Critical Point Drying

1. Biological specimens2. MEMS3. Aerogels4. Spices

Single film annealed: 16 stages

4 10 60

1

2

3

R0

increasing

R0 (K)

15.59 21.54 16.56 22.67 (R

c)

17.01 PowerLaw fit 18.30 23.77 18.57 25.01 19.08 26.11 19.68 28.90 20.63 30.83

xx /

L 0

Temperature (K)

xx

/ L0 = ( A + B T

p) fit

A = -0.01 (1)

B = 0.622 (9)

p = 0.390 (3)

L0= e2/ h

Single film annealed: 16 stages

5 10 15 20 25 30 35 40 45 50 55

0.3

0.4

0.5

0.6R

0(k) 15.59 19.08 23.77

16.56 19.68 25.01 17.01 20.63 26.11 18.30 21.54 28.90 18.57 22.67 30.83

W =

T{d

ln(

T)/

dT }

Temperature (K)

R0 increasing

ln( ( )( )

ln( )

ln( )

ln

p

p

p

d Tw T

d T

d A BT

d T

pBT

A BT

Scaling collapse for indicated values of R0 (metallic side, recall Rc=22.67 k)

Measurement of Irreversible magnetization

• Field cooled (MFC

) and Zero field cooled (MZFC

) magnetization.

• ∆M(H,T) = MFC

(H,T) – M

ZFC(H,T)

, is the irreversible magnetization

(Hm(T), ∆M

max (T))

Hm(T)

→ Magnetic field where maxima of ∆M

occures

∆Mmax

(T) → Maximum value of ∆M

Increasing T

Ni/AlOx multilayers

Field dependence of ∆M(H,T) isotherms

Scaling collapse of ∆M

∆M (H,T) = ∆M

max(T)F(H/H

m)

Ni/AlOx multilayers

(T)

T

(T)

Scaling collapse in other materials

0 2 4 6

0

1

2

3

Cu:Mn (1.5 at %) Spinglass

Gd thin film (50 nm thick)

Ni MD nanoparticles (12 nm)

FePt nanoparticles (6 nm)

LPCMO

M/

Mm

ax

H/Hm

Ni SD nanoparticles (3nm)

Zotev , Orbach et. all, PRB, 2002

Hydrogen Molecules(Quadrapolar Glass)

The combined 1The combined 1stst and 2 and 2ndnd Laws Laws

The 2nd law need not be restricted to reversible processes:

r r

dU đQ đW

đQ đW

TdS PdV

• đQ is identifiable with TdS, as is đW with PdV, but

only for reversible processes.

•However, the last equation is valid quite generally, even for irreversible processes, albeit that the correspondence between đQ & TdS, and đW & PdV, is lost in this case.

Heat reservoir at Heat reservoir at temperature temperature TT22 > > TT11

Heat reservoir at Heat reservoir at temperature temperature TT11 < < TT22

HeatHeatEngineEngine

QQ22

QQ11

2 1

2 1

W Q Q

Q Q

Q heatW workboth in Joules

Conversion of Heat to Work (a heat Conversion of Heat to Work (a heat engine)engine)

Efficiency (*):

2 2

output

input

WW

Q Q

2 1 1

2 2

1Q Q Q

Q Q

1

2

1Q

Q

**Don’t confuse with Joule-Don’t confuse with Joule-Thomson coefficientThomson coefficient

The combined 1The combined 1stst and 2 and 2ndnd Laws Laws

, , ,

, , ,

; ;

V n S n S V

V n S n S V

dU TdS PdV dn

U U UdS dV dn

S V n

U U UT P

S V n

If more than one type of particle (If more than one type of particle (constituentconstituent) ) is added, thenis added, then

1

, ,

1

m

j jj

jj S V nk

dU TdS PdV dn

Uk m j

n

Work and Internal EnergyWork and Internal Energy•Differential work đW is inexact (work not a state variable)

•Configuration work is the work done in a reversible process given by the product of some intensive variable (y) and the change in some extensive variable (X).

•đW is the work done on ‘the system’, e.g. đW is positive when a gas contracts.

•Dissipative work is done in an irreversible process and is always done ‘on the system’, i.e. đWirr > 0 always.

•Total work (configuration and dissipative) done in adiabatic process between two states is independent of path. This leads to the definition of internal energy (state variable).

ad ad adandb b

b aa a

dU U U đW W dU đW

Equation of State of an Ideal GasEquation of State of an Ideal Gas•In chapter 1, we used the zeroth law to show that a relationship always exists between P, V and T.

General form:General form: ff ((PP,,VV,,TT) = 0) = 0

Example:Example: PVPV = = nRTnRT (ideal gas law)(ideal gas law)

n quantifies the amount of the substance. The units of R and n are linked such that their product nR has the dimensions of joules/kelvin.

•If n measured in kilomoles, then R = 8.314 103 J/kilomoleK

•If n measured in moles, then R = 8.314 J/moleK

•Ideal gas law may also be written in intensive form PvPv = = RTRT

v is the specific volume in either m3/mole or m3/kilomole

Equation of State of a Real GasEquation of State of a Real GasVan der Walls’ equation in intensive form:

2

aP v b RT

v

0.1 0.2 0.3 0.4 0.5 0.60

10

20

30

40

50

60

70

80

90

100

12o 11o

10o 9.5o

9o 8.5o

8o 7o

6o 5o

4o 3o

Pres

sure

(ar

b. u

nits

)

Specific volume (arb. units)

CP

London – van der Waals’ interactionLondon – van der Waals’ interaction

EEtotaltotal = = KK + + VV = constant = constant•So, the addition of particles to a system results in the addition of kinetic energy to the system which almost immediately dissipates to the entire system.

•Therefore, the addition of particles changes the internal energy of the system. The change in the internal energy dU is proportional to the number of particles dn that are added. The proportionality constant is called the ‘chemical potential’.

dUdU = = TdSTdS – – PdVPdV + + dndn

Thus, we need to modify the combined 1Thus, we need to modify the combined 1stst and 2 and 2ndnd laws: laws:

Thermal (kinetic)Thermal (kinetic)Mechanical, chemical (potential)Mechanical, chemical (potential)

Properties of HeatProperties of HeatIt is the temperature of a body alone that determines whether heat will flow to or from a body,

“Heat energy is transferred across the boundary of a system as a result of a temperature difference only.”

•However, this does not necessarily imply that the transfer of heat to a body will increase its temperature. It may also undergo a change of state (phase) from e.g. a liquid to a gas, without a change in temperature.

•Also, if the temperature of a system increases, it does not necessarily imply that heat was supplied. Work may have been done on the system. Therefore,

“Heat is the change in internal energy of a system when no work is done on or by the system.”

Heat Capacity and specific heatHeat Capacity and specific heatThe heat capacity C of a system is defined as:

0limT

Q đQC

T dT

•Heat capacity is an extensive quantity.

The specific heat capacity c of a system is:

1 đQ đqc

n dT dT

•Specific heat is obviously an intensive quantity.

•SI units are J.kilomole-1.K-1.

Because the differential đq is inexact, we have to specify under what conditions heat is added.

• the specific heat cv; heat supplied at constant volume

• the specific heat cP; heat supplied at constant pressure andv P

v P

đq đqc c

dT dT

Using the first law, it is easily shown that:

vv v

đq uc

dT T

00 and

T

v vT

duc u u c dT

dT

•For an idea gas, the internal energy depends only on the temperature of the gas T. Therefore,

Heat Capacity and specific heatHeat Capacity and specific heat

The Gay-Lussac-Joule ExperimentThe Gay-Lussac-Joule Experiment

1

01 0

v

vu

TT T dv

v

30.001 K kilomole mu

T

v

•For an ideal gas: 0, so 0T

uu u T

v

The Joule coefficientThe Joule coefficient

vT u v u

u T u Tc

v v T v

•Suggests an experiment: measure temperature change of a gas as a function of volume whilst keeping u fixed; this will enable us to determine how u depends on v.

Two important measurable quantities:

Expansivity and CompressibilityExpansivity and Compressibility

1

P

v

v T

1

T

v

v P

Expansivity:

Compressibility:

The Second Law of ThermodynamicsThe Second Law of Thermodynamics

• Clausius’ statement: Clausius’ statement: It is impossible to construct a device It is impossible to construct a device that operates in a cycle and whose sole effect is to transfer that operates in a cycle and whose sole effect is to transfer heat from a cooler body to a hotter body.heat from a cooler body to a hotter body.

• Kelvin-Planck statement: Kelvin-Planck statement: It is impossible to construct a It is impossible to construct a device that operates in a cycle and produces no other effect device that operates in a cycle and produces no other effect than the performance of work and the exchange of heat than the performance of work and the exchange of heat from a single reservoir.from a single reservoir.

• Carnot’s theorem: Carnot’s theorem: No engine operating between two No engine operating between two reservoirs can be more efficient than a Carnot engine reservoirs can be more efficient than a Carnot engine operating between the same two reservoirs.operating between the same two reservoirs.

The Clausius Inequality and the 2The Clausius Inequality and the 2ndnd Law Law

Consider the following cyclic process:

1

2P

V

Irreversible

Reversible

The Clausius Inequality and the 2The Clausius Inequality and the 2ndnd Law Law

The Clausius inequality leads to the following relation between entropy and heat:

đQdS

T

This mathematical statement holds true for any process. The equality applies only to reversible processes.For an isolated system, đQ = 0, therefore

2 10 or 0dS S S S

• The entropy of an isolated system increases in any The entropy of an isolated system increases in any irreversible process and is unaltered in any reversible irreversible process and is unaltered in any reversible process. This is the principle of increasing entropy.process. This is the principle of increasing entropy.

This leads to the following statement:

The Carnot CycleThe Carnot Cycle1.1. aabb isothermal expansion isothermal expansion2.2. bbcc adiabatic expansion adiabatic expansion3.3. ccdd isothermal compression isothermal compression4.4. ddaa adiabatic compression adiabatic compression

1.1. WW22 > 0, > 0, QQ22 > 0 (in) > 0 (in)

2.2. WW'' > 0, > 0, QQ = 0 = 03.3. WW11 < 0, < 0, QQ11 < 0 (out) < 0 (out)

4.4. WW'''' < 0, < 0, QQ = 0 = 0

•Stirling’s engine is a good approximation to Carnot’s cycle.

1

2

1T

T

Environment at Environment at temperature temperature TT22 > > TT11

Refrigerator Inside,Refrigerator Inside,temperature temperature TT11 < < TT22

Refrig-Refrig-eratorerator

QQ22

QQ11

W

Cooling via Work (a Refrigerator)Cooling via Work (a Refrigerator)

Coefficient of Performance (c):

11

1

2 1

1

2 1

QQc

W W

Q

Q Q

T

T T

Available energyAvailable energy

1

2

1T đW

T đQ

MM

đđQQ

TT22 > > TT11

TT11 < < TT22

đđWW

Efficiency = Efficiency =

Available energy in a reversible cycle:

(1 – (1 – TT11//TT22))đđQQ

Unavailable energy:

TT11đđQQ//TT22

There exists no process that can There exists no process that can increase the available energy in increase the available energy in the universe.the universe.

Entropy changes in reversible processesEntropy changes in reversible processes

r

r

đq du Pdv

đq du Pds dv

T T T

Various cases:

• For an ideal gas quite generally:

2 22 1

1 1

ln lnv

T vs s c R

T v

Entropy changes in reversible processesEntropy changes in reversible processes

r

r

đq du Pdv

đq du Pds dv

T T T

Various cases:

• Adiabatic process: đqr = 0, ds = 0, s = constant. A reversible adiabatic process is isentropic. THIS IS NOT TRUE FOR AN IRREVERSIBLE PROCESS!

• Isothermal process:

2

2 1 1

r rđq qs s ds

T T

Entropy changes in reversible processesEntropy changes in reversible processes

r

r

đq du Pdv

đq du Pds dv

T T T

Various cases:

• Isochoric process: We assume u = u(v,T) in general, so that u = u(T) in an isochoric process. Therefore, as in the case for an ideal gas, du = cvdT. Thus,

22

2 1 11

ln ,v v

TdTs s c c

T T

provided cv is independent of T over the integration.

Entropy changes in reversible processesEntropy changes in reversible processes

r

r

đq du Pdv

đq du Pds dv

T T T

Various cases:

• Isothermal (and isobaric) change of phase:

2 1 ,l

s sT

where l is the latent heat of transformation.

Entropy changes in reversible processesEntropy changes in reversible processes

r

r

đq du Pdv

đq du Pds dv

T T T

Various cases:

• Isothermal (and isobaric) change of phase:

2 1 ,l

s sT

where l is the latent heat of transformation.

Entropy changes in reversible processesEntropy changes in reversible processes

r

r

đq du Pdv

đq du Pds dv

T T T

Various cases:

• Isobaric process: More convenient to deal with enthalpy

22

2 1 11

ln ,P P

TdTs s c c

T T

provided cP is independent of T over the integration.

rđq dh vdh du Pdv vdP ds dP

T T T

h = h(P,T) in general, so that h = h(T) in isobaric process.

The The TdsTds equations equations

The Joule-Thomson ExperimentThe Joule-Thomson Experiment

1

0

1 1 1 1vw Pdv Pv

2

2 2 2 20

vw P dv P v

1 2 2 2 1 1

1 2

w w w P v Pv

u u u

1 1 1 2 2 2 1 2oru Pv u P v h h •Thus, a throttling process occurs at constant enthalpy.

•The experiment then involves throttling the gas for different values of P2. If h depends on P, then T will change during a constant h throttling process (next slide).

•This contrasts the previous experiment where the search was for a change in T during a constant u expansion.

EnthalpyEnthalpyWhen considering phase transitions, it is useful to define a quantity h called ‘enthalpy’

h u Pv

•When a material changes phase (e.g. from a solid to a liquid) at constant temperature and pressure, latent heat l must be added. This heat of transformation is related simply to the enthalpy difference between the liquid and solid

2 2 1 1 2 1l u Pv u Pv h h

•Because h depends only on state variables, it too must be a state variable – hence its usefulness.

Enthalpy and specific heatEnthalpy and specific heat•The specific heat is not defined at any phase transition which is accompanied by a latent heat, because heat is transferred with no change in the temperature of the system, i.e. c = ∞.

•However, enthalpy turns out to be a useful quantity for calculating the specific heat at constant pressure

PP P

đq hc

dT T

00 and

T

P PT

dhc h h c dT

dT

•For an ideal gas, the enthalpy depends only on the temperature of the gas T. Therefore,

The Gibbs function And Chemical The Gibbs function And Chemical PotentialPotentialUsing Euler’s theorem, one can now define an Using Euler’s theorem, one can now define an

absolute value for the internal energy of a absolute value for the internal energy of a systemsystem

1

m

j jj

U ST PV n

ThermalThermal(kinetic)(kinetic) MechanicalMechanical

(potential)(potential)

ChemicalChemical(potential)(potential)

From this definition, one can easily show thatFrom this definition, one can easily show that

,1 1

andm m

j j j jT Pj j

G n dG dn

From the above, one readily sees that From the above, one readily sees that = = GG//nn for for a system with only one constituent, i.e. a system with only one constituent, i.e. = = gg. . This is not true for mixtures or multi-This is not true for mixtures or multi-constituent systems.constituent systems.

EnthalpyEnthalpyWhen considering phase transitions, it is useful to define a quantity h called ‘enthalpy’

h u Pv

•When a material changes phase (e.g. from a solid to a liquid) at constant temperature and pressure, latent heat l must be added. This heat of transformation is related simply to the enthalpy difference between the liquid and solid

2 2 1 1 2 1l u Pv u Pv h h

•Because h depends only on state variables, it too must be a state variable – hence its usefulness.

Enthalpy and PressureEnthalpy and Pressure

PT h P h

h T h Tc

P P T P

•Thus, we see the connection between the physics.

•The main difference is that enthalpy is relevant during constant pressure processes, whereas internal energy is relevant during constant volume processes.

0 for most gasesh

T

P

The Joule-Thomson coefficient:The Joule-Thomson coefficient:

•For an ideal gas: 0, so 0T

hh h T

P

The The TdsTds equations equations

,

,

,

v vv

P PP

vPP v

P v

P TTds c dT T dv c dT dv s s v T

T

vTds c dT T dP c dT Tv dP s s P T

T

ccT TTds c dv c dP dv dP s s v P

v P v

•heat transferred (đq = TdS) in a reversible process;

•the entropy by dividing by T and integrating;

•heat flow in terms of measurable quantities;

•difference in specific heat capacities, cP, , , T, etc..;

•relationships between coordinates for isentropic processes.

These equations give: