10.26.2011 - Lecture 21 Today in Chemistry 260 P0 Va...

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Unformatted text preview: Lecture 21 Today in Chemistry 260 P0 Va ,Pa Chemistry 260 Lecture 21 October 26, 2011 Pressure P 1 P2 Later this Week in Chemistry 260 P3 •  Reading: 12.5; 13.1, 13.2 (Oxtoby) •  Gases, thermochemistry, spontaneity and entropy Vb ,Pb P4 Volume 1 Classical Internal Energy of an Ideal Gas ATOMS: •  Review heat capacities •  Thermodynamics of gases Isothermic processes, Adiabatic processes 2 Molecular Interpretation of Heat Capacities 3 Translations per atom Almost classical ability to accept heat energy # of moles U=n U = N (3 × ½ kT) 3 2 RT # of atoms (or molecules) k=R/NA n=N/NA Little ability to accept heat energy !E < k BT Almost no ability to accept heat energy !E >> k BT !E > k BT 3 Translations per molecule LINEAR MOLECULES: 2 Rotations per molecule U=N 3 2 kT + N (2 × ½ kT) U=n 5 2 RT 3 Translations per molecule NON-LINEAR MOLECULES: 3 Rotations per molecule kT + N (3 × ½ kT) U = n 3RT 5 3 Heat Capacity of an Ideal Gas Constant Volume Molecular Interpretation of Heat Capacities N2 = e !(E2 ! E1 )/ kBT = e ! "E / kBT N1 Boltzmann Distribution kBT = 4 1. 10-21 J at RT qv = Cv ΔT ΔU = Cv ΔT Cv = ΔU/ΔT PV = nRT PΔV = nRΔT For one mole, 1 mole cv = ΔU/ΔT ATOMS: 3 U = n RT 2 ΔH = qp ΔU = qv Constant Pressure U=N 3 2 qp = Cp ΔT ΔH = CpΔT ΔU + PΔV = Cp ΔT ΔU + nRΔT = Cp ΔT For one mole, /n ΔU + RΔT = cp ΔT cp = ΔU/ΔT + R 12.5 J K-1/mol !U = 3 R!T 2 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 = 3R + R = 4 R !T 1 mole 2. LINEAR MOLECULES: 5 U = n RT 2 3. 6 !U = 5 R!T 2 NON-LINEAR MOLECULES: U = n3RT !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 7 1 Lecture 21 Heat Capacity of an Ideal Gas kBT ≈ 200 cm-1 at room temp 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 cv = LINEAR MOLECULES: 5 U = n RT 2 !U = U = n3RT !U 3 =R !T 2 cv = 2. !U 5 =R !T 2 Very good agreement cv = !U = 3R !T 9 Heat Capacity of a Monatomic Solid 12.5 J K-1/mol !U = for n = 1, !U = 3R!T !U = U = n3RT cv = At low T cv ≈ 0 cv = Low frequency vibrations accept some energy !U 5 =R !T 2 25.0 J K-1/mol !U = 3R!T cv = !U = 3R !T 10 ΔU = q + w Consider several pathways for expansion: !U = 3R !T • Isothermal (const T) • Adiabatic (no q) • Reversible Isothermic • Reversible Adiabatic (Reversible == continous equilibrium) -ΔE/kT At high T cv ≈ 3R !U 3 =R !T 2 20.8 J K-1/mol 5 R!T 2 NON-LINEAR MOLECULES: Quantized Energy Levels 24.9 cv = LINEAR MOLECULES: For a solid: d.o.f. = 3N ‒ 6 ≈ 3N U = 3nRT 3 R!T 2 • U – internal energy (state) • Q – heat • W – work For vibrational motion U = kBT per degree of freedom. U = 3 NkT SO2 519 cm-1 1151 cm-1 1361 cm-1 ATOMS: 5 U = n RT 2 3. 25.0 J K-1/mol !U = 3R!T H 2O 1595 cm-1 3651 cm-1 3756 cm-1 3 U = n RT 2 20.8 J K-1/mol 5 R!T 2 NON-LINEAR MOLECULES: Cv J K-1/ mol 3. 3 R!T 2 First a small quiz to warm up: Pn α e Energy 2. 1. 12.5 J K-1/mol !U = kBT ≈ 200 cm-1 at room temp Heat Capacity of an Ideal Gas Classical Behavior: ΔE << kT Non-Classical Behavior: ΔE >> kT T 11 12 WARNING: Most of the thermodynamics equations we re developing are only valid for ideal gases An Isothermal Expansion of an Ideal Gas (under a constant external pressure) No surprises work amount will differ upon the path Consider several pathways for expansion: First: Constant Temperature P1 = 15 atm •  Isothermal and •  Reversible Isothermal n, R, T constant  P1V1 = P2V2 assuming an ideal gas V1 = 1 L If the pins are removed what happens? What is the change in internal energy? ΔU = U2 ‒ U1 αΔT 15 ΔU = 0 What is the change in enthalpy? ΔH = H2 - H1 = (U2 + P2V2) - (U1 + P1V1) ΔH = H2 ‒ H1 = (U2 - U1) + (P2V2- P1V1) For an ideal gas, PV=const, DH = 0 P2 = 1 atm V2 = 15 L Note: this does not mean that q or w is 0, only that q=-w. In fact w = -PexΔV. 1 atm = 101.325 kPa w = -101.325 kPa (14 L) (10-3 m3/L) w = -1.42 kJ q = 1.42 kJ 16 2 Lecture 21 Vb w = −∫ P(V )dV Work is the integral of the PV curve Work is pathway dependent Va Pressure M Pa = 15 atm, Va = 1 L Pa = 15 atm Va = 1 L T = 298 K Pb = 1 atm, Vb = 15 L Pb = 1 atm Vb = 15 L T = 298 K M M w = - Pb DV = -1.42 kJ Pa = 15 atm Va = 1 L T = 298 K Pa = 15 atm, Va = 1 L Pressure Cooled to Pb w = - Pb ΔV = -1.42 kJ Pb = 1 atm, Vb = 15 L Pb = 1 atm Vb = 15 L T = 298 K Volume M Volume Vb w = ! " PdV Va A reversible process is in equilibrium at each step. The integral is the area below the PV curve A truly reversible process would take an infinite amount of time to complete. 17 Work done in an irreversible expansion Va ,Pa P0 Vb w = ! " PdV Va P2 P3 P4 Wo r on k do the ne su by s rro ys un tem din g Vb ,Pb e quilib rium s Volume Pressure P 1 P2 P3 P4 Vb P2 P3 Wo r by k do the ne su o n s rro ys un tem din g M Vb |wblue| < |wpurple| Volume tates 20 w = - Pb DV = -1.42 kJ Pa = 15 atm Va = 1 L T = 298 K Va quilibri rium s Pa = 15 atm, Va = 1 L w = ! " PdV Vb ,Pb e Vb ,Pb A reversible process is in equilibrium at each step. Work is pathway dependent Va ,Pa equilib Volume 19 Approaching reversibility P0 Va w = ! " PdV P 1 P4 tates Va ,Pa Pb = 1 atm Vb = 15 L T = 298 K M Pressure P 1 18 Work done in an irreversible compression Pressure Pressure P0 We can estimate the integral by essentially the same thought experiment (using number of smaller scale).. Pb = 1 atm, Vb = 15 L Volume b # b& dx dx ) x = ln( x ) + const ! ) x = ln(b) " ln( a) = ln % a ( $' a um sta tes 21 dV V " Vb % w = ! nRT ln $ ' = !4.12 kJ # Va & Vb Vb Va Va w = ! " PdV = ! nRT " A reversible process represents the maximum possible system s PV work. 22 3 Lecture 21 Heat in a reversible isothermic expansion Heat in a reversible isothermic compression ⎛ Vf ⎞ U (T ) ⇒ ΔU = 0 ⇒ q = − w = nRT ln ⎜ ⎟ ⎝ Vi ⎠ ⎛ Vf ⎞ U (T ) ⇒ ΔU = 0 ⇒ q = − w = nRT ln ⎜ ⎟ ⎝ Vi ⎠ Isothermal (constant T) Isothermal (constant T) First law w<0 q = -w heat output equal/opposite to work First law w>0 q = -w heat output equal/opposite to work The underlying physics: The underlying physics: When the gas expands, it does work, loses energy When work is done to compress the gas it gains energy, and would heat up relative to the surroundings. heat flows from the surroundings into the system to maintain thermal equilibrium. q>0 23 As a result, heat flows into the surroundings from the system to maintain thermal equilibrium. q<0 24 Adiabatic expansion of an Ideal Gas (under a constant external pressure) w<0 Consider several pathways for expansion: Next: w = -PextΔV •  Adiabatic (no heat exchanged) and •  Reversible Adiabatic For an ideal gas, ΔU = ncv ΔT q=0 Tf Ti 25 Reversible Adiabatic expansions (Pext = Pi = P) w<0 q=0 T2 T1 PVg=const g =Cp/Cv>1 ΔU = q + w Adiabatic q = 0 ΔU = w ncvΔT= -PextΔV ΔT ∝ -ΔV As Vf ↑, Tf ↓ The underlying physics: When the gas expands, it does work, loses energy, and therefore its temperature drops. 26 It must lose energy due to the work as no heat can be supplied (adiabatic) Adiabatic q = 0 ΔU = w For an ideal gas, ΔU = Cv ΔT ΔU = q + w Work done in an infinitesimal step wi = − PdV dU = ncv dT = − PdV nRT ncv dT = − dV V dT nR ncv =− dV T V Tf Vf dT dV ncv ∫ = −nR ∫ T V Ti Vi Tf Vf V cv ln = − R ln = R ln i Ti Vi Vf Tf ∝ 1/Vf As Vf ↑, Tf ↓ The underlying physics: When the gas expands, it does work, loses energy, 27 and therefore its temperature drops. The internal energy is the only available source of energy for the work 4 ...
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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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