THERMODYNAMICS Lecture 2: Evaluating properties Pierwsza strona
Macroscopic properties of the matter Properties of the matter will change if V, p and T will change. Substances can exist in different phases, in case of water: p Critical point A solid fusion sublimation liquid evaporation B vapour gas T The line AB indicates that water at the same pressure can exist in three phases (varying temperature) Triple point
Other macroscopic properties are ability for change of geometrical shapes. These are defined by relevant coefficients. The coefficient of volumetric thermal expansion: ( ) p T T p V V =, 1 α The coefficient of compressibility: ( ) T p T p V V =, 1 κ The coefficient of expansion β: ( ) V T T p V p =, 1 β
Multiplying RHS by V, and both sides by p we get: κ α β p 1 = If α and κ are known from experiments then β can be calculated. For ideal gas we get also: p T 1 1 = = κ α Above three coefficients are related to each other (α, κ, β). That results from relation between partial differentials of three variables consisting for the function. If we write, based on the ideal gas law, that V is dependent on pressure and temperature as V(p,T), then: T p V p V T V T p =
The Periodic Table Mendelejew and Meyer (1869)
The Periodic Table Look carefully proton neutron electron Atomic number 1 Atom H Atomic mass 1.00797 u 1 u = 1 atomic mass unit = (mass of 12C atom)/12 approximately #neutrons + #protons
Energy vs Mass He (m=4.0026 u) O (M=15.9995 u) 4 x He = 16.01 u Mass difference = 0.01 u = binding energy energy is mass and vice versa: E = mc 2
Molecules, Atoms and Moles Avogadro (1776-1856): All gases at same pressure, volume and temperature contain the same number of gas particles. One mole of a substance contains as many particles as they are atoms in 12 g of C i.e. 12, 6.022 x 10 23 atoms. N A = 6.022 x 10 23 1/mol is Avogadro s Number number of moles = number of particles/n A n = N/N A and number of moles = mass (in gram)/mass per mole (g/mol) Mass of 1 mol of a substance in grams = molecular mass in u e.g., 1 mol of N 2 has mass of 2x14=28 grams m particle = mass per mole/n conversion factor: 1u = 1.6605 x 10-27 kg
Concept Question Which contains the most molecules? 1. 1 mol of water (H 2 O) 2. 1 mol of oxygen gas (O 2 ) 3. Same correct H 2 O O 2
Concept Question Which contains the most atoms? 1. A mol of water (H 2 O) 2. A mol of oxygen gas (O 2 ) 3. Same correct H 2 O (3 atoms) O 2 (2 atoms)
Concept Question Which weighs the most? 1. 1 mol of water (H 2 O) 2. 1 mol of oxygen gas (O 2 ) 3. Same correct H 2 O (M = 16 + 1 + 1) O 2 (M = 16 + 16)
The Ideal Gas Law Ideal gas: low density => the only interaction that occurs between gas particles (and surrounding walls) is elastic collision => An ideal gas is highly compressible. If V decreases P increases! Boyle and Mariotte (1662,1667) found experimentally that P V = constant if T=constant (and N=const.) Experiments also revealed that P is proportional to T (V and N constant) P is proportional to N (V and T is constant) => P V is proportional to N T!
The Ideal Gas Law P V = N k B T P = pressure in N/m 2 (or Pascals) V = volume in m 3 N = number of molecules T = absolute temperature (K) k B = Boltzmann s constant k B = 1.38 x 10-23 J/K note: pv has units of N m or J (energy!)
The Ideal Gas Law P V = N k B T k B is Boltzmann s constant: k B = 1.38 x 10-23 J/K Alternate way to write this N = number of moles (n) x N A molecules/mole P V = N k B T n N A k B T n (N A k B )T n R T P V = n R T R = ideal gas constant = N A k B = 8.314 J/(mol K)
Phase transitions In most our considerations we assumed ideal gas, where particles do not interact. Recalling the gas equation: pv = nrt. In case of constant temperature T = const the distribution of a product pv with p should be following; pv 0 o -103 o -130 o p In real that is rather different.
Dla rzeczywistych gazów wykresy te wyglądają następująco; pv pv 16 o -35 o C 200 o C -75 o 100 o -103 o air -130 o 10 p CO 2 0 o 1 5 10 7 Pa p For two gases different results are obtained. In case of higher temperatures the discrepancies from ideal gas are smaller.
Rariefied real gas is similar to ideal gas. Molecules of different gases have different dimensions and different types of forces acting on them. Gases where inter-particle forces are present are named non-ideal gases. There are many approaches to define a non-ideal gas. Van-der-Waals (1879) postulated the following definition. Van der Waals Model reads; p a + Vm b V ( ) 2 m = R m T (7.27) The constant b is related to volume occupied by particles. The term 1/V 2 results from existence of inter-molecular forces, whereas a is a simple gas proportionality constant. The index m denotes the molar quantities.
Van der Waals isotherms for CO 2 : pressure p Ideal gas For temperatures above critical isotherm the curves follow ideal gas isotherms. For a decreasing volume the pressure varies not along the curve EDCA, but a straight line ECA. coexistence region molar volume Thatisdueto thefactthatin case of real gases in E condensation commences, which terminates at point A. Pressure at E is a vapour pressure. It reaches maximum value at a critical point K.
At point K the quantities p,v and T have corresponding critical values p k, V k and T k. Knowingthatatthatpoint thereisa point of bending we can, basing on mathematical principles for the existence of such point, find for the Van-der-Waals curve the critical values expressedas constantsa andb. Hence: p V T = 0 2 p = 2 V k T k Then we obtain:: V T mk k = = 3b 8a 27b R m. Substituting these values to the VdW equation we get:
p k = a 2 27b Finally we arrive at: pkv T k mk = 3 8 R m From experiments and observations we know that same substance can be found in different states with completely differentiated physical properties. Such states are known as phases. These are the systems which are properly spatially separated in homogeneous throughout. The simplest phases are the states of matter. We recall from the Van der Waals curve that phases can form equilibrium. The conditions presented here for equilibrium are simplified here.
Other Equations of State Redlich-Kwong Peng-Robinson RT p= V A ( V 1/2 m B T Vm m + B) RT p= V β V m m α ( V + β) + β( V β) m m Both are quantitative in region where gas liquefies Berthelot, Dieterici and others with more than ten parameters can give good fits!!! with seven free parameters, you can describe an elephant
Principle of Corresponding States All gases have the same properties if they are compared at corresponding conditions Define reduced variables P = P / P, T = T / T ; V = V / V R c R c R c For homework you will write the VdW equation in terms of the reduced variables Compression factor plotted using reduced variables. Different curves are different T R
Virial Equation of State Most fundamental and theoretically sound Polynomial expansion Viral Expansion PV 2 Z = = 1+ B2 P( P) P+ B3 P( P) P +... RT Z = PV RT = B V( T) B3 V( T) + + V V 2 1 + 2... B V 2 ( T) = 0 at Boyle temperature Used to summarize P, V, T data Also allow derivation of exact correspondence between virial coefficients and intermolecular interactions
A reversible process is a process which can be executed in both directions infinitely without losses. An irreversible process is a process where energy losses are present. Reasons for such losses are: friction, drop of voltage, temperature, pressure or concentration. An example of irreversible process is the burst of the latex bubble. X Irreversible process, unless energy is supplied One of examples for the process to be reversible or irreversible is money exchange. It is reversible if the exchange rate is constant and no commission fee is charged, but irreversible if the fee is charged.
1.2 Relations between classical mechanics and thermodynamics Problems of classical mechanics encompass such problems as: force, mass, distance and others. Force can be regarded as something which is pulling or pushing, but mathematically is represented as a vector. Mechanics is based on a Newton s II law: r F = d dt r ( mv) In description of mechanics the free body is used, which is influenced by forces in accordance to 2nd Law of Dynamics. Mechanical system is definced by spatial coordinated and velocity.
Work Work is defined as amount of energy supplied by the acting force on a specified distance and is defined as: W = c r F r ds (1.1) F θ c A linear integral defines the work in the direction of action of force W ds r r = ds c c F ds = F cosθ That expression contains the determination of the sign of work. Work is positive when the force and displacement have the same direction. Work is negative when thesehavetheopposite directions.
Work in the gravitational field m Force F which has to be applied to lift the mass m is equal to the weight P of a body. F = P = mg h W=P h Work W = h F ds = Pds = o r r h 0 mgh (2.1) F P
Work of electrical current The cell I(A) V R Completion of work means that there has to exist the force displacing the charge in adequate direction. Work of electrical current: W = V I Δt (2.2) How the sign of work can be determined? If the cell will be our thermodynamical system then the work done on the coil will be negative. If the coil will be our thermodynamical system then it will be positive.
i r Work of the magnetic field k r j r z ds B v y Its known that: r r r F = q v B (2.3) I X hence r F For the element of conductor length ds we have: r r ds = = r qv I dt I ds dt = l r r r I ds B = I B l j 0 (2.4)
Work performed per unit of time (power) is equal to: W hence W t t = r r l r r r = F v = I ds B v 0 r ( I Bl j) ( v j) = IlBv r (2.5) A negative sign of work is obtained in the case when the conductor is regarded as a thermodynamical system. We calculate work performed on the conductor by the magnetic field. In order to move to conductor there ought to be applied external force equal in the extent and directed in the other direction. The work performed by that force would be equal to; W t = I l Bv (2.5a)
Work of gas compression or expansion in the piston pa F = pa Displacement of the piston by Δx causes execution of work dw = F Δx Knowing that the piston has a surface A, after rearrangement dv dw = p A = p dv or A W = 2 p dv (2.6) V V 1 Executed work is positive when dv is positive. That is a work executed on the piston by the gas pressure. Gas in the piston is a closed system.
Work is positive when the piston is our thermodynamical system. If our system is gas, then work supplied to that system will be negative. Then; W V = 2 p dv V 1 (2.6a) The sign of work depends on selection of thermodynamical system!!! Graphical presentation of work performed by expanding gas is presented aside. The paths from point 1 to point 2 can be different (different processes). Work is denoted as area below the relevant path. p 1 Friction and other non quasistatic processes (where the system does not pass through subsequent equilibrium states) have not been considered. 2 V 1 V 2 V
Internal energy Recapitulating the work is a form of energy, which can pass the boundaries of the system. In order for the work to be acomplished there must exist the interaction between the system and its surroundings. The work depends on the kind of process and can be traced when the system passes from one state to another. We ought to recall that in mechanics, electromagnetism, etc. We came across with the term potential energy. Potential energy of gravitation is a work required to lift the weight above the reference level. Kinetic energy can be calculated by determination of work required to provide it with some speed.
Work carried out by the force imposing some velocity on some distancecanbe writtenintheform: W = = r r F ds = m v dv= dv m ds= dt 1 2 mv 2 = EK (2.7) Acceleration from one velocity to another renders cumulation of executed work in the form of kinetic energy. EK 2 EK 1 = v v 1 2 m v dv = 1 2 2 2 v1 m( v 2 ) Identical situation is with potential energy. EP EP = Δ EP = 1 2 W (2.8)
In equation (2.8) the change of potential energy is possible only by theconservativeforces. First law of thermodynamics enables to generalise the meaning of potential energy. Let s consider the system under transition from state I to state II. We assume that the system is absolutely isolated so there is no heat transfer. The only interaction with surroundings is through the work which does not change the potential energy. Q=0 I W ad II E I E II
The change of energy of the system is through the supply of adiabatic work by all forces acting on the system. E E = Δ E = W (2.9) II I It stems from experiments that adiabatic work between two states is always the same. If the system performs work than is exhausts the internal resources. We hence assume that energy E is a property of the system and depends only on the state of the system. If we allow same transition from state I to state II by removing of insulation then the change of internal energy is the same as these states are the same. However, in that case the system can exchange heat with surroundings. W Q = W ad ad + (2.10)
The Laws of Isolated systems: ΔE = 0 The total internal energy of isolated system remain the same. Internal energy consist of kinetic and potential energy. 1. Law of (energy conservation) change in heat added to/subtracted work done internal energy = from the system - by/on the system of the system D U = Q - W Q > 0 : heat is added to system Q < 0 : heat is subtracted from system W > 0 : work done by system on surroundings W < 0 : work done on system by surroundings