to oppose the flow of heat energy through a conductor 19

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) Heat Current  Temperature difference Thermal Resistance H T (1) RT (2) We have H K A T  H  [ T / KA] Thermal Resistance RT   / KA Unit of Thermal Resistance H T RT RT  T  Kelvin H Watt Combination of Thermal Conductors Series Consider two rods of length 1 and 2 connected in series  RT 1  1 RT 2  2 RT1  1  2 RT  KA K1A K2A K ' A s R1T  TT1  RT2 1  2  1  2 1  2  1  2 KSA K1A K2A KSA K1 K2 KS  1  2 KS  K1K2 [1  2 ] 1  2 1K2  2K1 K1 K2 If 1   2 KS  2K 1K2 K1  K2 20

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) Parallel Consider two rods connected in parallel  RT1  RT  KA  K1 A1 RT2   R'T  K2  A2 A2] KP [A1  11 1   RT' RT1 RT2    KP [A1  A2 ] K1A1 K1A2 KP  K1A1  K2A2 A1  A2 if A1 = A2 = A KP  K1  K2 2 Convection It is the transfer of heat energy along with actual transport of matter. Gravity plays an important role in natural convection. Natural Convection Forced Convection With the help of gravity. With some mechanical support In upward direction only Possible in all directions Eg : Sea breeze Eg : Human circulatory system Land breeze Automobile cooling system Trade winds House hold heating system Boiling water furnaces 21

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) Growth of ice ponds Ice starts forming in a pond at sub zero temperature 0C in winter season. To find time taken for growth of ice upto a thickness y, we will consider a small thickness dy The heat energy released when ice of thickness dy is formed is dQ. Heat energy is transferred to the environment dQ  KA [0   ] dt y For melting dQ  mL   A  dy L KA dt  A dy L y dt  L y dy K for total time taken to grow from o to y y L y dy t 0 K t  L y y dy K 0 t  L  y2  y   density of ice K  2  0   L Latent heat of fusion   Atmospheric temperature t  1 L y2 K  Thermalconductivity of ice 2 K y  Thickness of ice 22

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) Time taken by ice to grow up to thickness y, 2y and 3y from O is given by t1 : t2 ; t3  1 : 4 : 9 t y2 t1, t2, t3 are time taken by ice to grow upto y,2y and 0  y t1  y2 3y respectively 0  2y t2  (2y)2 0  3y t3  (3y)2 t1  y2 t2  4y2 t3  9y2 Time taken by ice ice to grow from o to y y to 2y and 2y to 3y is 0y t1  t1  0  y2  0  y2 y 2y t2  t2  t1  4y2  y2  3y2 2y 3y t3  t3  t2  9y2  4y2  5y2 t1 : t2 : t 3 1 : 3 :5 Radiation Radiation is the fastest mode of heat transfer Properties of heat radiations It travels in straight lines It is universal, invisible All bodies above zero kelvin will emit radiations It belongs to IR Region It shows reflection, refraction, interference, diffraction and polarisation They are em waves with speed 3 × 108 m/s Let Q be total incident energy, then part of it may absorbed. Some part will be reflected and the rest is transmitted QA RT ing by Q a  A  absorptance (fraction of energy absorbed) Q A R  T Q QQQ Q a  r  t 1 r  R  reflectance (fraction of energy reflected) Q t = T = transmittance (fraction of energy transmitted) Q 23

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) Absorptive Power (absorptance) [a] a  Radiant energy absorbed Total incident energy a A It has nounits and dimensions Q for a perfect black body a = 1 for a non black body a < 1 Emissive Power [E] Radiant energy emitted per unit area per unit time E  Q  Energy E  Power unit : Watt / m2 At area  time area Emissive power × area = Power Spectral Emissive Power (E ) Emissive power of body in unit wavelength range is called spectral emissive power. Represented as E E  dE d dE  Ed Emissive power  E  Ed 0 = area of E   graph Kirchoff’s Law for a given wavelength and temperature E1  E2  E3  ......  EB  EB a1 a2 a3 aB  a B  1 Emissive power = Constant Absorptive power E =Constant a Ea  A good absorber is a good emitter A bad absorber is a bad emitter 24

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) Emissivity / Relative Emittance (e) e  Emissive power of abody Emissivepower of black body e E E  e  EB EB for a perfect black body e = 1 for a non black body e < 1 Stefan’s Law Emissive power of a blak body EB  T4 T  absolute temperature EB   T4   Stefan's constant   5.68 108 wm2k 4 Power P  Q   AT4 t for a non black body E  e T4 e  emissivity Power P  e  AT4 Stefan - Boltzmann’s Law When the temperature difference between body and surrounding is very large, Stefan Boltzmann’s law is applicable. for black body EB   (T4  T04 ) for non black body E  e (T4  T04 ) Wiens Displacement Law “Wavelength corresponding to maximum spectral intensity is inversely proportional to absolute temperature” m 1 T T b aconstant b = Wien’s constant b = 0.29 cm K 25

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) 1T1  2T2 As wavelength is inversely related to temperature a Blue Star is more hotter than a Red Star Newton’s Law of Cooling When temperature difference between the surrounding is not very large, rate of cooling is proportional to temperature difference between body and surroundings R  Body  Surroundings Q   1  2   0 1 Initial temperature of body t  2  2 Final temperature of body 0  Surroundingtemperature mcd   1  2   0 t  time of cooling t  2  mc(1  2 )   1  2   0 t  2  mc(1  2 )  K  1  2   0 t  2  Derivation of Newton’s law of cooling from Stefan-Boltzmann’s law Consider a body of temperature T placed in a surrounding of temperature T0 so that T  T  T0 T is very small T  T0  T Taking 4th power T4  T0  T4  T  4 1  T4  T04  T0   T4  T04  4T  1    T0  26

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) T4  T04  4T03T T4  T04  4T03T.........(1) From Stefan Boltzmann’s law  P  eA T4  T04  dQ dt  eA T4  T04  mCdT  eA dt T4  T04 dT  eA  4T03T substituting (2) dt mC Substituting (1) in (2) dT  4eAT03  T dt mC dT  k T dt dT  k T  T0  dt Hence Newtons law of cooling To find time of cooling (t) Newton’s law of cooling dT k [T  T0 ] dt dT   Kdt T  T0  T2 dT   K dt T1 T  T0  n[T T0 ] T2   Kt T1 27

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) n[T2  T0 ]  n[T1  T0 ]   Kt n  T2  T0    Kt  T1  T0    t 1 n  T2  T0  K  T1  T0    t  1 n  T1  T0  K  T2  T0    t 1 n Initial temperature difference  K temperature  Final difference    Solar Constant It is the amount of solar radiant energy received by unit area of earth surface in unit time Solar constant S = 1400 W/m2 Graph of Newton’s Law of Cooling temp along y axis time along x axis Graph connecting loge [2  1] and time taken (t) 28

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) We have  d K [2  1] dt d   K dt 2  1 Integrating loge [2  1]   Kt  C y = mx + c Graph : Straight line with negative slope Kinetic Theory of Gases Postulates 1. Gases consists of small spherical particles called molecules 2. Molecules in a gas are in a random irregular motion. 3. During the motion they collide with each other and also with the walls of the container. 5. These collisions are perfectly elastic in nature. 5. Size of the molecules is negligible compared to average separation between the molecules. 6. Molecules behave like perfect spheres. 7. Molecular dynamics is governed by Newton’s Laws of motion Pressure exerted by an ideal gas on the walls of the container is given by P 1 C 2 3 RMS   density of the gas CRMS  Root mean square velocity Pressure exerted by an ideal Gas Consider an ideal gas enclosed in a cube of side ‘a’. Consider a gas molecule of mass ‘m’ moving  with velocity vx , vy , vz hitting a planar wall of area A  a2 parallel to y–z plane 29

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE)   vxˆi  vyˆj  vzkˆ v v  v2x  v2y  vz2 vx  vy  vz v  3v2x vQ  3vx vx  v 3 Change in linear momentum of the molecule = mvx  mvx  2mvx Momentum imparted on the wall P  2mvx Time taken for the collision 2a T  vx Force exerted by this molecule on the wall F  P t F  2mvx  mvx2  mv2 2a a 3a vx Total force F  nm v12  v22  v32  ...v2n  v12  v 2  ...  v2n 3a n 2 F  nm  v12  v22  v32  ...  v 2  CRMS  n 3a  n n    C2  v12  v22  ...  vn2 RMS n 1 nm C2 F  3 a RMS Pressure P  F  1 nm C2 A 3 a  a2 RMS 30

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) P  1 nm C2 3 a3 RMS P  1 nm C2 nm  M 3 v RMS P  1 M C2 3 v RMS P  1 C2RMS M  3 v Root mean Square Speed (CRMS) CRMS  C12  C22  C23  ...  C2n n We have P 1  C2 R / NA  K Boltzmann's constant 3 RMS K = 1.38  1023 C2  3P M / NA  m, mass of molecule RMS  CRMS  3P /  CRMS  3P CRMS  3PV CRMS  3RT M/ V M M CRMS  3R / NA CRMS  3 KT M / NA m CRMS  T Average Speed (Cavg) Cavg  C1  C2  C3  C4 ... Cn n Cavg  8RT Cavg  T M Cavg  8KT m Most Probable Velocity (Cmp) Velocity possessed by maximum fraction of molecules Cmp  2RT M 31

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) Cmp  2KT Cmp  T m CRMS : Cavg : Cmp  3 : 8: 2  CRMS  Cavg  Cmp Average Translational KE of an ideal gas We have P 1  C2 3 RMS P 1 M C2 3 V RMS PV  1 M C2 3 RMS 3 PV  1  2M C2 2 RMS 1  M C2  3 PV 2 RMS 2 KE  3 PV  3 RT KE  3 KT for a molecule 22 2 KE T Mean free path () It is average distance travelled by molecules between two successive collisions 1 n  no.of molecules / volume(number density)  P Pr essure K  Boltzmann' s cons tan t 2  d2n T  Temperature KT  2  d2P Collision frequency (f) f  CRMS  Specific Heat of Gases Molar Specific Heat Capacity It is the amount of heat energy required to rise the temperature of 1 mole of an ideal gas by 10C. It can be supplied by two ways. 32

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) CP  CV R Mayers Relation for n moles and t rise in temperature Q  nCvt at constant volume Q  nCpt at constant pressure Ratio of Specific Heats ()   CP CV  1 2 f f  degree of freedom (No. of independent possible ways in which a system can have energy) Monoatomic molecule Diatomic molecule A monoatomic molecule has 3 translational A diatomic molecule has 3 translational and degree of freedom 2 rotational degree of freedom. (along x,y,z axis) f = 5 (3 translational + 2 rotational) f=3  1 2 1 2  5  1 2 1 2  7 f 33 f 55  1.67  1.4 Note : At high temperature a diatomic molecule will have vibrational mode also. A vibrational mode will have 2 degree of freedom. One vibrational KE and one vibrational PE. So at high temperature for diatomic f = 7 (3 trans + 2 Rot + 2 vib)  1 2  9 77 33

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) Mayer’s Relation CP  CV  R dividing by CV CP  CV  R CV CV CV   1 R CV CV R CP   CP  C V  1 CV CP  R  1 for monoatomic gas for diatomic gas Generally CV  R  R 3 R CV  R  R  5 R CV  f R 53 1 2 2 75 1 2 2 2 3 5 CP    CV 53 5 R CP  CV 75 7   R   R  R 32 2 52 2 molecule f  CV CP monoatomic 3 diatomic gas 5 5/3 3/2 R 5/2 R 7/5 5/2 R 7/2 R Law of Equipartition of Energy An ideal gas divides its total energy equally among all degrees of freedom KE of amolecule / degree of freedom  1 KT 2 TotalKE of a molecule  f  1 KT 2 KE of n molecules  n f  1 KT 2 KE of 1 mole / degree of freedome  1 RT K  Boltzmann’s constant 2 Total KE of 1 mole  f  1 RT 2 Total KE of n moles  n f  1 RT 2 34

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) Specific Heat of Solids Consider 1 gram mole of a solid containing NA (Avogadro No.) of atoms. Solids do not have translational and rotational degrees of freedom. It has energy due to oscillation / vibration Average energy associated with an atom due to its oscillation in one dimension = 2  1 KT  KT 2 in 3 dimensions U 3  KT for 1 gram mole of solid U  3KT NA  3RT According to first law of thermodynamics Q  U  W  U  PV for solid V is negligible Q  U C  Q  U  3RT  3R T T T C  3R C  3 8.314  24.93 Jmol1 K1 Specific Heat Capacity of Water We treat water like a solid made up of 3 atoms [2 Hydrogen + 1 Oxygen] Total energy of 1 mole of water U = 3 × 3 KT × N4 = 9 RT C  U  9RT  9R  V  0  T T  Q  U  C  9R C  98.314 C  75.22 Jmol1 K1 35

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) Mixture of Gases when n1 molecules of an ideal gas is mixed with n2 molecules of another gas Mmixture  n1M1  n2 M2 n1  n2 Tmixture  n1T1  n2 T2 n1  n2 CV mixture  n1CV1  n2 CV2 n1  n2 n1CP1  n 2C P2 n1CP1  n2 CP2  mixture  n1CV1  n 2 C V2 n1  n2 CP mixture   mixture  CPmixture CVmixture Thermodynamics Zero’th Law of Thermodynamics It defined the existence of thermodynamic temperature. It states that “when two systems A and B are separately in thermal equilibrium with a third system C then the systems A and B will also be in thermal equilibrium with each other”. Consider 3 systems A, B and C Case 1 Now walls are interchanged Case 2 36

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) First Law of Thermodynamics It is another form of “Law of conservation of Energy” It states that Heat supplied to  change in + work a system internal energy done  Q  U  W d  pdV P F dQ  dU  dw dW F dx A dQ  dU  pdV d  p  Adx d  pdV Heat Energy (dQ) When heat is added to a system dQ  ve When heat is removed from a system dQ  ve Internal Energy (du) Internal energy is a state variable. It is independent of the path. If initial and final po int s are same In a cyclic process when a system returns to its initial status du  0 U 0 Eg : here initial and final points of all process are same at constant volume dV = 0 pdV = 0 du  nCV dT 37

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) Work done (W) dW  pdV V2 W   pdV V1 On cyclic process Work done in clockwise direction = +ve Work done in anticlockwise direction = –ve Expansion Compression Volume increases Volume decreases Work done by the system Work done on the system W = +ve W = –ve Thermodynamic Process Isobaric Process Constant Pressure V T V =constant T V1 = V2 T1 T2 38

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) fraction of energy for internal energy change du  nCVdT  CV  1 dQ nCPdT CP  fraction of energy for work done dw 1 1 dQ  dQ  nCpdT P-T graph dU  nCvdT dW  p d V P-V graph In isobaric heating temp  u  ve volume  W  ve Q  nCpt Q  ve In isobaric cooling temp  u  ve volume  W  ve Q  nCpt Q  ve Work done W P(V2  V1) W  nR (T2  T1) Isochoric Process Constant volume In isochoric heating temp  u  ve pressure  W  0 Q  nCvT Q  ve 39

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) In isochoric cooling temp  u  ve pressure  W  0 Q  nCvT Q  ve P T P =constant P1  P2 T T1 T2 dQ  nCVdT du nCVdT dw  0 dQ  du dV  0 PdV  0 Isothermal Process Constant temperature dT = 0 PV = a constant du  nCVdT p 1  du  0 V P1V1 = P2V2 P1  V2 P2 V1 Conditions 1) Walls of the container must be perfectly conducting to allow free exchange of heat 2) Process must be slow 40

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) Slope of isothermal graph PV = a constant du = 0 PdV + VdP = 0 dw  nRT n V2 Isothermal elasticity V1 B  P  P VP PdV = –VdP Work done W  nRT n P1 P2 dP P Specific heat capacity    dV V Slope   P V Adiabatic Process dQ  0 Work done Equation of state du  dw W  nR1[T1  T2 ] PV  cons tan t dw   du W  P1V1  P2V2 TV 1  cons tan t P1 T  a cons tan t  1 Conditions 1) There should not be any exchange of heat 2) Walls of the container must be insulated 3) Process is sudden Eg. Sudden bursting of tyres Slope of Adiabatic Graph dP     P dV V pv  a cons tan t pv  1dv  v   0 dp p dv  dp v dp   p dv v slope  tan     P v 41

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) Specific heat capacity C = 0 Since   1 Slope of adiabatic  slope of isothermal To find general equation for adiabatic process (Not in curriculum) In adiabatic process Q = 0 U  W W   U W   nCVdT (1) dU = -dw = –PdV nCVT   pdV (2) d(PV) = d [nRT]  PV nRT PdV + VdP = nRdT ndT  PdV  vdP (3) R Sub (3) in (1) CV [PdV  VdP]   PdV R [CV  R] dV  CV dP 0 V P  CP  dV  dP  0  CV  V P   Integrating CP n [V]  n[P]  n[C] CV The ratio CP  CV n V  nP  a constant nPV  a constant PV  a constant 42

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) Work done in Isothermal process P  nRT P1V1  P2V2 V V2 V2 nRT dV  W  PdV  VV1 V1  nRT V2 dV VV1  nRT n VV2 PV  aconstant V1 P1V1  P2V2 V2  P1 =nRT n V2 V1 P2 V1 W  2.303 nRT log V2 V1 W  2.303 nRT log P1 P2 Work done in Adiabatic Process In adiabatic process d  0 dW   dU dW   nCVdT W   n  R 1 [T2  T1]  W  nR [T1  T2 ]  1 W  P1V1  P2 V2  1 43

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) Comparison between Isothermal and Adiabatic Adiabatic Curves of Different Gases Polytropic process A process that can be expressd in the form PVx  a const. when x  1 is called a polytropic process Specific heat capacity C R  R  1 1 x C  CV  R 1 x Heat energy dQ  nCdT Internal energy change du  nCvdT 44

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) Work done W  xnR1T1  T2  W  P1V1  P2V2 x 1 Slope of graph = x  P V Free expansion Sudden expansion of gas against vacuum. Its an adiabatic process where isothermal conditions are valid. here W = 0 since wall are rigid Q = 0 since walls are insulated U  Q  W  0 Ui  Uf Cyclic Process A cyclic process consists of series of changes which return the system back to its initial state In cyclic process Ui  Uf dU  0 T  0 Q  W Work done in a cyclic process = area of cyclic loop W = area of ABCD Work done in clockwise direction is taken as positive Work done in anticlockwise direction is taken as negative Efficiency of a cyclic process In a cyclic process heat is supplied during some stage. While heat is released during other stage Efficiency   net work done per cycle heat absorbed per cycle 45

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) Reversible process A process which is reversed in such a way that all changes occuring in the direct process are exactly repeated in the opposite direction and no change is left in any of the bodies taking part in the process or in the surrounding Conditions 1) No dissipative forces are present 2) Process must be slow 3) System will be in equilibrium Perfect Reversible process are ideal Heat Engines Heat Engines are the devices that converts heat energy to mechanical work. Essential Parts 1. Source - A high temperature reservoir (T1 Kelvin) 2. Sink - A low temperature reservoir (T1 Kelvin) 3. A working substance Working substance absorbs an amount of heat energy Q1 from the source and will undergoes a series of charges and will release an amount of heat energy Q2 to sink. Useful work is transferred to external environment. Since working substance undergo cyclic process. du = 0 dQ=dW W Q1  Q2 Thermal Efficiency ()   net work done / cycle heat absorbed from source / cycle W   Q1  Q2  Q1 Q1   1 Q2 ................. (1) Q1 46

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) Carnot’s Heat Engine It is an ideal heat engine designed by Zadi Carnot. According to Carnot’s theorem “no reversible heat engine can claim more efficiency than a reversible Carnot’s heat engine” working in the same temperature range. From Carnot’s theorem Q2  T2 Q1 T1 Substitution in (1)  1 T2 T1 For  1 % of  100% T2  0 Kelvin It is practically unattainable The value of  can never be unity Carnot’s Cycle A Carnot’s cycle involves 4 stages 1. Isothermal expansion 2. Adiabatic expansion 3. Isothermal compression 4. Adiabatic compression AB (Isothermal Expanison) Cylinder is placed on source. Gas expands slowly. Q1 heat is absorbed. Temperautre is kept constant (T1) W1  2.303 nRT log v2 U=0 v1 W=Q Q1  2.303 nRT1 log v2 v1 47

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) BC (Adiabatic Expansion) Cylinder is placed on insulating pad. Gas expands further. No heat is absorbed. Temperature falls to T2 W2  rnR1T1  T2  T1v  1  T2 v 3 1.............(1) 2 CD (Isothermal Compression) Cylinder is placed on the source. Gas is compressed slowly. Q2 heat is rejected to the sink. W3  2.303 nRT2 log v4 v3 Q2  2.303 nRT2 log v4 v3 DA (Adiabatic Compression) Cylinder is again placed on insulating pad. Gas is compressed further. Reaches initial pressure and volume. Temperature of Gas raises to T1 W4  nR1T2  T1 W4  nR1T1  T2  T1v11  T2v41...............(2) 1  r 1 r1 2  v2   v3   v2    v4    v2  v3 v1 v 4 Net work done W  W1  W2  W3  W4  2.303nRT1 log v2  2.303 nRT2 log v2 also we get v1 v1 Q1  T1 Q2 T2 W  2.303nR T1  T2 log v2 v1 48

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) Second law of Thermodynamics Kelvin plank statement : It is impossible to construct a heat engine which would absorb heat from a reservoir and convert 100% of heat absorbed into work. Clausius Statement It is impossible to design a self acting machine unaided by an external agency to transfer heat energy from low temperature reservoir to high temperature reservoir. Refrigerator It is an ideal heat engine working in reverse order. It is used for cooling Co-efficient of performance ()   heat removedfrom sink / cycle Energy spent / cycle   Q2   Q2   T2 W Q1  Q2 T1  T2 The value  can be  For carnots refrigerator Q2 Q2  T2   Q2  Q1 Q1 T1 Q1  Q2 1 Q2 Q1 T2   T1  T2 1 T2 T1  T2 T1   T2 T2  T1 49

Brilliant STUDY CENTRE RESIDENTIAL PHYSICS -2023M (ONLINE) Relation between  and  We have   Q2 Q1  Q2 Q2 Q1  1  Q2   1 Q1 1   50