Chapter 2. First Law of Thermodynamics

Chapter 2. First Law of thermodynamics

Important Topics You should review • First law of thermodynamics • Law of Conservation of Energy

• Work, heat and internal energy • Heat Capacity Cv, Cp • Expansion work • Standard State • Heat capacity • Newton, Joule units • Reversible and irreversible process • Enthalpy • Math • Isothermal & adiabatic changes • Vectors • Joule-Thomson expansion • Scalar products •Thermochemistry • Properties of ln System and environment

• System The part of world we want to describe thermodynamically e.g. matter, reaction vessel, engine; completely enclosed within a well-defined boundary. • Anything outside of the system are the surroundings. Type of system Universe surroundings Open system Closed system Isolated system

system wall wall can be either diathermic (heat can flow) or adiabatic (insulating).

Open for Closed for Closed both matter but for both matter open for matter and energy and energy energy Processes

• Process: What is done to or by the system to cause a change of state (gas expansion, heating, mixing, reaction) • Isothermal process: the temperature does not change • Adiabatic process: no heat enters or leaves system; q = 0

PHYSICAL CHEMISTRY: QUANTA, MATTER, AND CHANGE 2E| PETER ATKINS| JULIO DE PAULA | RONALD FRIEDMAN ©2014 W. H. FREEMAN D COMPANY Work Work is associated with a thermodynamic process. Transfer of Energy Some kinds of work: Example: 1. mechanical work compression of spring, turn of shaft 2. electrical work charging of battery 3. magnetic work magnetizing a system 4. gravitational work raising of a weight Will concentrate on special case of mechanical work called PV work.

(arises from compression/expansion) x2 Mechanical work is defined as w   F(x)dx work = force x displacement in direction of force x1

Sign convention work done on system, heat supplied to system : + work done by system, heat removed from system : - Internal Energy • State: Described when we specify values for enough intensive variables such that all additional properties are fixed • State Function: Physical property that has a specific value once the state is defined e.g. Volume V(T, P); independent of path (i.e. how we got to that state) • Path function: a property of a system that depends on the previous history of the current state of the system. (heat, work) • Internal Energy: State variable, U Don't confuse with potential E or intermolecular potential Total energy of system

• Only changes in energy, U, are calculable   final  initial First law of thermodynamics (Conservation of Energy) • The total amount of energy is the universe is constant. • Energy may be neither created nor destroyed, but it may be converted from one form to another. U  U2 U1  q  w q: heat; w: work dU = dw + dq • Change in U is the sum of the heat and work supplied to a system in going from state 1 to state 2 • Energy is conserved only for system + surroundings; system & surroundings can exchange energy • In terms of the operation of machines, it is impossible to construct a device which will work continuously without the input of energy to the device. • The best that you can do is break even; you can't win. PV Work x2 Gas expanding against a piston w    Fdx (- sign b/c work done by gas) x2 F x1 Rewrite as w    A Adx x1 F Opposing pressure is Pex  A

Pressure of gas= P; and volume change is dV  Adx

V2 Therefore, work : w    Pex dV V1 px P  Pex gas will expand; gas will be compressed P  Pex

P  Pex equilibrium. No change in volume of gas. CHAPTER 2: FIGURE 2A.6

V2 w    Pex dV V1

• Gas expansions are technologically important for steam engine, internal combustion engine, refrigerator, air conditioner • Isothermal Expansion the gas is in thermal contact with its surroundings; temperature is constant; T = 0 • Adiabatic Expansion No heat enters gas during expansion; q = 0 V2 • Work done by gas expanding w   Pex dV V1 Larger the opposing pressure, the more work done. (But Pex < P, or 0 work done)

• How to maximize work? Consider ideal gas with P1 = 10 atm and V1 = 1 L. Pex = 10 atm (1 atm from air and 9 atm from weights)

If remove all weights at once, If move weights in 2 steps w  521  5L atm gas will expand until P1=Pex; step 1: Pex = 5, gas expands to 2 L final volume of 10 step 2: Pex = 1, gas expands to 10 L w  110 2  8L atm More work! w  1atm10 1L  9L atm  912J w 13L atm 1318J Reversible Process

• Can maximize work if P = Pex + dP at each stage of expansion nRT • In the limit of dP  0 with Pex  P  V for reversible expansion or compression of ideal gas V2 V2 nRT w   p dV   dV  nRT ln(V /V )  ex  V 2 1 V1 V1

• This limiting case is an example of a reversible process. Reversible change: a change that occurs through a series of equilibrium states; has to occur slowly enough to establish an equilibrium. • Occurs in the limit of an infinite number of infinitesimal steps. • Tells us the maximum work that can be obtained from a process. • Important because it simplifies thermodynamic calculations

-- Can use P in place of Pex -- State Variables depend only on final and initial state and not on path. Can calculate changes along reversible path; will be the same for an irreversible path between the same points. Irreversible expansion: external pressure and the pressure of the gas are not in in equilibrium during the expansion e.g. expansion into a vacuum Other Work

U = q + wexp + we

Temperature (T, K) Heat (q, J) Intensive Extensive Based on ideal gas in limit of low P All forms of energy transfer not due to work. (true for real gases) Ind. of material used Associated w/ a thermodynamic process. Scale chosen so entropy at 0 K is 0 ; The transfer of energy by heat to a pure 푇 = 푘 lim 푃푉 substance will change its T, depending on its 푃→0 heat capacity. Heat Capacity Heat Capacity Proportionality between the q and the T- change it produces CT  dq C not constant, depends on T dT Depends on whether material is allowed to expand Extensive property

dq Heat capacity at constant Cv  dT  volume v 1 dq Cvm  Molar Heat Capacity n dT v

Specific Heat 1 dq cv  (w = mass) w dT v Internal Energy and Heat Capacity at Constant Volume U U Consider U(T,V); Use slope formula dU  dT  dV T V V T Change state infinitesimally: dU=dq+dw U (1) only PV work: dU=dq-PexdV dU  dT  dq (2) If constant volume dV = 0 T V dq U CVm  C  dTV Vm T V

U What about second slope?   V T Called Internal Pressure, π. (will derive later) U P  T  P V T TV nRT Find internal Pressure of Ideal Gas P  V Internal energy of an ideal gas depends only on temperature! Equation of State for U

Putting the slopes together, we get  P  dU  CvdT  T  P dV  TV  1. Ideal gases: second term is 0. 2. Real gases at low to moderate P: second term is small. "near ideal" 3. Constant Volume (dV=0): Second term is 0. 4. Condensed Phases: Internal Pressure is large, but dV is small. Often neglect second term. So, in many cases, consider U(T):

T2 U  CvdT T1 Shows that internal energy is useful for constant volume processes CHAPTER 2: FIGURE 2E.1

Definition of enthalpy H = U+PV

Changes in enthalpy of the system is equivalent to the heat exchanged between the system and surroundings under constant pressure.

dH = dU+d(PV) = dq – PdV + PdV + VdP

Thus, dH = dq +VdP dH= dq (at constant P)

Enthalpy is convenient to describe heat transfer in and out of the system during the physical and chemical changes under constant pressure conditions. e.g. under atmospheric pressure. Equation of State for H(T,P)

H H Slope equation dH  dT  dP TP PT • Take this at constant P (dP = 0) H dH  dT  dqP TP H • First term is heat capacity at constant P CP  TP H V • Second term is (will derive later)    V  T  P T TP  V  • Overall equation of state dH  CPdT  V  T dP  T P  1. Ideal gases: second term is 0. For ideal gases, enthalpy depends only on T. 2. Constant pressure (dP=0): Second term is 0. First term tends to be more important T 2 H   C dT Change in enthalpy when T is changed from T1 to T2 at constant P P T1 CPm depends on T for polyatomic gases, liquids and solids Comparison of internal energy and enthalpy

V1 • U - used in constant V processes V1 P2 > P1 • H - used in constant P processes P1 • C - constant volume heat capacity, related U = q v constant V to U • C - constant pressure heat capacity, gas p Heat related to H (q) • If V = 0, w = 0, so U = q V2 > V1 • If P = 0, H = q P For an ideal gas, if    gas 1 • T = 0, H = U = 0!!! U = q + w

= q – p1V constant Pex

gas H = U +  (PV) Internal energy is lower = q – P1V + P1V than the heat transferred = q under constant P Why does C change w/T?

-1 -1 -1 3 3 • Average translation energy is (per molecule; kb (JK ) or permole R (JK mole )) tr  2 kbT tr  2 RT • Energy increase upon addition of heat is reflected in an increase in temperature 3 q  2 RT2  T1 q/T=C 3 • Therefore, molar heat capacity is Cvm  2 R • Only correct for closed shell atoms at low densities. (He, Ne, Ar, Xe...)

Molecules have types of energy that atoms do not. • Energy forms of molecules: translation; rotation; vibration 1 • Each translational term (x,y,z) contributes 2 kbT Equipartition of Energy - Each term contributes the same amount Equipartition Theorem to Calculate Energy in Diatomic Molecules translation: 3 rotation: 2 (2 angles) vibration: 2 (1 for kinetic , 1 for potential ) 7 Total molecular  2 kbT 7 • Predicted heat capacity: Cvm  2 R • In general vibrational modes 3N-5 linear; 3N-6 non-linear Fails in 2 ways: Does not predict variation of C with T & Incorrect values at low T Failure of Equipartition Theorem

• Energy is quantized. • The translation and rotational energy levels are close together. Even at room temperature, molecules have enough energy to occupy excited states. Vibrational energy levels are farther apart. At room temperature, the excited vibrational energy levels do not contribute to energy of molecule. Translation Rotation Vibration Only at high T does the vibrational energy contribute, Cvm= 7/2 R

Cp(T) given in Table 2.B1 and Resource Section

푐 퐶 = 푎 + 푏푇 + 푝,푚 푇2

Note that CP depends on P too!

C  2V P  T   P  T2 T  P Volume Changes: Thermal Expansion and Compressibility • Volume is a function of only two variables, e.g. P and T (slope equation) dV  V dT  V dP T p PT • Slopes themselves are a function of T and P, but for small changes in condensed phases, they can be considered to be constant. dV 1 1 • Better to consider fractional volume change  V dT  V dP V V T p V PT 1 • Coefficient of thermal expansion   V VT P

• Isothermal compressibility (not the same as compressibility factor!) 1    V T V PT

• Coefficient of thermal expansion and isothermal compressibility are related P    T V  T The Relationship between Cv and CP U PV U V • Given H and H=U+PV, C    P CP  P       TP T P T P T P U U V V U U U V C    P Using   P T V VT TP TP TP TV VTTP • Substituting in for C and factoring out: V U  v CP  CV   P TP VT 

U V nR CP  CV  nR  0 and  So, For an ideal gas   CPm  CVm  R V T TP P

U P For a real gas, using the formula for internal pressure  T  P VT TV V P CP  CV  T TP TV 2 • Can rewrite in terms of  and  TV T CP  CV   T CHAPTER 2: FIGURE 2B.3

PHYSICAL CHEMISTRY: THERMODYNAMICS,QUANTA, MATTER, AND STRUCTURE, CHANGE 2E|AND PETER CHANGE ATKINS| 10E | JULIO PETER DE ATKINS PAULA | |JULIO RONALD DE PAULA FRIEDMAN ©2014©2014 W. H. FREEMAN ANDD COMPANY COMPANY Adiabatic Reversible Expansion

• q = 0 so dU = dw or dU = -PexdV • With the assumptions 1) Expansion is reversible P = Pex 2) Gas is ideal so P  nRT , and dV V dU  CVdT nCVm dT  nRT V dT dV Rearranging CVm  R gas gas T V T2 V2 dT dV For a change of (T1, V1) to (T2, V2) integrate  CVm T  R  V T V each side 1 1 V1, P1, T1 V2, P2, T2

-- If it is reasonable to assume C does not vary with T (monatomic gases) T2  V2  Vm CVm ln  Rln T1  V1  -- Otherwise need to get T dependence of CVm • What if know P of the final state, not V? H  U  PV dq  0 dw  PdV dH  dU  PdV  VdP  dq  dw  PdV  VdP dH  VdP If Ideal gas, dH = CPdT and V = nRT/P dT dP nC dT  nRT dP Rearranging: CPm  R P Pm P T C  Vm   V  T  R p1 V2 T2 P2 2  2     CPmln  Rln     =Cp/CV -- If CP doesn't vary with T,   p2 V1 T1 P1 V1  T1    Adiabat and isotherm of ideal gas

p1V1  p2V2 For isotherm

T2 P2 CPmln  Rln   T1 P1 p1V1  p2V2 For adiabat c c c=Cvm/R T2  V2  CVm ln  Rln V1T1 V2T2  =C /C T1  V1  p V

ln(x y) = y ∙ ln(x) ln(x/z) = - ln(z/x)

Quantity Isothermal Adiabatic T2 U 0 Cv(T2-T1) U  CvdT T1 w -nRTln(V2/V1) Cv(T2-T1) w  nRT ln(V /V ) q +nRTln(V2/V1) 0 2 1 -R/Cvm T2 T1 T1 (V2/V1) CHAPTER 2: FIGURE 2D.1

• A gas at pressure P1 is expanded through a throttle valve into a region of lower pressure P2.

q=0, U = U2-U1 = w = p1V1 - p2V2 U1+p1V1=U2+p2V2 , (H =H ) isenthalpic process f i  T      Joule-Thomson coefficient    p H

T T  H 1 H     C Use Cyclic Rule PH HPPT P PT H V Remember,  V  T PT TP Then, TV   V   T P For real gas,  can be either positive (cooling on expansion) CP or negative (heating on expansion). For ideal gas, =0. Joule-Thomson inversion temperature: temperature above which  is negative and below which  is positive. Joule-Thomson Coefficients

 = Final State - Initial State rxn(property) = (property of prods) - (property of reacts) aA + bB → cC +dD Enthalpy change: ∆푟푥푛퐻 = 푐퐻푚,푐+푑퐻푚,푑-푎퐻푚,푎 − 푏퐻푚,푏

To simplify, define ni, to be stoichiometric coefficient of reactant i ( + for prods, - for reactants) Entropy change becomes: rxnH   niHm,i i • Formation reaction of a compound is the reaction by which a compound is formed from the elements in their standard states. e.g. Formation reaction for liquid water: 1 H2g 2 O2gH2O  • Elements must be in their most stable form and phase at the specified T and standard P. • At 25 ºC, most elements are solids. Must specify form: e.g. C graphite vs diamond S rhombic vs monoclinic

• Only two elements are liquids: Br2 and Hg • Gases: Rare gases and H2, N2, O2, F2, and Cl2

• fH commonly tabulated for 25 ºC (Table 2C.4) • fH = 0 for elements in their most stable states Heats of Reaction and Hess’s Law not a practical equation to use because only relative enthalpies (i.e. changes in rxnH   niHm,i i enthalpy) can be measured The practical equation to use:  rxn H   ni f Hi i Example: What is the heat of combustion of C3H8? Write the rxn:

Write equation with fH: (fH for elts in std state = 0) 1. C3H8g5O2g3CO2g4H2Oliq 2. 3Cs,graphite3O2g3CO2g 3HfCO2g 3. 4H2g2O2g4H2O  4HfH2O  4. 3Cs,graphite 4H2 gC3H8g HfC3H8g  H  3 HCO  4 HH O rxn1 rxn2 rxn3rxn4 rxn f 2 f 2 Hess’s Law: The standard enthalpy of an overall  f HC3H8  5 f HO2  reaction is the sum of the Substitute in data from tables: standard enthalpies of the rxnH  3393.509 4285.830103.850 individual reactions into which a reaction may be rxnH  2220.0kJ divided. Temperature Dependence of Reaction Enthalpies

If want rxnH at a different temp: T    rxn H  rxn Href   rxnCpdT Tref

Kirchoff’s Law (Follows from dH=CpdT)

휃 휃 ∆푟퐶푝,푚 = ෍ 휈푖퐶푝,푚

• Assume Cp constant • Formula for Cp (T) Enthalpies from other changes Calorimetry Heats of reaction are measured with Calorimetry (Ch. 2B) A rxn is carried out in contact with a water bath. The heat released by the rxn raises the T of the bath. The T rise can be related to H (constant P) or U (constant V). H=U+PV

For reactions with solids and liquids, small changes in volume, so H ≈ U.

∆퐻 = ∆푈 + ∆푛푔RT

Exothermic: A reaction in which heat is released Endothermic: A reaction that requires heat to be supplied The heat added to the system is positive, so an exothermic reaction has a negative H or U. Differential Scanning Calorimetry is used in modern research (phase transitions e.g. glass phase transitions in polymers; protein denaturation) Major Concept Review • Cp(T) • First Law of Thermodynamics, U = q+w • Equipartition theorem • Heat (q) -- All forms of energy transfer not due to work.• Relationship between Cv and Cp • Work (w) is the transfer of energy; many types defined V2  • Reversible w  nRT ln  • PV work defined as force•displacement: V2 max V w  PexdV work (ideal gas):  1  • Sign convention for work and heat V1 • Adiabatic, reversible expansions: w done on system, q supplied to system : + T2  V2  CVm ln  Rln w done by system, q removed from system : - T1  V1  • State variables depend on final & initial state; not path. T P C ln 2 R ln 2 • Reversible change occurs in series of equilibrium states Pm T P • Adiabatic q = 0; Isothermal T = 0 1 1 P C V • Equations of state for enthalpy, H and internal energy, U ln 2   Pm ln 2 P CVm V  P   V  1 1 dU CvdT T  PdV dHCPdTVT dP • Adiabats vs. isotherms  TV   TP  • Joule-Thomson expansion (H=0); Joule-Thomson coefficient, inversion U - used w/ constant V H - used w/ constant P temperature Cv - related to U Cp - related to H • Formation reaction; enthalipies of If V = 0, U = q If P = 0, H = q reaction, other changes • For an ideal gas, if T = 0, H = U = 0!!! • Calorimetry