Lecture 4: THERMODYNAMICS of GASES - School of …gja/thermo/lectures/lecture4.pdf ·...

...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


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.


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

�

? ?


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 .


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

- -

- -

�


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

- -

- -

�


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.


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.


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


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


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).


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


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.


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 ;


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).


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).


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


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