Thermodynamics Final
Clausius-Clapeyron equation for first-order phase changes
(δP/δT)_PB = (s2-s1)/(v2-v1) _PB means we are constraining this to the phase boundary
Work done during reversible change ON a volume
-∫P dV from v1 to v2 (work done on a system is -, work done by system on surroundings is +) This is undefined with dissipative/irreversible processes
Heat capacity at constant pressure
=dH/dT_P =(TδS/δT)_P
Availability
A = U-T_oS+P_oV ∆A=∆U-To∆S+Po∆V≤0 The availability of a given system is defined as the maximum useful work that can be obtained in a process in which the system comes to equilibrium with the surroundings or attains the dead state. It lets us derive conditions on system potentials that meximise total entropy of universe dWuseful≤-dA = -dU+TodS-PodV = (To-T)dS+(P-Po)dV
Macro-state
A group of microstates which correspond to the same thermodynamic quantities: P, V, T, phase, etc.
Natural variables
A set of appropriate variables that allow to compute other state functions by partial differentiation of the thermodynamic potentials
Micro-state
A way the particles could be arranged at one instant in a given phase Requires 6 coordinates per particle: 3 peositional and 3 for momentum Gas in equl, molecules are distributed evenly , momentum directions are uniformly distributed in "momentum space", the dist for momentum maximum has p as a maximum, is known through Maxwell-Boltzmann distribution
Joule-Kelvin expansion
Air flowing through porous plug from region of high pressure to low pressure Adiabetic so ∆Q=0, work = vi_∫^0 Pi dV W = Uf-Ui = Pi (Vi-0) - Pf(Vf-0) = PiVi -PfVf, H=U+PV, so Hi=Hf (H held constant, not U) ∆T = Pi_∫^Pf µ_JK dP = for ideal gas, for real gas can be +/0/- For free, adiabatic expansion ∆S=0 for surroundings, exerts 0 external pressure, so enthalpy H is the relavent potential TP is best indicator diagram
Carnot's corollary
All Carnot engines operating btwn same 2 reservoirs have the same efficiency (independent of working substance)
Murnagan equation
Allows bulk modulus to increase with pressure (K=K0+K0'P): P(V) = K0/K'0 [(V0/V)^K'0 -1]
Kinetic theory
Assumes the number of molecules is large, but separation >>> size of molecules. Molecules move randomly with unchanged distribution of speeds Undergo elastic collisions but do not exert forces on each other, obey Newtonian Laws of motion Energy = moles of gas * 3/2 * Avagadro's number * Boltzmann constant * Temperature
Lennard-Jones interactions
Attraction until electron fields overlap and repel each other
Thermal expanisvity
B = 1/V dV/dT
Magnetic force and MR
B=µo(H+M)≈µoH (δS/δB)_T=(δM/δT)_B
How to derive Maxwell rltns?
Begin with potential (U H, F, G). Differentiate it (may need to sub in another potential variable). Simplify, then look at your final derivative. This will be in terms of 2 variables (dX, dY where x,y = V, T, P, S, etc). Use multi to find derivative of function of these two variables, then set equal to the non-derivative variables in eqtn.
Joule coefficient for Irreversible adiabetic expansion, free expansion coefficient
Breaking glass partition btwn gas an a vacuum If gas is ideal U=U(T), T=T(U) so Ui=Uf, Ti=Tf T=T(U,V), differentiate, dU = 0 so dT = (δT/δV)_u dV = µ_J dV µ_J = (δT/δV)_u, constant value For ideal gas µJ=0 Van der Waals fluid for 1 mole: µJ= (1/Cv) [P-RT/(V-b)] Viral expansion: eqtn of state of real gas can be written as power series bc Pv=RT(1+B2/v+B3/(V^2)...) µJ =P-(R/v)(1+B2?v + B3?x^2+...) µJ = -1/Cv * RT^2?v^2 * dB2/dT Recent: µJK = 1/Cp [T(δV/δT)_p-V] ∆Tthrottling = µjk 8∆P
Coefficient of efficiency
Calc for machine, multiply result by this to determine how a machine would do. Usually around 4 or so, if negative means process occurs spontaneousl
Isolated system Closed system
Cannot exchange energy or matter with its surroundings Can exchange energy but cannot exchange matter with its surroundings
Gas equation of state: Virial Expansion
Correct ideal gas law using a power law expansion of 1/V or P: PV/nRT = (1+ B(T)*n/V+C(T)*(n/V_^2) +...) Or PV = n(RT + B(T)P + C(T)P^2+...) recover ideal gas law in low density limit
Reversible adiabetic expansion of ideal gas
DQ = dU + PdV = Cv dT + PdV, sub P=nRT/V, replace nR = cp-cv, integrate =nRTln(v2-v1)= CvT Therefore, TV^γ-1 = costant, TP^1-1/γ = different constant Work = Cv(Tf-Ti) ∆U = Cv(Tf-Ti) <-can use with irreversible bc U is a state function
Third Law of Thermodnynamics
Defines behavior of entropy at absolute zero. Any sustem ungoing a process between equilibrium states as a result of external influences, such as pressure experiences a change in entropy. THis change in entropy tends to zero as the temperature characterizing the process tends to absolute zero. The entropy of all perfect crystals is the same at absolute zero and may be taken to be zero. the contribution to entropy from each aspect of a system which is in thermodynamic equilibrium disappears at absolute zero. Consequences: as T->0 thermal expansion coefficient, heat capacity go to zero, slope of phase boundary -> 0 BUT It is impossible to reach absolute zero using a finite number of processes
Second order phase transition
Entropy is continuous, s1=s2 so heat capacity is not continuous. Volume is continuous, v1=v2, sl expansivity and compressability are discontinuous (δs/δP)_T = (δv/δT)_p
Heat
Exchange of energy between the system and the surroundings that cannot be identified as work (for systems that are not thermally isolated and W =/ U2-U1. Represented as Q
Helmholtz Free energy F
F = U-TS dF = -PdV -SdT Maxwell relation: (δP/δT)_V = (δS/δV)_T natural variables: T, V Has a minimum value for stystem at constant T, V, so good for treating mixture of 2 phases of a single sugnstance. dF=-PdV-SdT=0 Phase has different specific volumes, relative amount of phsae given by specified volume: V=N1v1+N2v2, N is number of molesin each phase
Force of piston on area A with pressure P
F=PA
Force-energy relation
F=PXA = -(δU/δx)_T+ATβK_t
Latent heat
For phase change at temp T from phase 1 to phase 2: latent heat = L = T(s2-s1) If sx>s1 l is positive so heat must be put into system (δP/δT)_PB = L/T(V2-V1)
Gibbs Free Energy G
G = H-TS dG = VdP -SdT Maxwell relation: (δV/δT)_P = -(δS/δP)_T natural variables: T, P
Enthalpy, H
H = U+PdVf dH=dQ +v dP dH=TdS + VdP (δ/δ)_P = Cp Useful for analyzing Isobaric (constant P) Jf-Hi = Q = Pi_∫^Pf VdP ≠ W Joule-Kelvin/Throttling, constant flow process Hf=Hi Adiabetic reversible processs : Hf-Hi = Pi_∫^Pf Vdp ≠ W Using Joule's law and PV=nRT Hf-Hi = Ti_∫^Tf Cp dT Equilibrium minimized for isobaric, adiabatic boundary Maxwell rltn: (δT/δP)_S = (δV/δS)_P Natural variables: S, P
Latent head
Heat required to convert between different phases
Bernoulli equation
Horizontal flow of fluids, point of higher fluid speed will have less presuure than point of lower fluid speed
Reciprocal rule
If 1/partial, can flip numerator/denominator of partial
Zeroth Law of Thermodynamics
If 2 thermally isolated systems are in equilibrium with a third, then they are in equilibrium with one another
Clausius statement Kelvin-Planck statement
Impossible to construct a device that operates in a cycle and: Does something besides transferring energy from hot to cold Extracts heat, E, and converts it into work and energy such that Qin = E = W =Qout ∫dq/T <= 0
Entropy
In an irreversible process the change of entropy is greater than the integration of contributions: heat supplied from surroindings / temp of contributing part of the system dq=0, therefore S>=0, entropy of a thermally isolated ensemble comprising of system plus surroundings can never decrease Entropy of surroundings = 1/(T_0)∫dQ
Principle of increasing energy
Irreversible process in thermal islation: dS > 0 (Sf-Si =∆S > 0 for finite process) entropy of thermally isolated system increases in any irreversible process and is unaltered by reversible process.
Ergodicity
It is possilbe to move from any microstate to another
Compressability and bulk modulus
K = -1/V dV/DP BulkM = -V dP/DV = 1/K
Second Law
K-P statement, Clausius statement. Entropy increases in the universe as a whole, so it must increase in any isolated system.
Isotherm
Line through indicator diagram (P vs. V). Has only one temperature, isotherms cannot cross
Young's modulus/elastic modulus
Measures stuffness of a solid material under stress/strain
Equipartition Theorem
Molecules in equilibrium have the same average energy associated with each independent degree of freedom 2-D energy = 1/2 kt per molecule, 1/2 RT per mole 3-D energy = 3/2 kt per molecule, 3/2 RT per mole k is the Boltzmann constant, R is gas constant Boltzmann distribution: KE = 3/2 kT
Ideal gas
Molecules take up negligible space, have no interactions, obey gas laws exactly, perfectly elastic collisions, no attractions PV=nRT
Carnot's theorem
No engine operating btwn 2 reservoirs can be more efficient than a Carnot engine operating btwn same 2 reservoirs
Adiabetic
No transfer of heat energy, thermally insulating wall PV^γ = constant so W=vi_∫^vf dV/(V^γ)
Equilibrium condition for 2 phases at fixed temperature and pressure
P and T are constant throughout system, so G must be at a minimum When 2 component phases of a system coexist at constant P and T, their specific Gibbs functions are equal There is an area of phase coexistence on a V-T diagram and on a P-V diagram (fixed vol boundary, eql requires P,T t obe the same and g1=g2. Vol constrant V=v1M1+v2M2 can be met my adjusting amounts of 2 phases)
Conjugate variables Critical point Phase transitions
P,V are conjugate bc they appear together in free energy expressions. If δP/δV> 0 free energy can be spontaneously reduced by volume collapse, so volume is not in equilibrium Point where liquid-vapor transition vanishes, dense gas is indistinguisalbe from liquid, called "fluid" Stable phase or comination of phases is aloways the one with the lowest free energy
Steffan-Boltzmann Law
P/A = σ T^4
Reversible isothermal expansion of ideal gas: What does pressure look like, describe work
PV=nRT, so P=constant/V, will be family of hyperbolas Work = ∫PdV = nRT ln(v2-v1)
Van der Waals equation
PVRNT, but adjusted for imperfections of a non-ideal gas. Correction to pressure to account for attraction btwn gas molecules, to volume for being lower due to compression
Cyclic rule
Partial = - partial with denominator switched with constant * numerator switched with constant
Approach to tackle thermo problem
Pick a system which contains constant amount of material. In steady flow problem, use constant mass, well defined P/T?V, but not always the same set of atoms. Pick something where boundaries have well-defined P, T, etc, need not be a physical boundary. Define an equivalent process: may be irreversible, but need state variables. First law always holds, so integrate btwn state points analyze equivalent reversible process Select coordinate system: Typcially 1 state variable can be expressed wrt ano 2 others. If integrating process at constant X chose X as one of the independent variables. Pressure of work: What is P in PdV? If reversible/quasistatic P is the same on either side of the piston, but if irreversible not in eql and cannot be defined. Work done on and by sytem is equal and opposite so we can use pressure of the surroundings. If boundary chosen correctly can regard surroundings as in eql with well-defined values of T, P, etc. Work can be calculated from PdV with P = pressure of surroundings, even if system undergoes non-eql process
Power of engine Efficiency of an engine
Power = Work per cycle x cycles per second η = W/Q1 = 1-Q2/Q1 Where Q1=heat supplied, Q2 = heat rejeced by engine, W
Reversible cycle
Process where cycled system returns to its initial equilibrium state on the completion of each cycle. All processes are quasi-static and reversible. During the cycle the values of the state variables change and the system exchanges heat and mechanical energy with its surroundings ∫dq/T = 0
Phase diagram Fixed volume Triple point Sublimation
Projection of equation of state surface onto the PT, shows which pase is stable for given T, P. Phase boundaries where coexistance is allowed are lines. Projection onto the TV plane also shows phase boundaries, but coexistence at fixed vol uccurs for range of volumes Where all three phases coexist, is a unique point on a PT diagram. Heating/work done at triple point changes proportion of phases ithout changing P,T so very stable heat/pressure baths at can be made accurately, useful for defining standard temp points Substance has liquid-solid-vapor triple point for a lower temperature and pressure than at the triple point no liquid can be formed, solid -> gas
State variables
Properties that can be quantified at equilibrium : internal energy, V, P, T They have path-independent integrals (doesn't order in which order you differentiate) Work is not a state function
Intensive variables Extensive variables
Property independent of how much of a material is present (density, temperature, pressure, specific heat capacity) Proportional to the amount of material (mass, energy, volume, heat capacity.) Usually in terms of x per mole, kg, or atom
Temperature
Property that determines whether or not that system would be in thermal equilibrium with others Thermal equilibrium: have no net exchange of material, chemical reactions, unbalanced force, charge flow
Efficiency of engine
Put heat in, get work out η = 1-Q2/Q1 ηc = 1-T2/T1 (Carnot only)
Efficiency of heat pump
Put work in, get heat out ηHP = Q1/W = Q1/(Q1-Q2) ηHPC= T1/(T1-T2)
Efficiency of refrigerator
Put work in, take heat out ηR=Q2/W = Q2/(Q1-Q2) ηC= T2/(T1/T2) Refrigerator cycle: Isenthalpic expansion Isobaric warming of working substance (extracting heat from cool box) Compression pump (external work done) Isobaric cooling of working substance (back to room temp)
Unification of Temperature Definitions
Q1/Q3 = f(T1, T3)
Gibbs-Helmholtz equation
Relates calculating free energy and entropy changes (δG/δT)_p=-S and (G2-G1)/(T2-T1) = -T1_∫^T2 H/(T^2)dT Once G is known you can calculate S
Maxwell relations
Relations involving, P, V, T, S, establish links btwn different measurable properties of substances (specific heat, comprehensibility) Derived from requirement that 2nd derivatives of U, H, F, G wrt natural variables do not depend on the order of differentiation Natural variables of potential from which each Maxwell rltn is derived appear in denominators of relation Cross multiplicatoin of numerators and denominators yield products of pairs of conjugate variables, δS, δT, δP, δV Negative signs with anything relating δT and δV
Reversible v. irreversible processes
Reversible process can be represented by a continuous line on an indicator (P vs. V) diagram - quasistatic, so values of state variables are uniform Irreversible process must be represented by circles
Statistical entropy for fixed energy
S=k_B*ln(Ω) k_B = Boltzmann constant Ω=number of possible microstates, microstate = fully specified set of momenta and positions for all molecules, is a function of E energy, N particles, V volume
Gibbs-Duhem relation
SdT -VdP + ∑Ni d µi = 0
Otto cycle
Simplified version of 2-stroke petrol engine with a single working substance, an ideal gas rather than air and petrol mix. No chemical changes are considered, source of heat is assumed to be external rather than internal from the combustion of fuel. Simplified cycle that remains is a 4 stage process of 2 aidabatic and 2 isochoric (constant V). Heat exchange takes place in the irreversible isochoric process, which can be done faster than isothermal, meaning the power output is high (although still lower than Carnot engine)
Equilibrium
State in which macroscopic properties (P, V, density, magnetisation) do not change. Disturbing the system in any place causes same results if disturbed in another place
Thermal expansivity
Tendency to change shape in response to a change in temperature
Diathermal
Thermal interactions allowed (exchanges of energy), thermally conducting wall
First order phase transition
Transitions have a discontinuous change of first derivative of free energy (entropy, volume, magnetic moment) g1=g2, s1≠s2, v1≠v2
Equation of vaporization curve
Using equation above, dP/dT_PB ≈ L/T * P/RT Assume L is constant, integrate to get hase boundary formula: lnP = ∫L/(RT^2) dT = -L/RT + constant
Fractional changes and absolute values
Variables with less-obvious zero values (temperature, internal energy), must talk about fractional changes or
Carnot Cycle and engine
Working substance (fluid) is taken around the cycle. Surroundings consist of 2 reservoirs with T2<T1, piston that allows the exchange of mechanical energy with other devises. Cycle: Q1 -> T1, Q2 -> T2, Work W being delivered. If working substance is not an ideal gas, shapes of isotherms will be different from graph Efficiency is independent of the nature of the working substance
Curie-Weiss law
X=Xo(Tc/(TCc-T))^γ, T>Tc
Magnetic refrigeration
ab: Isothermal magnetization: applied field aligns magnetic moments, reduces magnetic entropy bc: field is switched off. adiabatic/isenthalpic demagnetization increaases, so thermal entropy reduces and temperature drops ca: Isomagnetization: heat exchanger extracts heat from cold box, paramagnet warms up
Conditions for quantity dc = A da + B db to be path independent
dA/db _a = dB/da _b
Heat capacity
dQ/dT Specific heat capacity [ c*1/m =T(δS/δT)_at either v or p
Equilibrium conditions for different system variables (U, H, F, G)
dU = TdS -PdV U(S,V) dH=TdS+VdP H(S,P) dF=-SdT-PdV F(T,V) dG=-SdT+VdP G(T,P)
Chemical potential (mu)
dU = TdS-PdV+µdN for N particles µ=(δU/δN)_t,p G=Nµ At equilibrium, T, P, and the chemical potential are equal µ = u + Pv -Ts + mgz <-gravitational field z-height m = molar mass Raoult's Law: pi = PoNi/N = ciPo
Central equaition of thermodynamics
dU = dQ + dW -> dU = TdS-PdV Therefore dQ=TdS, dW = -PdV Valid for infinitesimal reversible processes (where T and P are well-defined), but involves only state variables and can always be integrated to relate changes of state variables even when changes occur via irreversible processes. Use the path-independence to replace actual irreversible path with equivalent reversible path along which integral can be taken; TdS=dU+PdV or Tds=du+Pdv For electrical charge (denoted by Z driven through system by emf ε) dU=TdS-PdV+εdZ
Internal energy, U
dU = dQ -PdV dU=TdS-PdV (δU/δT)_v = Cv Useful for analyzing isochoric (constant V): Uf-Ui = Q = Ti_∫^Tf Cv dT Associated with Joule flow, free expansion where Uf=Ui Adiabatic reversible process: Uf-Ui = -Vi_∫^Vf P dV = W Using Joul'els Law, adiabatic reversible processes for any gas: Uf-Ui = Ti_∫^Tf Cv dT Equilibrium minimized for isochoric, adiabatic boundary Maxwell relation: (δT/δV)_S = -(δP/δS)_v Natural variables: S, V
Heat capacity at constant volume
dU/dT _V = T(δS/δT)_V for an ideal gas: dU=cVdT = 3R/2
First Law of Thermodynamics
dU=-PdV +dQ Thermally isolated system brought from one equilibrium state to another, then the work necessary to achieve this is independent of the process used, so can just examine state function U at the endpoints: Wadiabetic = U2-U1 Work produced by the agency of heat, then quantify of heat is consumed which is proportional to the work done, expenditure of work equal to quantity of heat that is produced delta U = delta Q + delta W
Maxwell-Boltzmann velocity distribution
f(v) = sqrt(m/2pikt)^3* 4 pi v^2 e^(mv^2/2kt) Cannot have zero speed. Most probable speed at peak Area under graph = total molecules present - when heated it will shift right and decrease in amplitude bc more speeds availabe
Entropy of ideal gas
s = cv ln(T/To) + R ln(v/vo) + so AND s= cp ln(T/To) + R ln(P/Po) + so so-cvlnTo+Rln(vo) <- constant of integration this is entroly relative to a particular reference state Choosing To=0 and Po=0 is problematic, use STP = 273.15K , 101.3 kPa sf-si = cvlnTf+Rln(vf)+so-(cvln(Ti)+Rln(vi)+so)=Rln(vf/vi) And dS = Cv (δT?T) + (δP/δT)_V dV
Internal specific energy
u = U/V = 0_∫^∞ u dλ P = 1/3 u, U = uV, u=u(T) total energy density = integral of the energy density over all frequencies u=∫u_v(f, T)g(f)df where f=frequency, g(f) = probability distribution that photon will have frequency f u = (dU/dV)_T. Depends only on T, and is proportional to V (size of cavity) du=Tds-pdV
Trouton's rule
∆S_vap = 10.5 R applies reasonably well to large molecules that don't have a lot of structure
