chapter 3: interactions and implications. start with thermodynamic identities
TRANSCRIPT
Chapter 3: Interactions and Implications.Start with Thermodynamic Identities
, , ln , ,BS U V N k U V N
, , ,N V N U U V
S S SdS dU dV dN
U V N
TU
S
NV
1
,
,U N
S P
V T
,
?U V
S
N
Diffusive Equilibrium and Chemical Potential
0,
AA NVA
AB
U
S0
,
AA VUA
AB
N
S
B
B
A
A
N
S
N
S
VUN
ST
,
BA
Sign “-”: out of equilibrium, the system with the larger S/N will get more particles. In other words, particles will flow from from a high /T to a low /T.
Let’s fix VA and VB (the membrane’s position is fixed), but assume that the membrane becomes permeable for gas molecules (exchange of both U and N between the sub-systems, the molecules in A and B are the same ).
UA, VA, NA UB, VB, NB
TN
S
VU
,
For sub-systems in diffusive equilibrium:
In equilibrium, BA TT
- the chemicalpotential
0.2
0.4
0.6
0.8
0.2
0.4
0.6
0.8
-6
-5
-4
0.2
0.4
0.6
0.8
S
AN
AU
Chemical Potential: examples
Einstein solid: consider a small one, with N = 3 and q = 3.
10
!)1(!
!1)3,3(
Nq
NqqN 10ln)3,3( BkqNS
let’s add one more oscillator: 20ln)3,4( BkqNS
NdT
UdT
dS
1
SVN
U
,
SN
U
To keep dS = 0, we need to decrease the energy, by subtracting one energy quantum.
Thus, for this system
S
N
U
2
5ln
3
4ln),,( 2/5
2/3
2NU
h
mVkNUVNS B
N
VTmgTkTk
h
m
N
VTk
N
ST BBB
VU
2/32/3
2,
ln2
ln
Monatomic ideal gas:
At normal T and P, ln(...) > 0, and < 0 (e.g., for He, ~ - 5·10-20 J ~ - 0.3 eV.
Sign “-”: usually, by adding particles to the system, we increase its entropy. To keep dS = 0, we need to subtract some energy, thus U is negative.
The Quantum Concentration
At T=300K, P=105 Pa , n << nQ. When n nQ, the quantum statistics comes into play.
2/3
2
2
Tkh
mn BQ
- the so-called quantum concentration (one particle per cube of side equal to the thermal de Broglie wavelength). When nQ >> n, the gas is in the classical regime.
2/52/3
32/322/3
2 2ln
2ln
2ln
Tk
P
m
hTk
Tmk
hnTkTk
h
m
N
VTk
B
BB
BBB
n=N/V – the concentration of molecules
The chemical potential increases with the density of the gas or with its pressure. Thus, the molecules will flow from regions of high density to regions of lower density or from regions of high pressure to those of low pressure .
0
QB
BB n
nTk
Tmk
hnTk ln
2ln
2/32
when n nQ, 0
2/3
23
1
h
Tmkn
Tmk
h
p
h B
dB
Q
B
dB
when nincreases
Entropy Change for Different Processes
Type of interaction
Exchanged quantity
Governing variable
Formula
thermal energy temperature
mechanical volume pressure
diffusive particles chemical potential
NVU
S
T ,
1
NUV
S
T
P
,
VUN
S
T ,
The last column provides the connection between statistical physics and thermodynamics.
The partial derivatives of S play very important roles because they determine how much the entropy is affected when U, V and N change:
Thermodynamic Identity for dU(S,V,N)
NVUSS ,, if monotonic as a function of U (“quadratic” degrees of freedom!), may be inverted to give NVSUU ,,
dNN
UdV
V
UdS
S
UdU
VSNSNV ,,,
compare withTU
S
VN
1
,
SVSNVNN
UP
V
UT
S
U
,,,
pressure chemical potential
This holds for quasi-static processes (T, P, are well-define throughout the system).
NddVPSdTUd
shows how much the system’s energy changes when one particle is added to the system at fixed S and V. The chemical potential units – J.
- the so-called thermodynamic identity for U
Thermodynamic Identities
is an intensive variable, independent of the size of the system (like P, T, density). Extensive variables (U, N, S, V ...) have a magnitude proportional to the size of the system. If two identical systems are combined into one, each extensive variable is doubled in value.
With these abbreviations:
shows how much the system’s energy changes when one particle is added to the system at fixed S and V. The chemical potential units – J.
The coefficients may be identified as: TN
S
T
P
V
S
TU
S
UVUNVN
,,,
1
This identity holds for small changes S provided T and P are well defined.
- the so-called thermodynamic identity
The 1st Law for quasi-static processes (N = const):
The thermodynamic identity holds for the quasi-static processes (T, P, are well-define throughout the system)
dU TdS PdV dN
dU TdS PdV
1 PdS dU dV dN
T T T
, , ,N V N U U V
S S SdS dU dV dN
U V N
The Equation(s) of State for an Ideal Gas
2/,, NfNUVNgNVU Ideal gas:(fN degrees of freedom) / 2, , ln f
BS U V N Nk g N VU
V
kNT
V
STP B
NU
,
TkNPV B
UkN
f
U
S
T BNV
1
2
1
,
The “energy” equation of state (U T):
TkNf
U B2
The “pressure” equation of state (P T):
- we have finally derived the equation of state of an ideal gas from first principles! In other words, we can calculate the thermodynamic information for an isolated system by counting all the accessible microstates as a function of N, V, and U.
Ideal Gas in a Gravitational Field
Pr. 3.37. Consider a monatomic ideal gas at a height z above sea level, so each molecule has potential energy mgz in addition to its kinetic energy. Assuming that the atmosphere is isothermal (not quite right), find and re-derive the barometric equation.
NmgzUU kin
NddVPSdTUd
3/ 2
2,
2( ) lnB B
S V
U V mz mgz k T k T mgz
N N z h
In equilibrium, the chemical potentials between any two heights must be equal:
2/3
2
2/3
2
2
)0(ln
2
)(ln Tk
h
m
N
VTkmgzTk
h
m
zN
VTk BBBB
)0(ln)(ln NTkmgzzNTk BB Tk
mgz
BeNzN
)0()(
note that the U that appears in the Sackur-Tetrode equation represents only the kinetic energy
Pr. 3.32. A non-quasistatic compression. A cylinder with air (V = 10-3 m3, T = 300K, P =105 Pa) is compressed (very fast, non-quasistatic) by a piston (A = 0.01 m2, F = 2000N, x = 10-3m). Calculate W, Q, U, and S.
VPSTU
WQU holds for all processes, energy conservation
quasistatic, T and P are well-defined for any intermediate state
quasistatic adiabatic isentropic non-quasistatic adiabatic
JmPa 11010101
111
11
11
1
1)(
2335
1
1
11
x
xVP
x
xVP
V
VVP
VV
VP
dVV
VPconstPVdVVPW
ii
ii
f
iii
if
ii
V
V
ii
V
V
f
i
f
i
fi
V
V
VVPdVPWf
i
J2
m10m10Pa102 23225
The non-quasistatic process results in a higher T and a greater entropy of the final state.
S = const along the isentropic
line
P
V Vi Vf
1
2
2*
Q = 0 for both
Caution: for non-quasistatic adiabatic processes, S might be non-zero!!!
An example of a non-quasistatic adiabatic process
NfkUkNf
VkNNVUS BBB lnln2
ln,, Direct approach:
adiabatic quasistatic isentropic 0Q WU VPTkNf
B 2
VV
TkNTkN
f BB
2constTV f 2/
f
f
i
i
f
V
V
T
T/2
0ln2
2ln
i
fB
i
fB V
V
fkN
f
V
VkNS
adiabatic non-quasistatic
5 5 5
ln ln2 2
1
2 10 10 10 1J/K
300 300
f f f i f iB B B B
i i i i
i i i
i i i i
i
i
V U V V U Uf fS Nk Nk Nk Nk
V U V U
P U P V P V P VV U
T T T T
P P V
T
J 250m 10Pa 102
5
2
5
233-5 iiiBi VPTkN
fUJWU 2210
V
V
K/J300
2
K300
2J
T
Q
T
US
The entropy is created because it is an irreversible, non-quasistatic compression.
P
V Vi Vf
To calculate S, we can consider any quasistatic process that would bring the gas into the final state (S is a state function). For example, along the red line that coincides with the adiabata and then shoots straight up. Let’s neglect small variations of T along this path ( U << U, so it won’t be a big mistake to assume T const):
K/J300
1
K300
1J2J(adiabata) 0
T
QS
U = Q = 1J
For any quasi-static path from 1 to 2, we must have the same S. Let’s take another path – along the isotherm and then straight up:
1
2
P
V Vi Vf
1
2
U = Q = 2J
K/J300
1
K300
10m 10Pa 10
ln )(1
2-33-5
x
x
T
VP
V
V
T
VP
V
dV
T
VPdVVP
TS
ii
i
fii
V
V
ii
V
V
f
i
f
i
isotherm:
“straight up”:
Total gain of entropy: K/J300
1K/J
300
2K/J
300
1S
The inverse process, sudden expansion of an ideal gas (2 – 3) also generates entropy (adiabatic but not quasistatic). Neither heat nor work is transferred: W = Q = 0 (we assume the whole process occurs rapidly enough so that no heat flows in through the walls).
i
fBrev V
VTkNW ln
The work done on the gas is negative, the gas does positive work on the piston in an amount equal to the heat transfer into the system
J/K300
1ln0
ii
ii
f
iB
revrevrevrev V
V
T
VP
V
VNk
T
W
T
QSWQ
P
V Vi Vf
1
2
3 Because U is unchanged, T of the ideal gas is unchanged. The final state is identical with the state that results from a reversible isothermal expansion with the gas in thermal equilibrium with a reservoir. The work done on the gas in the reversible expansion from volume Vf to Vi:
Thus, by going 1 2 3 , we will increase the gas entropy by J/K300
2321 S
Systems with a “Limited” Energy Spectrum
1
,
NV
U
ST
The definition of T in statistical mechanics is consistent with our intuitive idea of the temperature (the more energy we deliver to a system, the higher its temperature) for many, but not all systems.
“Unlimited” Energy Spectrum
T > 0
the multiplicity increase monotonically with U : U f N/2
U
S
C
T
U
U
NVU
S
T ,
1
VV T
UC
S
U
T
U C
U
Pr. 1.55
0
0
,
2
2
,
NV
NV
U
S
U
S
ideal gas in thermal equilibriumself-gravitating ideal
gas(not in thermal
equilibrium)
T > 0
0VC0VC
Pr. 3.29. Sketch a graph of the entropy of H20 as a function of T at P = const, assuming that CP is almost const at high T.
dTT
CS PP
T
C
T
S P
P
At T 0, the graph goes to 0 with zero slope. At high T, the rate of the S increase slows down (CP const). When solid melts, there is a large S at T = const, another jump – at liquid–gas phase transformation.
S
T
ice water vapor
“Limited” Energy Spectrum: two-level systems
e.g., a system of non-interacting spin-1/2 particles in external magnetic field. No “quadratic” degrees of freedom (unlike in an ideal gas, where the kinetic energies of molecules are unlimited), the energy spectrum of the particles is confined within a finite interval of E (just two allowed energy levels).
-1.0 -0.5 0.0 0.5 1.00.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
T < 0T > 0
S / N
k B
U / NB
S
U
E
the multiplicity and entropy decrease for some range of U
in this regime, the system isdescribed by a negative T
S
U
T U
2NB
01
,
VNU
S
T
Systems with T < 0 are “hotter” than the systems at any positive temperature - when such systems interact, the energy flows from a system with T < 0 to the one with T > 0 .
½ Spins in Magnetic Field
The total energy of the system:
NNBNNBMBU 2
NNN
B
N - the number of “up” spins
N - the number of “down” spins
Our plan: to arrive at the equation of state for a two-state paramagnet U=U (N,T,B) using the multiplicity as our starting point.
- the magnetic moment of an individual spin
E
E1 = - B
E2 = + B
0an arbitrary choice
of zero energy
The magnetization:
NNNNM 2
(N,N) S (N,N) = kB ln (N,N) U =U (N,T,B)
1),(
U
UNST
BN
UNN
1
2
BN
UNN
1
2Express and with and , N N N U
From Multiplicity – to S(N) and S(U)
!!
!),(
NN
NNNThe multiplicity of any macrostate
with a specified N:
NNNNNNNNNNNN
NNNNNNN
N
k
NNS
B
lnlnlnln!ln
!ln!ln!ln!!
!ln
,
Max. S at N = N (N= N/2): S=NkBln2
BN
U
BN
UN
BN
U
BN
UNNk
BN
UN
BN
UN
BN
UN
BN
UNNNkUNS
B
B
1ln12
1ln12
2ln
12
ln12
12
ln12
ln,
-1.0 -0.5 0.0 0.5 1.00.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
T < 0T > 0
S / N
k B
U / NB
ln2 0.693
1 , 12 2 2
N N N N U N UN N
U B N N N B N B
From S(U,N) – to T(U,N)
2exp exp
B B
N E EB
N k T k T
expB
N E E
N k T
Energy
E
E
1
/
/ln
2
BUN
BUN
k
BT
B
BNUBNU
B
k
BBN
U
BBBN
U
Bk
BN
U
BN
UN
BN
U
BN
UNN
Uk
U
S
T
BB
BBN
1
1ln
22
11ln
2
1
2
11ln
2
1
1ln12
1ln12
2ln1
,
The same in terms of N and N : NNNNNNNNk
S
B
lnlnln
1 12 1
ln ln2 211
2
B B
N U UN N NN B k kN B
UN T B N B NN UN BN B
Boltzmann factor!
The Temperature of a Two-State Paramagnet
0
E1
E2
E1
E2
Tk
EE
N
N
B
12exp
Energy
E1
E2
T
Energy
E1
E2
Tk
EE
N
N
B
12exp
- N B
N B
U
11
/
/ln
2ln
2
BUN
BUN
k
B
N
N
k
BT
BB
E1
E2
T = + and T = - are (physically) identical – they correspond to the same values of thermodynamic parameters.
Systems with neg. T are “warmer” than the systems with pos. T: in a thermal contact, the energy will flow from the system with neg. T to the systems with pos. T.
The Temperature of a Spin GasThe system of spins in an external magnetic field. The internal energy in this model is entirely potential, in contrast to the ideal gas model, where the energy is entirely kinetic.
Tk
E
iB
i
en
B
E6
E5
E4
E3
E2
E1
B
At fixed T, the number of spins ni of energy Ei decreases exponentially as energy increases.
spin 5/2(six levels)
- lnni
Ei
- lnni
Ei
- lnni
Ei
T =
- lnni
Ei
T = 0
For a two-state system, one can always introduce T - one can always fit an exponential to two points. For a multi-state system with random population, the system is out of equilibrium, and we cannot introduce T.
Boltzmann distribution
no T
Tk
En
B
ii ln
the slope T
The Energy of a Two-State Paramagnet
Tk
BBN
e
eBNU
BTkB
TkB
B
B
tanh1
1/2
/2
BUN
BUN
B
k
TB
/
/ln
2
1
U approaches the lower limit (-NB) as T decreases or, alternatively, B increases (the effective “gap” gets bigger).
U
B/kBT
N B
- N B
U
T
- N B
1
The equation of state of a two-state paramagnet:
(N,N) S (N,N) = kB ln (N,N) U =U (N,T,B)1
),(
U
UNST
T < 0T > 0
S(B/T) for a Two-State Paramagnet
NNNNNNNN
NNN
N
k
NNS
B
lnlnln!!
!ln
,
S
U
0 N B - N B Problem 3.23 Express the entropy of a two-state
paramagnet as a function of B/T .
NNBU 2
Tk
BBNU
B
tanh
xTk
B
B
2
tanh1tanh2
xNNxBNNNB
2
tanh1ln
2
tanh1
2
tanh1ln
2
tanh1
2
tanh1ln
2
tanh1
2
tanh1ln
2
tanh1ln
,
xxxx
xN
xxN
xN
kN
xNS
B
x
e
x
xxx
x
e
x
xxx
xx
coshcosh
sinhcoshtanh1
coshcosh
sinhcoshtanh1
?)(TNN
NkBln2
xxxxx
ee
x
eex
xxx
exx
x
e
x
e
x
e
x
e
x
e
kN
xNS
xxxx
xx
xxxx
B
tanhcosh2lncosh2lncosh2cosh2
cosh2lncosh2
cosh2lncosh2
cosh2ln
cosh2cosh2ln
cosh2
,
Tk
B
Tk
B
Tk
BkN
Tk
BNS
BBBB
B
tanhcosh2ln,
B/T 0, S = NkB ln2
B/T , S = 0
S(B/T) for a Two-State Paramagnet (cont.)
0 2 40
0.5
10.692
0
fi
50.02 1
xi
S/NkB
ln2 0.693
kBT/B = x-1
high-T (low-B) limit
low-T(high-B)
limit0 2 4 6
0.0
0.2
0.4
0.6
0.8
1.0S / NkB
x = B / kBT
ln2 0.693
0 2 4 60.0
0.2
0.4
0.6
0.8
1.0S / NkB
x = B / kBT
Low-T limit
1221
1
11ln
22ln
tanhcosh2ln,
222
2
22
xekNexexkN
e
exexkN
ee
eex
eekN
Tk
B
Tk
B
Tk
BkN
Tk
BNS
xB
xxB
x
xx
Bxx
xxxx
B
BBBB
B
Which x can be considered large (small)?
1Tk
Bx
B
ln2 0.693
2
2
. ., 2, 0.018 1.8%
2 1 0.1
x
x
e g x e
e x
0 1 2 3 4 50.0
0.1
0.2
0.3
0.4
0.5CB / Nk
B
kBT / B
The Heat Capacity of a Paramagnet
TkB
TkBkN
T
UC
B
BB
BNB /cosh
/2
2
,
The low -T behavior: the heat capacity is small because when kBT << 2 B, the thermal fluctuations which flip spins are rare and it is hard for the system to absorb thermal energy (the degrees of freedom are frozen out). This behavior is universal for systems with energy quantization.
The high -T behavior: N ~ N and again, it is hard for the system to absorb thermal energy. This behavior is not universal, it occurs if the system’s energy spectrum occupies a finite interval of energies.
kBT
2 B
kBT2 B
compare withEinstein solid
E per particle
C
equipartition theorem(works for quadratic degrees
of freedom only!)
NkB/2
2cosh
xx eex
The Magnetization, Curie’s Law
B/kBT
N
M
1
xxx tanh1
Tk
BNM
B
2
The high-T behavior for all paramagnets (Curie’s Law)
Tk
BN
B
UNNM
B
tanh B/kBT
N
- N
The magnetization:
Negative T in a nuclear spin system NMR MRI
Fist observation – E. Purcell and R. Pound (1951)
Pacific Northwest National Laboratory
By doing some tricks, sometimes it is possible to create a metastable non-equilibrium state with the population of the top (excited) level greater than that for the bottom (ground) level - population inversion. Note that one cannot produce a population inversion by just increasing the system’s temperature. The state of population inversion in a two-level system can be characterized with negative temperatures - as more energy is added to the system, and S actually decrease.
An animated gif of MRI images of a human head.- Dwayne Reed
Metastable Systems without Temperature (Lasers)
For a system with more than two energy levels, for an arbitrary population of the levels we cannot introduce T at all - that's because you can't curve-fit an exponential to three arbitrary values of #, e.g. if occ. # = f (E) is not monotonic (population inversion). The latter case – an optically active medium in lasers.
E1
E2
E3
E4
Population inversionbetween E2 and E1
Tk
E
B
exp
Sometimes, different temperatures can be introduced for different parts of the spectrum.
Problem
A two-state paramagnet consists of 1x1022 spin-1/2 electrons. The component of the electron’s magnetic moment along B is B = 9.3x10-24 J/T. The paramagnet is placed in an external magnetic field B = 1T which points up.
• Using Boltzmann distribution, calculate the temperature at which N= N/e.• Calculate the entropy of the paramagnet at this temperature.• What is the maximum entropy possible for the paramagnet? Explain your reasoning.
(a)
K35.1J/K 101.38
T 1J/T 103.92223
24
B
B
k
BT
BEETk
EE
N
NB
B
2exp 1212
kBT
E1= - BB
E2 = + BB
B
spin 1/2(two levels)
B
- B
J/T 103.92
24e
B m
e1212
Tk
B
Tk
EE
B
B
B
N
N
Problem (cont.)
J/K012.009.0J/K1038.1101
tanhcosh2ln,
2322
Tk
B
Tk
B
Tk
BkN
Tk
BNS
B
B
B
B
B
BB
B
B
xx
xxxxxx
ee
eex
eex
eex
tanh2
sinh2
cosh
If your calculator cannot handle cosh’s and sinh’s:
0 2 40
0.5
1
fi
1
xi
kBT/ B
S/NkB
0.09
J/K1.0693.0J/K1038.1101, 2322
Tk
BNS
B
B
(b)
323
24
105.4K 300J/K 101.38
T 2J/T 103.9
Tk
B
B
B
the maximum entropy corresponds to the limit of T (N=N): S/NkB ln2
For example, at T=300K:
E1
E2
kBT
0 2 40
0.5
10.692
0
fi
50.02 1
xi
kBT/ B
S/NkB
T S/NkB ln2
T 0S/NkB 0
ln2
Problem (cont.)