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...Thermodynamics
First Law - Energy is Conserved
Sometimes redefine “energy” so that First Law is true.
Combined Work and Heat have the same effect
Add two inexact differentials - get an exact one
dU = d̄ Q + d̄ W
Two heat capacities (at least) Cv =(∂U∂T
)V
; CP =(∂H∂T
)P
Enthalpy: H= U+PV. For pressure boundary conditions
Graeme Ackland Lecture 4: THERMODYNAMICS of GASES September 28, 2017 1 / 19
Fridges, thermal expansion, Charles law
If you expand a gas, does it
(a)heat up (V ∝ T)
(b)cool down
(c)stay the same temperature
Graeme Ackland Lecture 4: THERMODYNAMICS of GASES September 28, 2017 2 / 19
SLOW IDEAL GAS PROCESSES - ISOTHERMAL
Reversible Isothermal Expansion of an Ideal gas ∆T = 0Equation of state PV = nRT → P(V ,T ) = nRT/V = constant/VConstant depends on temperature: isotherms are hyperbolaeThe work done on the surroundings is
∫PdV = nRT ln(V2/V1).
It is convenient to use the equationof state P(V,T) to analyse isothermsbecause T is constant, so P dependsonly on V.
Graeme Ackland Lecture 4: THERMODYNAMICS of GASES September 28, 2017 3 / 19
FAST IDEAL GAS PROCESSES - ADIABATS
Reversible Adiabatic Expansion (No heat transfer ∆U =work )
First Law: d̄ Q = dU + PdV = CvdT + PdV .
Adiabatic Work =∫
PdV = −∫
CvdT = −Cv∆TFind variation of P,V,T in a reversible adiabatic process.Substitute for P, replace nR by Cp − Cv and put d̄ Q = 0 (adiabatic).
−CvdT
T=
nRdV
V
−cv ln(T ) = (cp − cv )ln(V ) + ln(A)
T−cv = AV cp−cv
A−1/cv = TV (cp−cv )/cvDefine: γ = CP/CV = cP/cv
TV γ−1 = a constant, A
Equivalently: PV γ = a different constant
And also: TP(γ−1)/γ = yet another constantGraeme Ackland Lecture 4: THERMODYNAMICS of GASES September 28, 2017 4 / 19
Irreversible Free Expansion and Joule Coefficient
Model for expansion into vacuum: Container with rigid adiabatic walls.Inner rigid partition divides container into equal volumes: gas & vacuum
System: everything inside the container.Process: Partition breaks, and the gasmoves irreversibly into vacuum.
This mimics a free expansion.
On system: no work, no heat Ui = Uf
Ideal gas, U = U(T ) so Tf = Ti
For a real gas, U = U(T ,V ). gas everywhere
2V
gas vacuum
V V
breakpartition
partition
�
? ?
Graeme Ackland Lecture 4: THERMODYNAMICS of GASES September 28, 2017 5 / 19
For a real gas, U can be thought of as a function of T and, say, V .
So we can write T = T (U,V ), then
dT =
(∂T
∂V
)U
dV +
(∂T
∂U
)V
dU
=
(∂T
∂V
)U
dV
= µJdV
gas everywhere
2V
gas vacuum
V V
breakpartition
partition
�
? ?
µJ is the Joule coefficient, a material property (Ideal gas: µJ=0 ).For a free expansion of a gas from Vi to Vf , the temperature change is
∆T =∫ Vf
ViµJdV .
Graeme Ackland Lecture 4: THERMODYNAMICS of GASES September 28, 2017 6 / 19
Joule-Kelvin Expansion and Coefficient
Joule-Kelvin Expansion is a forced expansion, (throttling process).Gas is forced steadily through a porous plug held in a cylinder withadiabatic walls: work done, no heat transfer.
��������
�� ��������
��
��������
��
gas gas
gassample
gassample
Pi ,Ti Pf ,Tf
Pi ,Vi ,Ti Pf ,Vf ,Tf
Pi + dP Pf − dP Pi Pf
porous plug
before after
- -
- -
�
Graeme Ackland Lecture 4: THERMODYNAMICS of GASES September 28, 2017 7 / 19
Joule-Kelvin Expansion and Coefficient
Consider the gas as the system initial equilibrium state Pi , Vi and Ti .
Left chamber: Pushingpiston, maintains the gas at(Pi ).Right chamber: Other piston,maintains the expanded gasat Pf ,final equilibrium state(Pf ,Vf ,Tf ). ��
������
����
gas gas
gassample
gassample
Pi ,Ti Pf ,Tf
Pi ,Vi ,Ti Pf ,Vf ,Tf
Pi + dP Pf − dP Pi Pf
porous plug
before after
- -
- -
�
Graeme Ackland Lecture 4: THERMODYNAMICS of GASES September 28, 2017 8 / 19
Joule-Kelvin expansion
Ignore irreversibility.
Plug assumed to be in its initial state after the gas has passedthrough it.
The gas emerges in its final equilibrium state (Pf ,Tf ).
∆Q = 0, so first law is: Uf − Ui = PiVi − Pf Vf .
Enthalpy H = U + PV so Hi = Hf :
The Joule-Kelvin process is isenthalpic
Actual process is irreversible, BUT for state variables...We can calculate ∆T via equivalent reversible isenthalpic process.
Graeme Ackland Lecture 4: THERMODYNAMICS of GASES September 28, 2017 9 / 19
Choose to write T = T (H,P) because dH = 0. Then
dT =
(∂T
∂P
)H
dP +
(∂T
∂H
)P
dH =
(∂T
∂P
)H
dP = µJKdP
µJK is the Joule-Kelvin coefficient, whose value depends on the gas beingused, and can be predicted from the gas’s equation of state.
For a Joule-Kelvin expansion of a gas from Pi to Pf ,
∆T =∫ Pf
PiµJKdP.
If ∆T is negative and large, the process can be used to liquefy gases.Analysis is similar to free expansion process, but with constant enthalpyrather than internal energy held.
Graeme Ackland Lecture 4: THERMODYNAMICS of GASES September 28, 2017 10 / 19
Joule-Kelvin coefficients
∆T =∫ Pf
PiµJKdP.
All real gases have an inversion point where µJK changes sign. Below theinversion temperature, gas cools on expansion.Low T: as molecules get further apart, attraction reduces, PE increases soKE (temperature) reducesHigh T: as molecules collide, see attraction, PE reduced so KE(temperature) increases
Graeme Ackland Lecture 4: THERMODYNAMICS of GASES September 28, 2017 11 / 19
Constant flow processes
Consider a fluid flowingthrough A1 and A2.
The bounding surface of the system couldbe stream lines or the walls of a vessel.
Fluid at A1 does work on the system.P1dv = P1v1 ; or P1/ρ1 per unit mass.
Similarly −P2v2. for fluid leaving.
Energy includes internal, bulk kineticand gravitational.
First Law : ∆U = heat + work
Graeme Ackland Lecture 4: THERMODYNAMICS of GASES September 28, 2017 12 / 19
1st Law constant flow processes (consider unit mass)
Energy entering at A1 = u1 +P1
v1+∨212
+ gz1
= h1 +∨212
+ gz1
Energy leaving at A2 = u2 +P2
v2+∨222
+ gz2
= h2 +∨222
+ gz2
z1, z2 are the heights of A1 and A2, with g the gravitational field.
∨1,∨2 are the velocities of the fluid entering and leaving.
∨21/2,∨22/2 are the specific centre of mass kinetic energies.
u1, u2 are specific internal energy of the fluid entering and leaving
h1, h2 are specific enthalpy. ( H = U + PV =⇒ h = u + P/ρ).
Graeme Ackland Lecture 4: THERMODYNAMICS of GASES September 28, 2017 13 / 19
Jet engine: thermodynamics analysis
Steady flow, mass enters at A1 and leaves at A2.First Law : change in energy = heat + work
w + q =
(h2 +
∨222
+ gz2
)−(
h1 +∨212
+ gz1
)with w and q the work done and heat transferred per mass.If there is no work done and no heat transfer, this is the Bernoulli equation.For a jet engineThe flow is fast =⇒ insufficient time for heat transfer so =⇒ q ≈ 0.No work is done and A1 and A2 are the same height
h2 +∨222
= h1 +∨212
See hand in problem
Graeme Ackland Lecture 4: THERMODYNAMICS of GASES September 28, 2017 14 / 19
NON-IDEAL GASES: THE VAN DER WAALS GAS
1 The finite size of gas moleculesb the ‘hard core’ volume taken up by 1 mole of the gas molecules.Replace V by V − b
2 Intermolecular forcesa measures combined strength and range interactions.Replace P by P + a
(V /n)2,
3 Vn in the ‘pressure correction’ term ensures that, it is an intensivequantity,
4 (V /n)−2 implies the same r−6 dependence as the van der Waalsinteraction.
Graeme Ackland Lecture 4: THERMODYNAMICS of GASES September 28, 2017 15 / 19
Van der Waals (1837-1923)
The van der Waals equation of state is:(P +
a
(V /n)2
)(V − nb) = nRT
from which, dividing through by n and writingV /n = v , the van der Waals equation for 1 mole ofgas is (
P +a
v2
)(v − b) = RT ;
Graeme Ackland Lecture 4: THERMODYNAMICS of GASES September 28, 2017 16 / 19
Other equations of state: Virial Expansion
A power law expansions gives corrections to the ideal gas. In volume:
PV
nRT=(
B1(T ) + B2(T )n
V+ B3(T )
( n
V
)2+ . . .
)With B1 = 1 so as to recover the ideal gas law in the low density limit.A similar expansion also often referred to as a Virial expansion is:
PV = n(
RT + B(T )P + C (T )P2 + . . . )
For solids one often uses the Murnaghan equationExpand about equilibrium using bulk modulus K0 and K ′0 =
(∂K∂P
)P=0
P(V ) =K0
K ′0
[(V0
V
)K ′0
− 1
]All these equations only approximate the behaviour of real materials.Accurate equations of state are often presented as tables of P(V,T).
Graeme Ackland Lecture 4: THERMODYNAMICS of GASES September 28, 2017 17 / 19
Note on intensive/extensive variables
A sample of gas in equilibrium has the same pressure throughout.A small volume of the system has less mass but the same pressure.Intensive quantities are the same independent of the amount of material(e.g. P).Extensive quantities depend explicitly on the amount of material (e.g. V).
Graeme Ackland Lecture 4: THERMODYNAMICS of GASES September 28, 2017 18 / 19
Comparison of Internal Energy and Enthalpy
Internal Energy, U Enthalpy H
Defined by dU =d̄ Q − PdV dH = d̄ Q + VdP(∂U∂T
)V
= Cv
(∂H∂T
)P
= Cp
Useful for ISOCHORIC processes ISOBARIC processesstatic (constant volume) (constant pressure)
Uf − Ui = Q Hf − Hi = Q
Uf − Ui =∫ Tf
TiCvdT Hf − Hi =
∫ Tf
TiCpdT
useful for Free Expansion (Joule) Throttling (Joule-Kelvin)flow process one-off expansion constant flow
Uf = Ui Hf = Hi
Adiabatic reversible
(any gas) Uf − Ui = −∫ Vf
ViPdV = W Hf − Hi =
∫ Pf
PiVdP 6= W
Ideal gas (needs Joule’s law) (needs Joule’s law & PV=nRT)
any process Uf − Ui =∫ Tf
TiCvdT Hf − Hi =
∫ Tf
TiCpdT
Graeme Ackland Lecture 4: THERMODYNAMICS of GASES September 28, 2017 19 / 19