chap 02
TRANSCRIPT
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Chap. 2: Thermodynamics
2.2: P-V Work
2.3: Heat
2.4: The 1st Law of Thermodynamics
2.5: Enthalpy
2.6: Heat Capacities
2.7: The Joule and Joule-Thomson
Experiments
2.8: Perfect Gases and the 1st Law
2.9: Calculation of 1st Law Quantities
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2.2: P-V Work
1. Work:
2. P-V Work:
xFW
xdFdW
A
Expansion
dx
F = PA
PdVxdAPxdAPxdFdWsys
xdAdV
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For a reversible process in a closed system:
A reversible process is one where the system is
always infinitesimally close to equilibrium,
and an infinitesimal change in conditions can
reverse the process to restore both system and
surroundings to their initial states.
2
1 )(
)(
dVPW
dVPdW
Vrev
VrevEqu-2.26*
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For an isobaric process, P is a constant, then:
3. Line Integrals:
See Fig-2.3 on p. 44.
The P-V work is path dependent and it is not a
state function. Be careful of this!
VPW
dVPPdVW
rev
rev
2
1
2
1
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Exp-2.2 on p. 44: P-V work
Find the work Wrev for processes (a) and (b) of Fig-2.3
if: P1 = 3.00 atm, V1 = 500 cm3,
P2 = 1.00 atm, V2 = 2000 cm3.
Also, find Wrev for the reverse of process (a).
1) Wrev for the process (a):
This process can be splitted into two sub-processes:
[a,1] from the initial point, 1, to the turning point.
[a,2] from the turning point to the ending point, 2.
Note: [a,1] is an isochoric process and [a,2] is an
isobaric process.
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Because and dV = 0 for [a,1], we
obtain: Wrev,[a,1] ≡ 0. However, Wrev,[a,2] ≠ 0 in that
[a,2] is an isobaric process. For this change:
Wrev,[a] = Wrev,[a,1] + Wrev,[a,2] = 0 + (-152J) = -152 J
Similarly, for the reverse process (a): Vinitial = V2, Vfinal =
V1, and Pinitial = P2, and Pfinal = P1. Use these relations we
can get Wrev,[-a] = +152 J for the reverse process (a).
2
1 )( dVPW Vrev
J152
10.50010.2000Pa101325
2,,
3636
2,,
122,,
arev
arev
arev
W
mmW
VVPVPW
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2) Wrev for the process (b):
Process (b) is very similar to process (a) except
that:
[b,1] is an isobaric expansion process, and
[b,2] is an isochoric process.
Thus,
Wrev,[b] = Wrev,[b,1] + Wrev,[b,2]
Wrev,[b] = -P1(V2 – V1) + 0
Wrev,[b] = -(3 101325 Pa)(2000. – 500.) 10-6 m3
Wrev,[b] = -456J
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4. Irreversible P-V work:
The work W in a mechanically irreversible volume change
sometimes cannot be calculated with thermodynamics.
According to the Law of The Conservation of Energy:
It is very hard to figure out the change of the kinetic
energy of the piston, ΔKpiston, during its acceleration.
However, if we wait longer enough, Δkpiston → 0. We
then have:
pistonexternalirrev dKdVPdw
2
1dVPW externalrev
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2.3: Heat
Heat is a means of energy transfer due to the
temperature difference between two objects.
Heat Capacity:
Specific Heat Capacity:
Molar Heat Capacity:
Calorimetric Method: dqin = dqout
dq = mcdT
Calorie (cal): The amount of heat needed to
raise one gram of water from 14.5 ºC to 15.5 ºC
at 1 atm pressure.
dT
dqC
mdT
dqc
ndT
dqc~
2
1
T
TP dTTcmq
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2.4: The 1st Law of Thermodynamics
E = K + V + U
E: The energy of a body
K: Kinetic energy
V: Potential energy
U: Internal energy
If focus on thermodynamics: let K = 0, V =0,
then, E = U
Molar Internal Energy: Um = U/n
The 1st Law of Thermodynamics: ΔU = q + w
for closed system, at rest, no fields.
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For an infinitesimal process:
dU = dq + dw (for closed systems) Equ-2.40*
ΔU ≡ 0 for a cyclic process.
Example 2.3 on p. 50: Calculation of ΔU
Calculate ΔU when 1.00 mol of H2O goes from 25.0 °C
and 1.00 atm to 30.0 °C and 1.00 atm.
Note: U is a state function. You can always choose your
favorite path to do the calculation.
ΔU = q + w, figure out q and w separately and then
calculate the ΔU.
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2.5: Enthalpy
Enthalpy, H ≡ U + PV Equ-2.45*
Why we define enthalpy, H?
Expansion
Fixed V but
P increases
Fixed P but
V increases
q q
U H
Note: ΔH = qP , ΔU = qV (Equ-2.46*/2.49*, p. 52)
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At fixed pressure:
ΔH = Δ(U + PV) = ΔU + Δ(PV) = ΔU + PΔV
For solids and liquids, ΔV is very small when
temperature changes (incompressible), thus:
ΔH ≈ ΔU (for solids and liquids)
For gases, ΔV changes a lot when temperature
changes, the PΔV term cannot be overlooked,
thus: ΔH ≠ ΔU (for gases)
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2.6: Heat Capacities
For an infinitesimal process:
For a closed system, we can define two types of
capacities:
Heat capacity at constant pressure:
Heat capacity at constant volume:
dT
dqC
pr
pr
P
PP
T
H
dT
Hd
dT
dqC
V
VV
T
U
dT
Ud
dT
dqC
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CP(T, P) and CV(T,V) ↔ H(T,P) and U(T,V)
Molar Heat Capacity:
The relations between CP and CV
VV
mVPP
mP Cn
CCC
n
CC
~
~..
VPP
VP
VP
VP
VP
VP
T
U
T
VP
T
UCC
T
U
T
PVUCC
T
U
T
HCC
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How do we figure out ?
Thus
VP T
U
T
U
PTVP
P
P
P
TVP
P
P
T
P
V
P
TV
T
V
V
U
T
U
T
U
dT
dV
V
U
T
U
dT
dU
dVV
UdT
T
UdU
dVV
UdT
T
UdU
:P pressureconstant At the
PT
VPT
VP
V
UCC
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2.7: The Joule & Joule-Thomson Experiment
1. Joule Experiment (1843): Try to find
See Fig-2.6 on p. 55.
Actually, the Joule Experiment measures the
T change with change in V at constant U.
From the experiment, we can measure the
Joule Coefficient: Equ-2.62
How could we obtain from µJ?
Apply Euler’s Chain Rule to U(T,V)!
TV
U
U
JV
T
TV
U
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Because
Thus
U
TV
1TUV U
V
V
T
T
U
JV
UVT
CV
T
T
U
V
U
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2. Joule-Thomson Experiment (1853):
See Fig-2.7 on p. 56.
The Joule-Thomson Experiment measures
the T change with change in P at constant H.
From the experiment, we can measure the
Joule-Thomson Coefficient:
Equ-2.64*
How could we obtain from µJ?
Apply Euler’s Chain Rule to H(T,P)!
TP
H
H
JTP
T
TP
H
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Because
Thus
H
TP
1THP H
P
P
T
T
H
JTP
HPT
CP
T
T
H
P
H
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2.8: Perfect Gases and The 1st Law
1. Perfect Gas: see Eq-2.66*, Eq-2.69*, Eq-2.70*
This implies: U = U(T) → U only depends on
T for perfect gases.
Since , at constant V, this equation
becomes an ordinary derivative,
Thus, . Similarly, .
0 and TV
UnRTPV
V
VT
UC
dT
dUCV
dTCdU vdTCdH P
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Combine Eq-2.61* with Eq-2.66*:
Since ,
Therefore, for perfect gases: CP,m – CV,m = R
The 1st Law for perfect gases:
dU = CVdT = dq + dw = dq - pdV
nRTPV
T
VP
V
UCC
PT
VP
0 TV
UnR
T
VPCC
P
VP
0 ,0 0 ,0 0; ,0 JTJP
T
V
T
CP
HC
V
U
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Example 2.4, p. 59: Calculation of q, w and ΔU
Suppose 0.100 mol if a perfect gas having CV,m =
1.50R independent of temperature undergoes the
reversible cyclic process 1 → 2 → 3 → 4 → 1
shown in Fig-2.10, where either P or V is held
constant in each step. Calculate q, w, and ΔU for
each step and for the complete cycle.
Solution: see p. 59
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2. Reversible Isothermal Process in a Perfect Gas
Isothermal → dU = 0, thus dq = -dw
Because dw = -pdV, dq = pdV
Ideal gas: PV = nRT, thus, P = nRT/V
2
1
1
2
1
2
2
1
12
lnln
lnln
lnlnln2
1
2
1
2
1
2
1
P
PnRT
V
VnRTWq
P
PnRT
V
VnRTW
VVnRTVnRTW
V
dVnRTdV
V
nRTPdVW
V
V
V
V
V
V
V
V
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Example 2.5 on p. 60.
A cylinder fitted with a frictionless piston
contains 3.00 mol of He gas at P = 1.00 atm and
is in a large constant-T bath at 400 K. The
pressure is reversibly increased to 5.00 atm.
Find q, w, and ΔU for this process.
For an isothermal process, ΔU = 0, q = -w
3. Reversible Constant-P (or Constant-V)
Process in a Perfect Gas: See Example 2.4!
1
2
1
2 lnln2
1
2
1 P
PnRT
V
VnRTdV
V
nRTPdVw
T
T
V
V
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4. Reversible Adiabatic Process in a Perfect Gas
For an adiabatic process: dq ≡ 0
For a reversible process with only P-V work:
dw = -PdV
For a perfect gas: dU = CVdT
Thus, the 1st law becomes: CVdT = -PdV
For an ideal gas: P = nRT/V = RT/(V/n) = RT/Vm
Then we get:
V
dVR
T
dTC
dVV
RTdTC
mv
mV
,
,
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2211
,
,
,
,
,
,,
1
11
1
22
2
1
11
22
11
22
1
2
2
22
1
11
2
1
1
2
2
1
1
2
1
2,
,
:get we, Define
1 thus,, :Note
Hence,
gas,perfect aFor
lnlnln
11
,,
,
,
,
2
1
2
1
VPVPC
C
C
C
C
RRCC
VPVP
V
V
VP
VP
VP
VP
T
T
T
VP
T
VP
V
V
T
T
V
VR
V
VR
T
TC
dVV
RdTT
C
mV
mP
mV
mP
mV
mVmP
CR
CR
CR
mV
T
T
V
VmV
mVCR
mVCR
mV
mV
mV
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For an adiabatic process, ΔU = q + w = w. Since
ΔU = CVΔT and CV is a constant for a perfect gas
ΔU = CV(T2 – T1) = w
This formula is applied to a perfect gas,
reversible adiabatic process, and a constant CV.
See Fig-2.12 on p. 61.
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2.9: Calculation of 1st Law Quantities
Thermodynamic Processes
Cyclic: ΔT = ΔP = ΔV = ΔU = ΔH = 0
Reversible: dwrev = -PdV infinity small process
Isothermal: T is constant throughout the
process
Adiabatic: dq ≡ 0, q ≡ 0
Constant-V: dV ≡ 0, ΔV ≡ 0, ΔU = qV
(Isochoric Change)
Constant-P: dP ≡ 0, ΔP ≡ 0, ΔH = qP
(Isobaric Change)
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The q, w, ΔU, ΔH for various process:
1. Reversible phase change at const. T and P
A Phase change or phase transition
2. Const.-P heating with no phase change
3. Const.-V heating with no phase change
2
1
const.at
2
1
T
TPP
rev
PdTTCqH
VPPdVww
PVUPVUPVUH
VdTTCqUT
TVV
2
1
const.at
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4. Perfect gas change of state
5. Reversible isothermal process in a perfect gas
6. Reversible adiabatic process in a perfect gas
7. Adiabatic Expansion of a perfect gas into vacuum
q = 0 (adiabatic process)
w = 0 (w = -PdV, P = 0 for vacuum surroundings)
ΔU = q + w = 0, ΔH = ΔU + Δ(PV) = ΔU + nRΔT = 0
2
1
2
1
T
TP
T
TV dTTCHdTTCU
1
22
11
2 ln ln
0,0 thus,0or 0 implies 0
V
VnRTwq
V
VnRTPdVw
HUqqq PV
2211 implies 0 VPVPwUq
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Example 2.6: Calculation of ΔH
CP, m of a certain substance in the temperature
range 250 K to 500 K at 1 bar pressure is given
by CP, m = b + kT, where b and k are certain
known constants. If n moles of this substance is
heated from T1 to T2 at 1 bar (where T1 and T2
are in the range 250 K to 500 K), find the
expression for ΔH.
2
1
2
212
2
21
,
2
1
2
1
2
1
2
1
TTkTTbnH
kTbTndTkTbndTnCHT
T
T
T
T
TmP