{"id":35845,"date":"2022-11-25T11:13:16","date_gmt":"2022-11-25T05:43:16","guid":{"rendered":"https:\/\/tsboardsolutions.com\/?p=35845"},"modified":"2022-11-25T11:13:16","modified_gmt":"2022-11-25T05:43:16","slug":"ts-inter-1st-year-physics-notes-chapter-13","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.com\/ts-inter-1st-year-physics-notes-chapter-13\/","title":{"rendered":"TS Inter 1st Year Physics Notes Chapter 13 Thermodynamics"},"content":{"rendered":"
Here students can locate TS Inter 1st Year Physics Notes<\/a> 13th Lesson Thermodynamics to prepare for their exam.<\/p>\n \u2192 Thermodynamics: It is a branch of physics in which we shall study the process where work is converted into heat and vice versa.<\/p>\n \u2192 Thermodynamic variables: In thermodynamics the state of a gas is specified by macroscopic variables such as pressure, temperature, volume, mass and composition that are felt by our sense perceptions and are measurable.<\/p>\n \u2192 Thermal equilibrium: In general at thermal equilibrium the temperatures of the two bodies or systems are equal. \u2192 Zeroth law of thermodynamics: It states that if two systems say A & B are in thermal equilibrium with a third system ‘C’ separately then the two systems A and B are also in thermal equilibrium with each other.<\/p>\n \u2192 Internal energy: It includes only the energy associated with random motion of molecules of the system \u2192 First law of thermodynamics: The heat energy (dQ) supplied to a system is partly used to increase its internal energy (dU) and the rest is used to do work (dW) \u2192 Isothermal expansion: If a system is taken through a thermodynamic process in which \u0394U = 0 then it is called Isothermal process. <\/p>\n \u2192 Adiabatic process: In an adiabatic process system is insulated from the surroundings. So energy absorbed or released is zero (\u0394Q = 0). In adiabatic process temperature of the system may change. It follows the equation PV\u03b3<\/sup> = constant. Where \u03b3 = \\(\\frac{C_P}{C_V}\\) ratio of specific heats of a gas.<\/p>\n \u2192 Isobaric process: In isobaric process pressure P’ is kept constant, volume and temperature changes are permitted. Work done in isobaric process \u2192 Isochoric process: In isochoric process volume (V) of the system is kept constant. Work done by isochoric process is zero. In this process heat energy absorbed is totally used to increase the internal energy of the system.<\/p>\n \u2192 Cyclic process: In a cyclic process the system returns to initial state (P, V and T). Change in internal energy \u0394U = 0. Heat absorbed during cyclic process is equal to work done.<\/p>\n \u2192 Reversible process : A thermodynamic process is said to be reversible if the process can be turned back such that both the system and surroundings return to their original state, with no other change any where else in universe.<\/p>\n \u2192 Irreversible process : If a thermodynamic process cannot be reversed exactly in opposite direction of direct process then it is called irreversible process. \u2192 Quasi static process: In a quasi static process at every stage the difference on pressure and temperature of systems and surroundings is infinitesimally small. \u2192 Heat engines: A heat engine is a device by which a system is made to undergo a cyclic process. As a result heat is converted into work. Work done by heat engine W= Q1 \u2014 Q2; efficiency \u03b7 = 1 – \\(\\frac{\\mathrm{Q}_2}{\\mathrm{Q}_1}\\) (or) \u03b7 = 1 – \\(\\frac{\\mathrm{T}_2}{\\mathrm{~T}_1}\\)<\/p>\n Important parts of heat engine : every heat engine mainly consists of<\/p>\n \u2192 Refrigerators or heat pumps : A Refrigerator is a heat pump which is a reverse of heat engine. Here working substance extracts heat Q2<\/sub> from cold body at temperature T2<\/sub> and delivers it to hot reservoir at temperature T1<\/sub>. Coefficient of performance of refrigerator \u2192 Second law of thermodynamics : Second law of thermodynamics gives a fundamental limitation to the efficiency of heat engine i.e. heat released to a colder body will never become zero. So 100% efficiency of heat engine cannot be achieved.<\/p>\n \u2192 Carnot engine : Carnot engine operates between a hot reservoir of temperature T1<\/sub> and a coldreservoir of temperature T2<\/sub> through a cyclic process. In this cyclic process it absorbs heat energy from source and releases heat energy Q2 to cold reservoir efficiency of<\/p>\n Carnot engine \u03b7 = 1 – \\(\\frac{\\mathrm{T}_2}{\\mathrm{~T}_1}\\)<\/p>\n \u2192 Carnot theorem: (a) Any heat engine working between two given temperatures T1<\/sub> and T2<\/sub> cannot have efficiency more than that of carnot engine, (b) The efficiency of a carnot engine is independent of nature of working substance.<\/p>\n \u2192 Isotherm: The pressure (p) and volume (v) curve for a given temperature is called isotherm.<\/p>\n \u2192 adiabatic wall: An insulating wall that does not allow heat energy to flow from one side to another side is called “adiabatic wall.”<\/p>\n \u2192 diathermic wall : It is a conducting wall which transfers heat energy from one side to another side.<\/p>\n \u2192 Heat mechanical equivalent (J): In M.K.S system heat and work are measured with same unit ‘joule’. But in C.G.S system heat is measured in calorie and work in erg (1 joule = 107<\/sup> erg). \u2192 Calorie: The amount of heat energy required to rise the temperature of 1 gram of water through 1\u00b0C or 1 K is defined as “calorie.” Note: Magnitude of calorie slightly changes with the initial temperature of water.<\/p>\n \u2192 Mean 15 \u00b0C calorie : The amount of heat energy required to rise the temperature of 1 gram of water from 14.5 \u00b0C to 15.5 \u00b0C is called “mean 15 \u00b0C calorie.”<\/p>\n \u2192 Joule’s Law, work W \u221d Q \u21d2 W = JQ where J = mechanical of heat equivalent \u2192 From 1st Law of thermodynamics, dQ = dU + dW<\/p>\n \u2192 Heat capacity of a body = \\(\\frac{\\Delta \\mathrm{Q}}{\\Delta \\mathrm{t}}\\) = me (i.e., mass \u00d7 specific heat)<\/p>\n \u2192 Specific heat S or C = \\(\\frac{\\Delta Q}{m \\Delta t}\\) = \\(\\frac{\\text { Heat energy supplied }}{\\text { mass } \\times \\text { temperature difference }}\\)<\/p>\n \u2192 From method of mixtures, Heat lost by hot body = Heat gained by cold body<\/p>\n \u2192 When two spheres of radii r1<\/sub> : r2<\/sub> and ratio of specific heats S1<\/sub>: S2<\/sub> and densities p1<\/sub>: p2<\/sub> then their thermal capacities ratio (a) Specific heat of a gas, Cp<\/sub> = \u0394Q \/ m\u0394T \u2192 Work done in expanding a gas against constant pressure (P) is W = P dV.<\/p>\n <\/p>\n \u2192 Work done during ideal expansion \u2192 Relation between Cp<\/sup> and Cv<\/sup> \u21d2 CP<\/sup> – Cv<\/sup> = R<\/p>\n \u2192 Isothermal relation between P, V & T is PV = RT or PV = nRT.<\/p>\n \u2192 Adiabatic relation between P, V & T<\/p>\n \u2192 Work done in Isothermal process \u2192 Work done in adiabatic process, \u2192 Efficiency of heat engine, \u03b7 = 1 – \\(\\frac{\\mathrm{Q}_2}{\\mathrm{Q}_1}\\) or \u03b7 = 1 – \\(\\frac{\\mathrm{T}_2}{\\mathrm{T}_1}\\) \u2192 Heat energy supplied to heat a body within the same state is Q = mct.<\/p>\n \u2192 Heat energy supplied during change of state is Q = mL.<\/p>\n","protected":false},"excerpt":{"rendered":" Here students can locate TS Inter 1st Year Physics Notes 13th Lesson Thermodynamics to prepare for their exam. TS Inter 1st Year Physics Notes 13th Lesson Thermodynamics \u2192 Thermodynamics: It is a branch of physics in which we shall study the process where work is converted into heat and vice versa. \u2192 Thermodynamic variables: In … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":[],"categories":[27],"tags":[],"yoast_head":"\nTS Inter 1st Year Physics Notes 13th Lesson Thermodynamics<\/h2>\n
\nIn a thermally isolated system it is said to be in “thermal equilibrium” if the thermodynamic variables such as pressure, volume, temperature, mass and composition do not change with time and they have fixed values.<\/p>\n
\ni. e., internal energy is simply the sum of kinetic and potential energies of these molecules. Internal energy is denoted by ‘U’.<\/p>\n
\ni. e., dQ = dU + dW. (OR)
\nHeat energy supplied to a system (dQ) always equals to the sum of change in internal energy (dU) and workdone (dW).
\nThis law is a consequence of ”law of conservation of energy.”<\/p>\n
\nIn isothermal process change in internal energy \u0394U = 0 i.e., temperature of the system is constant. Isothermal process obeys gas equation PV = RT.<\/p>\n
\nW = P(V2<\/sub> – V1<\/sub>) = \u00b5R(T2<\/sub> – T1<\/sub>).<\/p>\n
\nAll spontaneous process of nature are irreversible.<\/p>\n
\ni.e., P + \u0394V \u2248 P and T + \u0394T = T .
\nIn this process the thermodynamic variables (P,V,T) will change very slowly so that it remains in thermal and mechanical equilibrium with surroundings throughout that process.
\nNote: Quasi static process is an imaginary concept only.<\/p>\n
\nWork done by heat engine W= Q1<\/sub> – Q2<\/sub>;<\/p>\n\n
\n\u03b1 = \\(\\frac{\\mathrm{Q}_2}{\\mathrm{~W}}=\\frac{\\mathrm{Q}_2}{\\mathrm{Q}_1-\\mathrm{Q}_2}\\)<\/p>\n
\n(a) Kelvin – Planck statement: No process is possible whose sole resultant is the absor-ption of heat from a reservoir and the complete conversion of heat into work.
\n(b) Clausius statement: No process is possible whose sole resultant is the transfer of heat from a colder object to a hotter object.<\/p>\n
\nThis cyclic process consists of<\/p>\n\n
\nSo in C.G.S system a conversion factor heat mechanical equivalent (J) is used to convert work into heat or vice versa.
\n1 Calorie = 4.2 Joules \u21d2 J = 4.2 Joule\/ cal. or J = 4200 joule\/kilocal.<\/p>\n
\nJ = \\(\\frac{\\text { Work }}{\\text { Heat }}\\) = 4.18 J\/Cal.<\/p>\n
\n= \\(\\frac{m_1 S_1}{m_2 S_2}=\\left(\\frac{r_1}{r_2}\\right)^3\\left(\\frac{\\rho_1}{\\rho_2}\\right)\\left(\\frac{S_1}{S_2}\\right)\\)<\/p>\n
\n(b) Molar specific heat, Cp<\/sub> = \\(\\frac{\\Delta \\mathrm{Q}}{\\mathrm{n} \\Delta \\mathrm{t}}\\) (n = Number of moles)
\n(c) Ratio of specific heats, y = Cp<\/sup>\/ Cv<\/sup>;
\nCv<\/sup> = \\(\\frac{\\mathrm{R}}{\\gamma-1}\\), Cp<\/sup> = \\(\\frac{\\gamma \\mathrm{R}}{\\gamma-1}\\)
\nCv<\/sup> = \\(\\frac{C_V}{M}=\\frac{1}{M} \\frac{R}{(\\gamma-1)}\\)
\n= \\(\\frac{\\mathrm{PV}}{\\mathrm{M}(\\gamma-1) \\mathrm{T}} \\frac{\\mathrm{P}}{\\rho \\mathrm{T}(\\gamma-1)}\\)J\/kg-K
\nor Cv<\/sup> = \\(\\frac{P}{\\mathrm{~J} \\rho \\mathrm{T}(\\gamma-1)}\\)k.cal\/kg.K<\/p>\n
\nW = P (V2<\/sub> – V1<\/sub>) or W = nR (T2<\/sub> – T1<\/sub>)
\nn = number of moles of gas;
\nR = universal gas constant.<\/p>\n\n
\n(a) W = RT loge<\/sub> \\(\\frac{V_2}{V_1}\\)
\n(b) W = 2.303 RT log10<\/sub> \\(\\frac{V_2}{V_1}\\)<\/p>\n
\n(a) W = \\(\\frac{1}{\\gamma-1}\\) (P1<\/sub>V1<\/sub> – P2<\/sub>V2<\/sub>) per mole (OR)
\n(b) W = \\(\\frac{\\mathrm{nR}}{\\gamma-1}\\) (T1<\/sub> – T2<\/sub>);
\nn = number of moles.<\/p>\n
\nT1<\/sub> = Temperature of source,
\nT2<\/sub> = Temperature of sink.<\/p>\n