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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 * 1021 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 K1/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 K1/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 NONLINEAR 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 NONLINEAR MOLECULES:
U = n3RT 28 !U = !U = 3R!T 20.8 J K1/mol
cv = !U 5
=R
!T 2 29.1 J K1/mol 25.0 J K1/mol
cv = !U
= 3R
!T 33.3 J K1/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 cm1
at room temp Heat Capacity of an Ideal Gas
Vibrational frequencies Vibrational frequencies CO2
667 cm1
667 cm1
1388 cm1
2350 cm1 4155 cm1
2340 cm1
1632 cm1
1. ATOMS:
3
U = n RT
2 2. 3. !U = 5
R!T
2 NONLINEAR MOLECULES:
U = n3RT 1. 12.5 J K1/mol
3
!U = R!T
2 LINEAR MOLECULES:
5
U = n RT
2 !U = 3R!T kBT ≈ 200 cm1
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 NONLINEAR MOLECULES:
U = n3RT SO2
519 cm1
1151 cm1
1361 cm1
12.5 J K1/mol !U = LINEAR MOLECULES:
5
U = n RT
2 3. 25.0 J K1/mol
cv = ATOMS:
3
U = n RT
2 20.8 J K1/mol H 2O
1595 cm1
3651 cm1
3756 cm1 !U = 3R!T cv = !U 3
=R
!T 2 20.8 J K1/mol
cv = !U 5
=R
!T 2 Low frequency
vibrations accept
some energy 25.0 J K1/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 K1/
mol Reading Quiz question: Classical Behavior: ΔE << kT NonClassical 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.
 Fall '08
 STAFF
 Equilibrium, Enthalpy

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