Recap of last lecture
The third law states that the entropy goes to zero as temperature goes to zero. This holds on a macroscopic level. The small amount of entropy calculated will not make a difference.
<p>
</p>
The Hemholtz function can be written as a simple expression of the partition function. The constants of <math>V, N,</math> and <math>T</math> are the same as is found in the canonical ensemble. The differential was written, and thermodynamic variables can be found from the equation of state.
<center>
<br>
<math>F(V, N, T)</math>
<br>
<math>F = -k_B T \ln Q</math>
<br>
<math>dF = -SdT - p dV + \mu dN</math>
<br>
</center>
There is a pattern in finding thermodynamic variables for all boundary conditions. The partition function plays a central role and is a function of the energy levels of the system. A fixed characteristic potential can be written in a simple relation with the partition function for the same boundary conditions. The partition function is found, and the characteristic potential is calculated. The same derivation structure holds with other boundary conditions. Through relations and equations of state, the thermodynamic state of the system is determined.
Summary
Calculation of entropy
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</p>
Hemholtz free energy is a function of <math>V, N,</math> and <math>T</math>
- Write differential
- Find thermodynamic variables from equation of state
<p></p>
Pattern in finding thermodynamic variables for all boundary conditions.
- Write relation between characteristic potential and partition function
- Find partition function
- Through relations and equations of state, determine thermodynamic state of system
Grand Canonical Ensemble
A difference between this ensemble and others is that the number of particles is not fixed. The terms <math>T, V,</math> and <math>\mu</math> are constant. The number of particles, <math>N</math>, is not constant. It can fluctuate, and there is a constant cost of each extra particle. In the canonical ensemble, there was an average energy and pressure. This can still fluctuate in the grand canonical ensemble. When is this ensemble useful?
Examples from research
Consider the adsorption of some molecule, such as oxygen on the surface of platinum. The number of oxygen molecules on the surface is not fixed. The system is under constant gas chemical potential.
<center>
Unable to render embedded object: File (Adsorption_of_oxygen_on_platinum.PNG) not found.
<br>
<math>\mu = \mu_0 + kT\ln P_{\mbox{gas}}</math>
<br>
</center>
Consider a battery. The voltage depends on the material and is set by materials in the anode and cathode. Lithium moves from one place to another, and choice of materials control the chemical potential, <math>\mu_{\mbox{Li}}</math>. The grand canonical ensemble is applied to systems that are open to species.
<center>
Unable to render embedded object: File (Battery_schematic.PNG) not found.
</center>
Equation for Grand Canonical Ensemble
Identify the same structure as in the Hemholtz. Identify a characteristic potential. Start with the energy as a function of <math>S, V,</math> and <math>N</math>. Natural variables that are wanted are <math>T, V,</math> and <math>\mu</math>. Use the Legendre transform to remove dependence of <math>S</math> and <math>N</math>. Differentiate to find relations.
<center>
<br>
<math>\Phi = E - TS - \mu N</math>
<br>
<math>\Phi = -pV </math>
<br>
<math>\Phi = \mbox
</math>
<br>
<math>d \Phi = -SdT - pdV -Nd \mu</math>
<br>
<math>S = - \left ( \frac
\right )_
</math>
<br>
<math>p = - \left ( \frac
\right )_
</math>
<br>
<math>N = - \left ( \frac
\right )_
</math>
<br>
</center>
Determine potential from partition function
Everything can be calculated from the equation of state. Get the potential, <math>\Phi</math>, in terms of the partition function for any boundary condition. Get the potential from the partition function appropriate for the correct boundary conditions, in this case, <math>T, V,</math> and <math>\mu</math> fixed.
<p>
</p>
This is the same procedure as in the canonical ensemble. Two steps are determining the probability to be in a state and finding a relation between mechanical quantities and compare with thermo. Probability, <math>P_
</math> is defined by <math>N</math> and <math>\nu</math>. The two terms are interrelated. Calculating the probability for a system to be in a state allows the calculation of mechanical properties and averages.
<center>
<br>
<math>\overline
= \sum_
P_
E_
</math>
<br>
<math>\overline
= \sum_
P_
p_
</math>
<br>
<math>\overline
= \sum_
P_
N </math>
<br>
</center>
The derivation is very similar but a little more tricky. See McQuarry (p. 52 - 54). The sum over <math>N</math> and <math>\nu</math> is not an easy sum. There is a relation between <math>N</math> and <math>\nu</math>. They are generally coupled. There is a sum over states, and the state is determined by energy.
<center>
<br>
<math>P_
= \frac{e^{- \beta (E_
- \mu N})}{\sum_N \sum_
e^{-\beta (E_
- \mu N)}</math>
<br>
<math>\theta = \sum_N \sum_
e^{-\beta (E_
- \mu N)</math>
<br>
</center>
Summary
Grand-Canonical Ensemble
- <math>T, V,</math> and <math>\mu</math> are constant
- The number of particles, <math>N</math>, is not constant
- Energy and pressure can fluctuate
- Examples
- Oxygen absorbing on platinum
- Battery
- Equation
- Identify a characteristic potential
- Start with energy as a function of <math>S, V,</math> and <math>N</math>
- Use the Legendre transform to remove dependence of <math>S</math> and <math>N</math>
- Differentiate to find relations.
- Determine potential from partition function
Relation between potential and partition function
There is always a relation between potential and partition function. It is very easy to figure out the potential, and there is a general structure that can be applied to extract probability with any set of thermodynamic boundary conditions. It is very easy to figure out potential, and the probability to be in any state can be calculated. The relation below is true with any boundary condition, and a goal is to find the box, <math>\box</math>.
<center>
<br>
<math>P_
=\frac{e^{\box \nu}}{\sum_k e^{\box k}}</math>
<br>
</center>
Below is a relation between the partition function and any characteristic potential, <math>\Omega</math> for a specified set of boundary conditions:
<center>
<br>
<math>- \beta \Omega = \ln Z</math>
<br>
</center>
Sum over all states <math>k</math> to find the partition function, and find the box, <math>\box</math> by using a Legendre transform of energy.
<center>
<br>
<math>Z = \sum_k e^
</math>
<br>
</center>
Start with the canonical ensemble
In the canonical ensemble, the terms <math>T, V,</math> and <math>N</math> are constant. The energy, <math>E</math>, is not constant but its conjugate, <math>\beta</math>, is.
<center>
<br>
<math>\beta = \frac
</math>
<br>
</center>
Take the differential of energy, <math>E</math> and divide by <math>kT</math>. Apply the Euler theorem and Legendre Transform to remove <math>E</math>
<center>
<br>
<math>dE = TdS - pdV + \mu dN</math>
<br>
<math>\frac
= \beta dE + \beta dV - \beta \mu dN</math>
<br>
<math>S(E, V, N)</math>
<br>
<math>\frac
= \beta E + \beta p V - \beta \mu N</math>
<br>
<math>\frac
[\overline{\underline{-\beta E}}] = \beta p V - \beta \mu N</math>
<br>
</center>
The term <math>-\beta E</math> is what is in the box, <math>\box</math>, in the expressions above. The probability, <math>P_\nu</math>, and partition function are written below. There is a simple relation between the Hemholtz free energy and the partition function.
<center>
<br>
<math>P_
= \frac{e^{- \beta P_
}}{\sum_
e^{- \beta P_{\nu}}</math>
<br>
<math>Z = Q</math>
<br>
<math>Q = \sum_
e^{-\beta E_{\nu}}</math>
<br>
<math>- \beta F = \ln Q</math>
<br>
</center>
Grand Canonical Ensemble
Do the Legendre transform for the grand canonical ensemble and see if get the right box. The terms <math>T, V,</math> and <math>\mu</math> are constant. The energy, <math>E</math>, is not constant, but its conjugate <math>\beta</math> is. The number of particles, <math>N</math>, is also not constant, but <math>\mu</math> is. Start with the differential for energy, use the Legendre transform, and Legendre transform to arrive at variables.
<center>
<br>
<math>dE = TdS - pdV + \mu N</math>
<br>
<math>\frac
= \beta dE + \beta p dV - \beta \mu dN</math>
<br>
<math>S(E, V, N)</math>
<br>
<math>\frac
= \beta E + \beta p dV - \beta \mu N</math>
<br>
<math>\frac
[\underline{\overline{- \beta E + \beta \mu N}}] = \beta p dV </math>
<br>
</center>
The terms in the box are used in the probability. There is a sum over states, and the number of particles is a function of state. There is a simple relation between the partition function and the characteristic potential. Legendre transform the entropy to find the probability. Sum over all allowed states, which is allowed to fluctuate.
<center>
<br>
<math>P_
= \frac{e^{- \beta E_
- \beta \mu N}}{\sum_
e ^{ + \beta E_
- \beta \mu N_{\nu}}</math>
<br>
<br>
<math>Z = \theta </math>
<br>
<math>Z = \sum_
e^{-\beta E_
+ \beta \mu N_
</math>
<br>
</center>
Microcanonical Ensemble
In the microcanonical ensemble, <math>E, V,</math> and <math>N</math> are constant. Write the differential and find that it is already in a final form. The partition funtion is found to be equivalent to a degeneracy of particular energy, and the entropy can be expressed in terms of the degeneracy.
<center>
<br>
<math>S(E, V, N)</math>
<br>
<math>\frac
= \beta dE + \beta dV - \beta \mu dN</math>
<br>
<math>\frac
[\underline{\overline{+ 0}}] = \frac
</math>
<br>
<math>P_
= \frac
{\sum_
e^0}</math>
<br>
<math>P_
= \frac
</math>
<br>
<math>Z = \sum_
e^0</math>
<br>
<math>Z = \Omega</math>
<br>
<math>\frac
= \ln \Omega</math>
<br>
<math>S = k_B \ln \Omega</math>
</center>
The final expression holds for any set of boundary conditions. There is a certain degeneracy. Sum over all the states, and the energy expression is from the Schrodinger equation. In the grand canonical ensemble, there is a summation over all the states that allow particles and corressponding energies.
Isothermal, Isobaric Ensemble
In the isothermal, isobaric example, the terms <math>T, P,</math> and <math>N</math> are constant. Try this at home. This general technique is a favorite to be asked on exams and is a good technique to know.
Summary
Second Law in an Isolated System
Consider an isolated system. The number of particles is fixed, and the constant terms are <math>N, E,</math> and <math>V</math>. There is no heat or work exchanged with the environment. An expression of entropy is below. The second law states that entropy increases until reaching equilibrium. The second law states that there is an increase with entropy associted with a spontaneous process from state one to two.
<center>
<br>
<math>S = k_B \ln Q</math>
<br>
<math>\Delta S > 0</math>
<br>
</center>
Example of a gas in vacuum
Analyze the statement about the increase in entropy from a statistical mechanics perspective. Consider a gas in a container. with a partition dividing the container. There is a vacuum in one half of the container and gas in the other.
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Unable to render embedded object: File (Box_filled_with_gas.PNG) not found.
</center>
Remove the constraint and consider the spontaneous process of the gas filling the second half of the chamber. There are <math>N</math> non-interacting particles. Consider the degeneracy, <math>\Omega (E)</math>.
<center>
<br>
<math>\Omega (E) \prop V^N</math>
<br>
<math>\Omega \prop \left (\frac
\right )^N</math>
<br>
<math>\Omega \prop \left (V \right )^N</math>
<br>
</center>
Calculate the increase in entropy. The result is greater than zero. This result is found without many assumptions. It is a microcanonical ensemble with energy fixed. The system is accessing more states. At equilibrium, there are more states available and used. The system spontaneously evolves to the thermodynamic macroscopic state consistent with the maximum number of states available. The process involves non-equilibrium thermodynamics. Consider the end state and beginning.
<center>
<br>
<math>\Delta S = k_B \ln \Omega_2 - k_B \ln \Omega_1</math>
<br>
<math>\Delta S = k_B \ln \left (\frac
\right )</math>
<br>
<math>N k_B \ln 2</math>
</center>
Entropy is considered to increase always. Imaging moving from state two to one. In an isolated system, the change in entropy is <math>-N k_B \ln 2</math>. There is a change of state with negative entropy in an isolated system. Thermodynamically, this cannot happen. How can these two pictures be resolved? Consider the probability, <math>P</math> that over measurement time all the gas particles spontaneously are on one half of the chamber. The term <math>\Omega</math> represents all states, and the probability is found to be very small.
<center>
<br>
<math>P = \frac
</math>
<br>
<math>P = \frac
</math>
<br>
<math>P = \frac
{2^{10^
}}</math>
<br>
</center>
There is a non-zero probability of occurrence, but it is very small. There are so many more states associated with filling the whole container. Consider a royal straight flush versus a bad hand. At high temperature, there is a chance of finding an ordered state, but there is a very small number of ordered states. At high temperature, disordered states are thermally accessible. There are many more states in a liquid. The second and third law of thermodynamics can be considered macroscopic laws. Consider a very small system. It is possible to see negative entropy in isolated systems. Thermodynamic laws are not strict in a mathematical sense.
Summary
Catalysis (isolated systems)
Consider a mixture of hydrogen and oxygen. There are <math>\Omega_1</math> states available. Add a catalyst to form water. After the reaction, there are <math>\Omega_2</math> microstates. There are more states available in the second state, <math>\Omega_2 > \Omega_1</math>, and there is an increase in entropy, <math>\Delta S > 0</math>. There are more states available and the system flows to them. The system sees more states with the addition of a catalyst. There is a removal of a constraint that making more quantum states available.
Summary
A spontaneous processs in an isolated system is the same as removal of a constraint which makes more quantum states of a system available. The system is going to "flow" to new states.
Fluctuations
Canonical Ensemble
In the canonical ensemble, <math>N, V,</math> and <math> T </math> are fixed. The energy and pressure can fluctuate from one microstate to the next.
<p>
</p>
There are two non-mechanical variables. They cannot be determined by one microstate. Temperature does not fluctuate. It is defined as an average. The entropy is non-mechanical. It can't be defined from just one microstate.
Grand-Canonical Ensemble
These variables fluctuate. The fluctuations are not very large, which means there is good meaning of the average. Consider the Gaussian distribution. The average of a variable with larger spread is less meaningful.