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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 NONLINEAR 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. 1021 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 K1/mol
!U = 3
R!T
2 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 = 3R + R = 4 R
!T 1 mole
2. LINEAR MOLECULES:
5
U = n RT
2 3.
6 !U = 5
R!T
2 NONLINEAR MOLECULES:
U = n3RT !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
7 1 Lecture 21 Heat Capacity of an Ideal Gas kBT ≈ 200 cm1
at room temp 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 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 K1/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 K1/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 K1/mol 5
R!T
2 NONLINEAR 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 cm1
1151 cm1
1361 cm1 ATOMS: 5
U = n RT
2 3. 25.0 J K1/mol !U = 3R!T H 2O
1595 cm1
3651 cm1
3756 cm1 3
U = n RT
2 20.8 J K1/mol 5
R!T
2 NONLINEAR MOLECULES: Cv J K1/
mol 3. 3
R!T
2 First a small quiz to warm up: Pn α e
Energy 2. 1. 12.5 J K1/mol
!U = kBT ≈ 200 cm1
at room temp Heat Capacity of an Ideal Gas Classical Behavior: ΔE << kT NonClassical 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) (103 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.
 Fall '08
 STAFF
 Thermochemistry

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