10.24.2011 - Lecture 20 Today in Chemistry 260 • ...

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Unformatted text preview: Lecture 20 Today in Chemistry 260 •  First law of thermodynamic. Thermal equilibrium. Review. •  Enthalpy •  Heat Capacities Later this Week in Chemistry 260 Chemistry 260 Lecture 20 October 24, 2011 •  Reading: 12.6 (Oxtoby) •  Thermodynamics of gases •  Problem set 7. 1 2 A State Function Example Is the distance between the Xs a state function? (answer Y or N) The key players in the first law of thermodynamics (conservation of energy): U=Q+W x ΔU = q + w • U – internal energy (state) • Q – heat • W – work x 3 A State Function Example Location is a state function (path independent) Distance traveled is NOT a state function (path dependent) 4 Work Definition: Work is defined as energy deposited into the uniform macroscopic motion of a large number of particles. NOT a state function ‒ path dependent x Heat x Definition: Heat is defined as energy deposited into the microscopic random motion of the particles. NOT a state function ‒ path dependent An adiabatic wall is one that DOES NOT ALLOW for heat transfer. 5 A diathermic wall is one that ALLOWS for heat transfer. 6 1 Lecture 20 Equilibrium Thermodynamics Internal Energy •  Open System mass, heat, energy flow freely System U = Sum Total Internal Energy •  Closed System heat, energy flow freely The sum of all of the kinetic and potential energy contributions to the energy of all the atoms, ions, molecules, etc. in the system. •  Isolated System no mass, heat, or energy flow System System HEAT He gas HEAT q>0 q<0 Exothermic Translational Energy 7 First Law of Thermodynamics/ energy conservation Vibrational Energy Nuclear Energy Bond Energy First Law of Thermodynamics ΔU = q + w for a closed system ΔU = 0 for an isolated system U: ‒ KE of the molecules -- PE between molecules -- Bond energies Rotational Energy Electronic Energy Endothermic ETotal = KE + PE + U Methanol Gas 8 Internal energy is a state function. ΔU = q + w Pex closed system constant volume closed system expansion against external pressure Conservation of Energy The change in internal energy (ΔU) of a closed system is equal to the sum of the heat (q) added to it and the work (w) done upon it. The internal energy of an isolated system is constant. q heat introduced T w = -PDV = 0 no work done Internal energy is a state function. - Quantity is independent of path. U 9 H =U + PV Enthalpy Enthalpy is a description of the thermodynamic potential of a system. It is frequently most convenient to carry out a process at constant pressure. H ≡ U + PV ΔH = Δ (U + PV ) ΔH = ΔU + Δ( PV ) At constant pressure: ΔH = ΔU + PΔV ΔH = ΔU + PΔV = qp •  synthesizing a compound in lab •  cooking dinner •  drying the laundry ΔU = q + w At constant pressure if only PV work is done: General statement Vf ΔU = q p − P ∫ dV Vi ΔU = q p − PΔV q p = ΔU + PΔV At constant pressure assuming only PV work: Change in Enthalpy is the heat transferred3in 1 the process ΔU = q + w = q - PexΔV Heat energy is stored as internal energy and released as work. ΔU = qv 11 H=U + PV Enthalpy Enthalpy is a description of the thermodynamic potential of a system. H ! U + PV U, P and V are all state functions  H is a state function In general: !H = !U + !( PV ) ΔU, P and V are all state functions  ΔH is a state function At constant pressure At constant pressure: !H = !U + P!V assuming only PV work: Change in Enthalpy is the heat transferred in the q p = !U " w process Amount of thermal energy being absorbed Change in internal energy Amount of energy going 14 into expansion work 2 Lecture 20 Energy, Heat, and Work Summary Work and Heat are both FLOWS of energy.  They are NOT state functions or properties of state.  They are not possessed by the system.  They are defined in terms of a process. 1. WORK: deposit of energy into the uniform macroscopic motions of matter ‒ flow of kinetic and potential energy surroundings work ‒ energy theorem: w = ΔKE + ΔPE m w = -PEX ΔV dx System HEAT Exothermic HEAT T1 •  initially T1 > T2 •  E flow will continue until T1 = T2 T2 q αΔ T q = C ΔT proportionality constant is the heat capacity, C C is the amount of heat energy it takes to change the temperature by 1oC. If heat flows out of the system: HEAT If heat flows into the system: ΔT= Tf - Ti > 0 ΔT= Tf - Ti < 0 q>0 q<0 Heat Capacity! Temperature change (ΔT= Tf - Ti ) is proportional to the heat (q) received. Note: Heat and temperature are NOT the same thing. system 2. HEAT: deposit of energy into the random microscopic motions of matter ‒ flow of thermal energy System More on heat Heat always flows from the hot object to the cool object, from high temp to low temp. q<0 15 Endothermic q>0 Heat Capacity Heat capacity q C= !T Heat capacity We will consider: • Physical effects (not a state function) • Chemical (molecular) perspective 16 Heat supplied Temperature change Specific heat capacity: Molar heat capacity: 18 Heat Capacity Physical effects T T C(J / K ) !J " = c# n(mol ) & mol % K $ ' q = C !T q = nc!T q = ms!T The actual values of the heat capacities depend on the specific physical and chemical properties of the system, as well as on pressure and temperature. 19 Calorimeters q = C ΔT The value of C depends on the process: Closed system Constant volume !J" !J" C(J / K ) = cs # = s# $ mass ( g ) g%K $ & ' &g%K ' Closed system Constant pressure Constant Pressure: Constant Volume: 1 atm Experimentally: qv = C v ΔT 0 ΔU=q+w ΔU=qv –PΔV ΔU=qv No heat wasted on work all is used for temp. increase qp = Cp ΔT ΔH = q p Some heat IS wasted on work need more heat to increase temp. Cp > Cv More energy must flow into (or out of) the system under constant pressure to change the temperature since some goes into (or comes 20 out of) expansion (compression). ΔU=qv ΔH = qp 21 3 Lecture 20 Internal Energy (reminder) heat capacity (chemical structure perspective) U = Sum Total Internal Energy Methanol Gas Translational Energy Rotational Energy Electronic Energy Vibrational Energy Nuclear Energy Bond Energy PV=nRT Connects microscopic properties (speed, # collisions, etc.) to macroscopic properties (volume, pressure, temperature, etc.) Based on ideal gas assumptions: •  Molecules very small, don't interact, have random motion, elastic collisions •  Pressure related to speed •  Temperature related to average kinetic energy The sum of all of the kinetic and potential energy contributions to the energy of all the atoms, ions, molecules, etc. in the system. He gas Recap of Kinetic Model (reminder) 22 Heat Capacity of an Ideal Gas Temperature is essentially a measure of the average kinetic energy of the atoms/ molecules in the gas Pressure ⇒ PV = Etrans and speed of 1 Nms 2 molecules related 3 3 = k BT 2 23 Molecular Interpretation of Heat Capacities Thermal energy state occupation (Boltzmann Distribution) Classically, energy can be stored in a gas as internal energy with: ROTATION kT per degree of freedom (includes also PE) VI BR AT IO N N2 = e !(E2 ! E1 )/ kBT = e ! "E / kBT N1 ½ kT per degree of freedom Boltzmann Distribution TRANSLATION ½ kT per degree of freedom •  Contributions to the heat capacity can be considered classically only for sufficiently large temp, where En << kT (discretization can be neglected). •  For energy levels with En ≥ kT contribute little, if at all, to the heat capacity. Since vibrational energies are generally comparable to or greater than kT, only rotations and translations store much energy under normal conditions. 24 Molecular Interpretation of Heat Capacities Almost classical ability to accept heat energy !E < k BT Little ability to accept heat energy !E > k BT 25 Molecular Interpretation of Heat Capacities Almost no ability to accept heat energy !E >> k BT N2 = e !(E2 ! E1 )/ kBT = e ! "E / kBT N1 Boltzmann Distribution kBT = 4 * 10-21 J at RT 26 27 4 Lecture 20 3 Translations per atom U = N (3 * ½ kT) Heat Capacity of an Ideal Gas # of moles U=n 3 2 RT Kinetic theory of gases k=R/NA n=N/NA N=# of atoms (or molecules) 1. 3 Translations per molecule LINEAR MOLECULES: 3 2 kT + N (2 * ½ kT) U=n 5 2 2. U=N kT + N (3 * ½ kT) 1 mole cv = ΔU/ΔT ATOMS: For one mole, /n ΔU + RΔT = cp ΔT cp = ΔU/ΔT + R 12.5 J K-1/mol !U = 3 R!T 2 U = n 3RT LINEAR MOLECULES: 5 U = n RT 2 3 Rotations per molecule 3 2 PV = nRT qp = Cp ΔT ΔH = CpΔT ΔU + PΔV = Cp ΔT ΔU + nRΔT = Cp ΔT cv = !U 3 =R !T 2 20.8 J K-1/mol cp = !U 3 5 +R = R+R = R !T 2 2 cp = !U 5 7 +R = R+R = R !T 2 2 cp = !U + R = 3 R + R = 4 R 29 !T 1 mole RT 3 Translations per molecule NON-LINEAR MOLECULES: ΔH = qp PDV = nRDT For one mole, 3 U = n RT 2 2 Rotations per molecule U=N qv = Cv ΔT ΔU = Cv ΔT Cv = ΔU/ΔT ΔU = qv Constant Pressure ATOMS: Considering only rotation and translational DOFs: Constant Volume Classical Internal Energy of an Ideal Gas 3. ConcepTest 5 R!T 2 NON-LINEAR MOLECULES: U = n3RT 28 !U = !U = 3R!T 20.8 J K-1/mol cv = !U 5 =R !T 2 29.1 J K-1/mol 25.0 J K-1/mol cv = !U = 3R !T 33.3 J K-1/mol ConcepTest Of the following gases, which would you predict to have the highest constantvolume molar heat capacity? Of the following gases, which would you predict to have the highest constantvolume molar heat capacity? A) He B) N2 C ) H 2O D) They would all be the same A) He B) N2 C ) H 2O D) They would all be the same 30 Heat Capacity of an Ideal Gas 31 kBT ≈ 200 cm-1 at room temp Heat Capacity of an Ideal Gas Vibrational frequencies Vibrational frequencies CO2 667 cm-1 667 cm-1 1388 cm-1 2350 cm-1 -4155 cm-1 -2340 cm-1 -1632 cm-1 1. ATOMS: 3 U = n RT 2 2. 3. !U = 5 R!T 2 NON-LINEAR MOLECULES: U = n3RT 1. 12.5 J K-1/mol 3 !U = R!T 2 LINEAR MOLECULES: 5 U = n RT 2 !U = 3R!T kBT ≈ 200 cm-1 at room temp !U 3 cv = =R !T 2 cv = !U 5 =R !T 2 2. Very good agreement !U = 3R !T 32 3 R!T 2 !U = 5 R!T 2 NON-LINEAR MOLECULES: U = n3RT SO2 519 cm-1 1151 cm-1 1361 cm-1 12.5 J K-1/mol !U = LINEAR MOLECULES: 5 U = n RT 2 3. 25.0 J K-1/mol cv = ATOMS: 3 U = n RT 2 20.8 J K-1/mol H 2O 1595 cm-1 3651 cm-1 3756 cm-1 !U = 3R!T cv = !U 3 =R !T 2 20.8 J K-1/mol cv = !U 5 =R !T 2 Low frequency vibrations accept some energy 25.0 J K-1/mol cv = !U = 3R !T 33 5 Lecture 20 Heat Capacity of a Monatomic Solid For any system, Cp is greater than Cv. This difference can be quite large for gases. It is usually negligible for solids and liquids, because their only volume change at constant pressure is the small expansion or contraction on heating and cooling. 34 For vibrational motion U = kBT per degree of freedom. For a solid: d.o.f. = 3N ‒ 6 ≈ 3N U = 3 NkT U = 3nRT for n = 1, !U = 3R!T cv = !U = 3R !T Quantized Energy Levels 24.9 -ΔE/kT At high T cv ≈ 3R At low T cv ≈ 0 T Pn α e Energy •  Which of the following statements are TRUE about constant volume heat capacity (Cv) and constant pressure heat capacity (Cp)? (Choose ALL that apply.) •  Cv = (3/2)R for any ideal gas. •  Cp is ALWAYS larger than cv. •  Cp = Cv + R for all substances. •  Cp and cv are very similar for solids. •  None of the above statements are true. Cv J K-1/ mol Reading Quiz question: Classical Behavior: ΔE << kT Non-Classical Behavior: ΔE >> kT 35 6 ...
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This note was uploaded on 01/19/2012 for the course CHEM 260 taught by Professor Staff during the Fall '08 term at University of Michigan.

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