THERMODYNAMICS thermo – heat dynamics – changes Laws of Thermodynamics: 0th, 1st, 2nd, 3rd 

Note: Thermodynamics …….. ·     is not concerned about rate of changes (kinetics) but the states before and after the change ·     does not deal with time

Classical Thermodynamics Marcoscopic observables T, P, V, … Statistical Thermodynamics Microscopic details dipole moment, molecular size, shape

Joule’s experiment T  mgh adiabatic wall (adiabatic process)   T  mgh adiabatic wall (adiabatic process) U (energy change) = W (work) = mgh Thermometer w h

Conclusion: work and heat has the same effect Page 59 w T  time interval of heating U = q (heat) Conclusion: work and heat has the same effect to system (internal energy change) * FIRST LAW: U = q + W

U: internal energy is a state function [ = Kinetic Energy (K.E.) + Potential Energy (P.E.) ] q: energy transfer by temp gradient W: force  distance E-potential  charge surface tension  distance pressure  volume First Law: The internal energy of an isolated system is constant

Convention: Positive: heat flows into system work done onto system Negative: heat flows out of system work done by system Pressure-volume Work P1 = P2 V2 V1 M

If weight unknown, but only properties of system are measured, how can we evaluate work? M M V2 P P V1 Assume the process is slow and steady, Pint = Pext

Free Expansion: Free expansion occurs when the external pressure is zero, i.e. there is no opposing force

Quasi equilibrium process: Reversible change: a change that can be reversed by an infinitesimal modification of a variable. Quasi equilibrium process: Pint = Pext + dP (takes a long time to complete) infinitesimal at any time quasi equilibrium process

what we have consider was isobaric expansion Page 64-66 1 2 P V Example: P1 = 200kPa = P2 V1 = 0.04m3 V2 = 0.1m3 * what we have consider was isobaric expansion (constant pressure) other types of reversible expansion of a gas: isothermal, adiabatic

Isothermal expansion remove sand slowly at the same time Page 65-66 Isothermal expansion remove sand slowly at the same time maintain temperature by heating slowly * P T V1 V2 V area under curve

Ex. V1 = 0.04m3 P1 = 200kPa V2 = 0.1m3

Adiabatic Reversible Expansion For this process PV = constant for ideal gas (proved later) V1 P1 T1 (slightly larger than 1) V2 P2 T2

|Wa|>|Wb|>|Wc|>|Wd| Ex. V1 = 0.04m3, P1 = 200kPa, V2 = 0.1m3,  = 1.3 200kPa const a d b V P d: pressure drop without volume change |Wa|>|Wb|>|Wc|>|Wd| isothermal adibatic

State Function Vs Path Function   State function: depends only on position in the x,y plane e.g.: height (elevation) 300 1 B 2 200 100m Y X A

Path Function: depends on which path is taken to reach destination from 1  2, difference of 300m (state function) but path A will require more effort. Internal energy is a state function, heat and work are path functions P 3 2 481.13K 192.45K 1 200kPa 0.04 0.1 V/m3

5 moles of monoatomic gas

Joule’s second experiment Energy UU(V,T) ??? thermometer V2 V1 adiabatic wall At time zero, open valve

After time zero, V1 V1+V2 T=0  Q=0 W=0 no Pext  U=0 thermometer No temperature change adiabatic wall After time zero, V1 V1+V2 T=0  Q=0 W=0 no Pext  U=0

U=U(T) Energy is only a function of temperature for ideal gas * 0 (for ideal gas) CV is constant volume heat capacity

*Enthalpy Define H = U + PV state function intensive variables     state function intensive variables locating the state Enthalpy is also a state function H = U + PV + VP

H = QP constant pressure heating At constant pressure H = U+PV = U - W = Q H = QP constant pressure heating H is expressed as a functional of T and P for ideal gases

Thermochemistry   Heat transferred at constant volume qV = U Heat transferred at constant pessure qP = H Exothermic H = - Endothermic H = +

Standard states, standard conditions   do not measure energies and enthalpies absolutely but only the differences, U or H The choice of standard state is purely a matter of convenience

What is the standard state ? The standard states of a substance at a specified temperature is its pure form at 1 atmosphere

25oC, 1 atmosphere: the most stable forms of elements assign “zero enthalpy” Ho298 = 0 used for chemical reactions

Standard enthalpy of formation Standard enthalpy change for the formation of the compound from its elements in their reference states. Reference state of an element is its most stable state at the specified temperature & 1 atmosphere C (s) + 2H2 (g)  CH4 (g) Hfo = -75 kJ 289K, 1 atm From the definition, Hfo for elements  0

Hess’s Law The standard enthalpy of an overall reaction is the sum of the standard enthalpies of the individual reactions into which a reaction may be divided. Standard reaction enthalpy is the change in enthalpy when the reactants in their standard states change to products in their standard states.

Hess’ Law H1 H1 = H2 + H3 + H4 state function Hess’s law is a simple application of the first law of thermodynamics

e.g. C (s) + 2H2 (g)  CH4 (g) H1o = ? 298K, 1 atm C (s,graphite) + O2 (g)  CO2 (g) Ho = -393.7 kJ H2 (g) + ½O2(g)  H2O (l) Ho = -285.8 kJ CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (l) Ho = -890.4 kJ H1o = -393.7 + 2(-285.8) - (-890.4) = -75 kJ/mole

Heat of Reaction (Enthalpy of Reaction)   Enthalpy change in a reaction, which may be obtained from Hfo of products and reactants Reactants  Products I stoichiometric coefficient, + products, - reactants E.g. CH4 (g) + Cl2 (g)  CH3Cl (g) + HCl (g)

Hfo/ kJ CH3Cl -83.7 HCl -92.0 Cl2 0 CH4 -75.3 Hro = (-83.7-92.0) - (-75.3+0) = -100.4 kJ Reactants Products elements elements

* Page 83 2A + B  3C + D 0 = 3C + D - 2A - B Generally, 0 = J J J J denotes substances, J are the stoichiometric numbers *

Bond energy (enthalpy) Assumption – the strength of the bond is independent of the molecular environment in which the atom pair may occur. C (s,graphite) + 2H2 (g)  CH4(g)Ho = -75.4 kJ H2 (g)  2H (g) Ho = 435.3 kJ C (s,graphite)  C (g) Ho= 715.8 kJ  C (s,graphite) + 2H2 (g)  C (g) + 4H (g) Ho = 2(435.3)+715.8 = 1586.5 kJ C (g) + 4H (g)  CH4 (g) H = -75.4-1586.5 = -1661.9 kJ CH Bond enthalpy = 1661.9/4 kJ = 415.5 kJ

Temperature dependence of Hr HrT CP,P R Hro CP,R 298 T

H2 = Ho + CP(T2-T1) assume CP,i constant wrt T What is the enthalpy change for vaporization (enthalpy of vaporization) of water at 0oC? H2O (l)  H2O (g) Ho = -241.93 - (-286.1) = 44.01 kJmol-1 H2 = Ho + CP(T2-T1) assume CP,i constant wrt T H (273) = Ho(298) + CP(H2O,g) - CP (H2O,l)(273-298) = 44.10 - (33.59-75.33)(-25) = 43.0 kJ/mole

Reversible vs Irreversible Non-spontaneous changes vs Spontaneous changes Reversibility vs Spontaneity First law does not predict direction of changes, cannot tell which process is spontaneous. Only U = Q + W

Second Law of Thermodynamics Origin of the driving force of physical and chemical change The driving force: Entropy Application of Entropy: Heat Engines & Refrigerators Spontaneous Chemical Reactions Free Energy

Second Law of Thermodynamics No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work Hot Reservoir q w Engine

Direction of Spontaneous Change More Chaotic !!!

Entropy (S) is a measurement of the randomness Spontaneous change is usually accompanied by a dispersal of energy into a disorder form, and its consequence is equivalent to heating Entropy (S) is a measurement of the randomness of the system, and is a state function! S  Q S  1 / T

Page 122 Entropy S For a reversible process, the change of entropy is defined as (thermodynamic definition of the entropy) * Another expression of the Second Law: The entropy of an isolated systems increases in the course of a spontaneous change: Stot > 0 where Stot is the total entropy of the isolated system

Entropy S The entropy of an isolated systems increases in the course of a reversible change: Stot = 0 where Stot is the total entropy of the isolated system

Carnot’s Theoretical Heat Engine Heat flows from a high temperature reservoir to a low temperature body. The heat can be utilized to generate work. e.g. steam engine.

The efficiencies of heat engines Hot Reservoir q w Engine DS = - |q|/Th < 0 not possible! contrary to the second law

The Nernst Theorem The entropy change accompanying any physical or chemical transformation approaches zero as the temperature approaches zero. DS  0 as T0

Third Law of Thermodynamics If the entropy of each element in its most state is taken as zero at the absolute zero of temperature, every substance has a positive entropy. But at 0K, the entropy of substance may equals to 0, and does become zero in perfect crystalline solids. Implication: all perfect materials have the same entropy (S=0) at absolute zero temperature Crystalline form: complete ordered, minimum entropy

Statistical Interpretation of S S = 0 at 0K for perfect crystals S = k ln    Boltzmann number of arrangements postulate of entropy Boltzmann constant

Entropy Change of Mixing one distinguish arrangement number of arrangement increased

In general mixing NA, NB Entropy change of mixing Stirling’s approximation: ln N!  N ln N + 0(N) for large N

Extensions of 2nd Law   TdS dq Clausius Inequality For adiabatic process, Entropy will always attain maximum in adiabatic processes. A similar function for other processes?

Define Helmholtz free energy A = U - TS Thermodynamic State Function dA = dU - TdS - SdT Substitute into Clausius Inequality for isothermal, isochoric (constant volume) process,

change in Helmholtz free energy = maximum isothermal work Example of isothermal, isochoric process: combustion in a bomb calorimeter O2 + fuel Temp. bath O2, CO2, H2O Higher P heat given out

Gibbs Free Energy Define Gibbs free energy G = H - TS Thermodynamic state function = U + PV -TS dG = dU +PdV + VdP -TdS -SdT

constant pressure, constant temperature G will tend to a minimum value equilibrium spontaneous change More applications since most processes are isothermal, isobaric chemical reactions at constant T, P Reactants  Products endothermic H is positive exothermic H is negative

Change of Gibbs free energy with temperature (constant pressure)

In P 1/T

Example: What is the change in the boiling point of water at 100oC per torr change in atmospheric pressure?   Hvap = 9725 cal mol-1 Vliq = 0.019 l mol-1 Vvap = 30.199 l mol-1