✤ Thermal equilibrium :-

        => When two conducting body comes in physical contact then their temperature become equal after some time i.e. net heat transfer is zero.

                                

✤ Zeroth law :- It states that if two systems AB are separately in thermal

                                        If A ↔ C equilibrium with 3rd system

                                        B ↔ C Adiabatic wall

                                          then A ↔ B C, the A&B are also in the

                          Formulated by R.H.Fowler Thermal equilibrium in 1931. each other

✤ System of Dynamics :-

✤ Isothermal: In this system we have to study to change in Volume (V) & Pressure (P) at constant temp (CT).

✤ Conditions for Isothermal:-

          (i) Wall of a container should lee perfectly conducting.

          (ii) It should occur slowly so that it get time to exchange the heat with Surrounding.

          (iii) It obey the Boyle's Law. 

           (iv) Eq. of Isothermal dP/dv= -P/V

          (v) P dv/dv+V dP/dv= V.dP/dv= -P

✤ Specific heat -- At infinite

✤ Adiabatic Process

              In this system, we have to study the change in volume & pressure at constant heat.

              Properties :-

                         1) Walled container should be insulated i.e exchange of fund should be zero.

                         2) It should occur quickly, so that it get no time to exchange the heat

                         3) Equation of Adiabatic  PVγ = k Here, γ = Cp / Cv γ = log

                         4)       

                                  We have, PVγ = k

                                  PγVγ + Vγdp/dV = 0

                                  Vγdp/dV = -γP.V γ-1

                                  dP/dV = -γp.V γ-1 / V. γ = -γp.V -1

                               Slope of adiabatic Course  dP/dV = -γP / V

                         5) Specific heat is zero. That is exchange of temperature in terms of heat is zero.

                                            S = ΔQ / mΔT in adiabatic ΔQ = 0, so S = 0.

                                                   in adiabatic ΔQ = 0 , so S=0

✤ ISOBARIC PROCESS :-

          We have to study the change in volume & temp. at constant pressure.

                         VαT P = Constant

                             V / T = k

✤ ISOCHORIC PROCESS :-

         We have to study the change in pressure & temp. at constant volume.

                             P / T = k

✤ CYCLIC PROCESS :-

          THERMODYNAMICS :- Thermodynamics is the branch of science that deals with the concepts of heat & temperature into converging heat & other                                                 forms of energy.

          DIATHERMIC WALL'S : A conducting that allows heat to flow through it.

          THERMODYNAMIC EQUILIBRIUM :- A system is said to be in the state of thermodynamic equilibrium. If the macroscopic variables describing the                                       thermodynamic state of the system do change with time.

✤ Work done in Isothermal Process :-

          IN EXPANSION :-

                  Let a monatomic gas of volume Vi is filled in a Container (conducting wall). Let area of Piston = A

                  We know, do work is done in displacement

                             dw = F . dx

                                    = P.Adx

                                = P(Adx)

                             dw = P.dv

                      Now, By the equation of Isothermal

                             PV = K

                         P = K / V

                             ⇒ dw = K / V . dV

                            0∫w dw = k v1∫v2 1/v dv

                            (w)w0=K [log V]v2v1

                            W - 0 = k [log V2 - log V1]

                            w = k log(V2/V1)

                      We know by Gas Eq.

                            PV=RT & PV=k

                            K=RT (isothermal)

                            W = RT loge |V2/V1|

                           W= 2.303 RT log |V2/V1|

                     We know, P1V1 = P2V2

                             P1/P2 = V2 / V1

                            W= 2.303 RT log |P1 / P2 |

                              Heat = Q = W/4.2

                              Q = 2.303 / 4.2 RT log |V2 / V1|

            Case 2: In Compression:-

                         We know volume of the gas decreases

    ❆ W=2.303RT log |V2 / V1|

       Or

       W = 2.303 RT log |V2 / V1 |

✤ WORK DONE IN ADIABATIC :-

              Let a monoatomic Gas is expanded adiabatically.

  we know

    dw = Pdv

  In Adiabatic

                PVγ = k

            P= K . Vγ

            P= K. V-γ

             dw= k V-γ.dv

            w∫0 dW = k v1∫v2 V-γ . dV

            (w)w0= K(V-γ + 1 / V1-γ+1)

     we have P1V1γ = k P2V2γ = k

             W = 1 / (1-γ) X (-P1V1 + P2V2)

       By gas eq. P₁V₁ = RT₁

                 P₂V₂ = RT₂

              W= 1/1-r (RT2 – RT1)

              W = R / 1-r (T2 – T1)

 Notes : IN ISOTHERMAL :-

                        P1V1 = P2V2

              IN ADIABATIC :-

                        1) P1V1γ = P₂V₂ γ

                        2) P1V₁ = P₂V₂ = R (T1-T₂)

✤ CYCLIC PROCESS:-

          Those dynamic process in which by applying initial conditions & system get into initial state work done in Cyclic Process

                     = ar(ABCDA)

                     = ar(CABEPA) - ar (CEPA)

✤ INTERNAL ENERGY: -

          It is the energy possessed by the gas due to motions of its atoms molecules & Configuration

                        du X dt

  ❆ First law of thermodynamics :-

                It states that when heat is given to a dynamic system, then some parts is used to increase internal energy & other part is used to do the work.

                 dQ = heat supplied

                         ie dQ=du+dw dw = work done

                            du = change in internal energy

  ❆ SIGN CONVENTION :-

            (1) When heat is supplied to the system, the dQ is + ve.

            (2) du is +ve, when temperature increases.

            (3) If dV is +ve, dw is +ve

  NOTE :-

          (1) In Isothermal :- We know dT=0.

               => dU=0 so, in isothermal process.

                 dQ=dw work is maximum

          (2) In Adiabatic:- dQ=0

                0 = dq + dw

                du=-dw

          (3) In Isochoric :- dV=0 dW = PdV dW = 0 ❈ dQ = du

  ❆ LIMITATIONS :-

              (1) In this law there is no information given about source of heat.

              (2) We can't get how extent work is done.

  ✤ HEAT ENGINE: - [CARNOT ENGINE]

                    

              * It is based on the 1st law of thermodynamics.

              * It is used to convert Heat energy into mechanical energy

            AB represent, Isothermal Exp.

            W1 = 2.303 RT1 log |V2 / V1|

            BC represent adiabatic Exp.

            W2 = R / 1-r (T2 – T1)

                                 CD represents, Isothermal Compression

 W2 = -2.303 RT2 log |V4 / V3|

 PA represent, adiabatic comps.

 W4 = R / 1 -r (T2 – T1)

 Total Work Done = W1 + W2 + W3 + W4 => W = W1 + W3

 W3 = -2.303 RT1 log |V2/V1| - log |V4/V1| X 2.303 RT2

✤ Efficiency of heat engine: -

          It is defined as the amount of work done per cycle by an engine per unit Heat supply to the engine per cycle.

                                   N = work done per cycle / Heat supplied per cycle

                                    

         Let Q, Heat is supplied to the engine per cycle & heat ejected to the sink per cycle is Q2.

                   ❃ Work done per Cycle = Q1-Q2

                                η = ω / Q1

                            η = Q1-Q2 / Q1

                            η = 1-Q2 / Q1 It shows that η can’t be 100%

                     Note: Q∝T

                                    Q2 = T2

                                Q2 / Q1 = T2 / T1

                                    η = 1-T2 / T1

✤ REFRIGERATOR: - It is device, which work in the reverse processes & Carnot engine i.e transfers the heat from lower temp. to higher temp.             

                    

    Working of Refrigerator

              → The Gas is allowed to expand suddenly (adiabatically) from high to low pressure

              → The cold fluid is allowed to absorb isothermally from the cold reservoir. This converts the mixture into vapor.

              → Then the vapor is adiabatically compressed till it heats up to the temp. of surroundings.

              → The vapor is compressed isothermally in contact with the surrounding, The vapor releases heat (Q1 =Q2+w) to the                    surround & returns to the initial state. Here w is the work done on the Gas per cycle.

 

COEFFICIENT OF PERFORMANCE:-

          It is defined as the ratio of heat absorb per cycle to the work done per cycle.

                    Let heat absorb per cycle = Q2

                            heat ejected = Q1

                        work done= -(Q2 - Q1)

                                W = Q1-Q2

                          ß= Q2 / N ß= Q2 / Q1- Q2

                         ß = (Q2/Q1)/(1- Q2/Q1) { Q2/Q1= T2/T1 }

                         ß = (T2/T1)/(1- T2/T1) = T2/(T1-T2)

                         ß = T2/(T1-T2)

✤ SECOND LAW OF THERMODYNAMICS

      ❆ KELVIN-PLANCE'S STATEMENT: It is impossible to construct an engine which will produce no effect other than extracting heat from a reservoir &                                                                 performing an equivalent amount of work.

                                                                A reservoir & performing an equivalent amount of work...

      ❆ CLAUDIUS STATEMENT: It is impossible for a self-acting machine acted by any external agency to transfer heat from a body at a higher temperature.

✤ FOR YOUR KNOWLEDGE :-

          -> The efficiency of a Carnot engine:

          -> depends upon the temperature of the source & the sink.

          -> is independent of the nature of the working substance.

          -> is the same for all reversible engines working between the same two temperatures.

          -> is directly proportional to temp. difference (T1-T2)

          -> is always less than 100%, because Q2 < Q1.

          -> The efficiency of a Carnot engine will be unity or 100% if T1= ∞ or T2=0K or at OK or infinite temp. Cannot be realized hence a Carnot engine                 working on reversible cycle cannot have 100% efficiency.

          -> If T1=T2, then η = 0. This means that the conversion of heat into mechanical work is impossible without having source & sink at diff. temp.

          -> According to first law of thermodynamics, the heat (Q) is absorbed by a system is capable of doing mechanical work equal to the internal

               energy & additional work done by the system.

                                  ΔQ = ΔU + ΔW = ΔU + PV.

          -> The whole mechanical energy can be absorbed by the molecules in the form of kinetic energy. But the whole of hand energy cannot be converted                 into work as a part of (i) actual energy retained by internal energy.

          -> Massey the rarely change when it absorbs energy or loses energy & with mass decreases or increases in accordance with Einstein's equation                         E=mc². And the values of c2. Very large the change in mass is very low.

          -> The relations PV=a constant & PV= constant another relations blue the pressure & volume in the isothermal & adiabatic processes respectively.                             They are derived from the eqn. PV=RT

           -> The area of a closed curve on a PV diagram gives the work done in a cyclic process.

           -> Cp>Cv, when a gas is heated at a constant volume, all the heat is used do increase temperature or ΔU. When a gas heated at constant pressure,                 the Gas expands. It does work against external pressure.

                       The heat supplied is used in (ΔU) partly to increase internal energy (U) partly do the work against pressure.

           -> Internal energy Can be changed by heating or cooling vessel.

 ✤ MAYER'S FORMULA

               Cp-Cv=R

             At Constant Pressure:-

               We know by 1st law, thermodynamics

               dQ = dU + dW - ①

               Here dQ = mSΔT

           For monoatomic gas & 1 gm gas

                   dQ = Cp dT

                   & dW = P.dV

                  from ① Cp dT = dU + AdV

                  Cp dT - PdV = dU - ②

           At Constant Volume

                   dQ' = dU' + dW'

                     Here dQ' = Cv dT

                  dU' = dU & dW' = 0

                  Cv dT = dU + 0

                   Cv dT = Cp dT - PdV

                  Cp dT - Cv dT = PdV

                   [Cp - Cv] dT = PdV - ③

            By gas Eqn. PV = RT

                    PdV/dT = R

                   PdV = RdT Put in ③

                   (Cp - Cv) dT = RdT

                    Cp - Cv = R