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Lecture 5: Temperature, Adiabatic Processes Lecture 5: Temperature, Adiabatic Processes Chapter II. Thermodynamic Quantities A.G. Petukhov, PHYS 743 September 20, 2017 Chapter II. Thermodynamic Quantities Lecture 5: Temperature, Adiabatic Processes A.G. Petukhov,September PHYS 20, 743 2017 1 / 8 Thermodynamic quantities are this which describe macroscopic states of bodies. They can have (e.g. E; V ) or cannot have (e.g. T;S) mechanical meaning. Relations between thermodynamical quantities are called thermodynamic relations Let us consider two bodies in thermal equilibrium forming a closed system. Also V1 = const and V2 = const. We write: E = E1 + E2 ) ∆Γ = ∆Γ1(E1)∆Γ2(E2) / g1(E1)g2(E − E1) S = kB ln ∆Γ1(E1) + kB ln ∆Γ2(E2) = S1(E1) + S2(E2) ) S = S(E1;E2) = S(E1;E − E1) = S1(E1) + S2(E − E1) In equilibrium the total entropy reaches its maximum. Therefore dS dS dS dE dS dS = 1 + 2 · 2 = 1 − 2 = 0 dE1 dE1 dE2 dE1 dE1 dE2 Chapter II. Thermodynamic Quantities Lecture 5: Temperature, Adiabatic Processes A.G. Petukhov,September PHYS 20, 743 2017 2 / 8 Temperature cont'd We define temperature as: 1 dS = T dE This is a statistical definition of temperature. Since V is also a variable we have to use partial derivatives: @S −1 @E T = = @E V @S V From the maximum of entropy condition we have: T1 = T2 Chapter II. Thermodynamic Quantities Lecture 5: Temperature, Adiabatic Processes A.G. Petukhov,September PHYS 20, 743 2017 3 / 8 Temperature cont'd Consider now the two bodies that are not in equilibrium with each other, i.e. the entropy S(t) = S1(t) + S2(t) should increase until equilibrium will be established. Therefore dS dS dS dS dS dE dS dE > 0 or = 1 + 2 = 1 1 + 2 2 dt dt dt dt dE1 dt dE2 dt Since dE dE E = E + E = const ) 1 + 2 = 0 1 2 dt dt Therefore dS dS dS dE 1 1 dE = 1 − 2 1 = − 1 > 0 dt dE1 dE2 dt T1 T2 dt dE1 dE2 Suppose T2 > T1 then dt > 0 and dt < 0 Clausius statement of the second law of thermodynamics: Energy passes from bodies having higher temperatures to bodies having lower temperatures (at constant volume) Chapter II. Thermodynamic Quantities Lecture 5: Temperature, Adiabatic Processes A.G. Petukhov,September PHYS 20, 743 2017 4 / 8 Temperature cont'd Later we will show that from canonical distribution and from the fact that the energy E0 ≤ En < 1 it follows that T > 0 If we set kB = 1 the dimension of T will be the same as of E because S is now dimensionless, [T ] = [E] Most of the time we will use −23 kB = 1. When needed we will use kB = 1:38 × 10 J/K. We will do the replacements T ) kBT and S ) S=kB in all formulas. For instance: 1 1 dS = kBT kB dE Chapter II. Thermodynamic Quantities Lecture 5: Temperature, Adiabatic Processes A.G. Petukhov,September PHYS 20, 743 2017 5 / 8 Adiabatic Process A body is said to be thermally isolated if it is subject to no interactions other than changes in external fields λ (i.e. E; H;V ) etc. In contrast to a closed system its Hamiltonian (or total energy) depends explicitly on time: E = E(p; q; λ(t)) because external fields λ vary. If a body is thermally isolated and if external field λ varies slowly (quasi-statically) such that jλj τ jdλ/dtj relaxation (i.e. at each given moment of time it can be considered in thermodynamic equilibrium) the process is called adiabatic. During the adiabatic process S = const. Obviously this process is reversible because neither the entropy of the system nor of the environment are changing. Chapter II. Thermodynamic Quantities Lecture 5: Temperature, Adiabatic Processes A.G. Petukhov,September PHYS 20, 743 2017 6 / 8 Adiabatic Process cont'd Let us prove that S indeed a constant. Consider the time derivative: dS dλ dλ2 = A + A + A dt 0 1 dt 2 dt The coefficients A0 and A1 must be zero. Indeed, if λ = const then dS=dt = 0. Also if A1 6= 0 then dS=dt / dλ/dt. It implies that dS=dt < 0 if t ! −t. Therefore dS dλ2 dS dS dλ−1 dλ = A or = = A dt 2 dt dλ dt dt 2 dt Thus dλ dS ! 0 then ! 0 then S = const dt dt Example: In a compression of a gas in a cylinder by a thermally isolated piston the speed of the piston must be smaller than the speed of sound in the gas, i.e. it can be quite fast. In practice it is always assumed that the adiabatic processes must be "rapid" enough to prevent heat exchange with the environment. Chapter II. Thermodynamic Quantities Lecture 5: Temperature, Adiabatic Processes A.G. Petukhov,September PHYS 20, 743 2017 7 / 8 Adiabatic process In adiabatic processes a body interacts with an external fields (other bodies) that are not statistical objects (objects with zero entropy Sext = 0). More generally, if not only λ but also T varies slowly the process is called quasistatic. It is reversible because the system and environment are in equilibrium (∆S = ∆Ssys + ∆Senv = 0) but the entropy of the system will depend on time (either decrease or increase) because the system is open. The thermodynamic energy by definition E = hE[p; q; λ(t)]i Therefore for an adiabatic process d hEi dE @E(p; q; λ) dλ @ hEi dλ = = · = · dt dt @λ dt @λ dt @E(p; q; λ) @ hEi ) = = −Λ(Generalized Force) @λ @λ S Chapter II. Thermodynamic Quantities Lecture 5: Temperature, Adiabatic Processes A.G. Petukhov,September PHYS 20, 743 2017 8 / 8.
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