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 PHYS20, 7432017 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 V 1 = const and V 2 = const. We write: E = E 1 + E 2 Γ = Γ 1 (E 1 ) Γ 2 (E 2 ) g 1 (E 1 )g 2 (E E 1 ) S = k B ln Γ 1 (E 1 ) + k B ln Γ 2 (E 2 ) = S 1 (E 1 ) + S 2 (E 2 ) S = S(E 1, E 2 ) = S(E 1, E E 1 ) = S 1 (E 1 ) + S 2 (E E 1 ) In equilibrium the total entropy reaches its maximum. Therefore = 1 + 2 de 2 = 1 2 = 0 de 1 de 1 de 2 de 1 de 1 de 2 Chapter II. Thermodynamic Quantities Lecture 5: Temperature, Adiabatic Processes A.G. Petukhov, September PHYS20, 7432017 2 / 8
Temperature cont d We define temperature as: 1 T = de This is a statistical definition of temperature. Since V is also a variable we have to use partial derivatives: T = ( ) S 1 = E V ( ) E S V From the maximum of entropy condition we have: T 1 = T 2 Chapter II. Thermodynamic Quantities Lecture 5: Temperature, Adiabatic Processes A.G. Petukhov, September PHYS20, 7432017 3 / 8
Temperature cont d Consider now the two bodies that are not in equilibrium with each other, i.e. the entropy S(t) = S 1 (t) + S 2 (t) should increase until equilibrium will be established. Therefore Since > 0 or = 1 + 2 E = E 1 + E 2 = const Therefore ( = 1 ) 2 de1 de 1 de 2 = 1 de 1 + 2 de 2 de 1 de 2 de 1 ( 1 = 1 T 1 T 2 Suppose T 2 > T 1 then de 1 > 0 and de 2 < 0 + de 2 ) de1 = 0 > 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 PHYS20, 7432017 4 / 8
Temperature cont d Later we will show that from canonical distribution and from the fact that the energy E 0 E n < it follows that T > 0 If we set k B = 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 k B = 1. When needed we will use k B = 1.38 10 23 J/K. We will do the replacements T k B T and S S/k B in all formulas. For instance: 1 k B T = 1 k B de Chapter II. Thermodynamic Quantities Lecture 5: Temperature, Adiabatic Processes A.G. Petukhov, September PHYS20, 7432017 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 λ dλ/ τ 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 PHYS20, 7432017 6 / 8
Adiabatic Process cont d Let us prove that S indeed a constant. Consider the time derivative: ( ) = A dλ dλ 2 0 + A 1 + A 2 The coefficients A 0 and A 1 must be zero. Indeed, if λ = const then / = 0. Also if A 1 0 then / dλ/. It implies that / < 0 if t t. Therefore = A 2 ( dλ ) 2 or dλ = ( ) dλ 1 dλ = A 2 Thus dλ 0 then 0 then S = const 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 PHYS20, 7432017 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 S ext = 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 = S sys + S env = 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 = E[p, q, λ(t)] Therefore for an adiabatic process d E de E(p, q, λ) = = dλ λ = E λ dλ ( ) E(p, q, λ) E = = Λ (Generalized Force) λ λ S Chapter II. Thermodynamic Quantities Lecture 5: Temperature, Adiabatic Processes A.G. Petukhov, September PHYS20, 7432017 8 / 8