Advanced Thermodynamics
Introduction
This section will delve deeper into thermodynamics. Multivariable calculus will be used to expand upon topics covered earlier as well as introduce new topics.
The
Fundamental Thermodynamic Relation will be covered as an extension of the first law.
Chemical Potential
Previously, we did not consider chemical reactions in our thermodynamic calculations.
The chemical potential, denoted by \(\mu\), is a fundamental concept in thermodynamics and statistical mechanics.
It represents the change in the system's free energy as the number of particles changes, all other variables being held constant.
In a multicomponent system, the chemical potential of each component is crucial for understanding phase equilibrium, reaction equilibrium, and various other phenomena.
Definition
The chemical potential of a species i in a system is defined as the partial derivative of the system's energy, enthalpy, or free energy with respect to the number of moles (\(n_i)\) of that species, holding everything else constant.
$$\mu_i=(\frac{\partial E}{\partial n_i})_{S,V,n_{j\neq i}}$$
In terms of the chemical potential, the differential of the energy is given by:
$$dE=\sum_i \mu_i dn_i$$
Physical Significance
The chemical potential is a measure of the 'chemical drive' or tendency of species to enter or leave a phase or to react in a chemical reaction. It plays a vital role in:
- Phase Equilibria: Dictating the distribution of species between phases at equilibrium.
- Reaction Equilibria: Determining the direction and extent of reactions.
Phase Change
At phase equilibrium, the chemical potential of each component is equal in all phases:
$$\mu_i^\alpha=\mu_i^\beta$$
where alpha and beta denote different phases.
Chemical Potential and Reaction Equilibrium
At reaction equilibrium, the total chemical potential of reactants equals that of products, leading to the condition:
$$\sum_i\nu_i\mu_i=0$$
where \(\nu_i\) is the stoichiometric coefficient for specie i.
Laws of Thermodynamics Revisited
Total derivatives
While the derivation is outside the scope of this textbook, the first law can be expressed as the following total derivative:
$$dE=(\frac{\partial E}{\partial S})_{V,N}dS+(\frac{\partial E}{\partial V})_{S,N}dV+\sum(\frac{\partial E}{\partial N_i})_{S,V}dN_i$$
$$=TdS-p_{ext}dV+\sum\mu_idN_i$$
where T is temperature, \(p_{ext}\) is external pressure, V is volume, \(\mu_i\) is the chemical potential for a chemical species i, and \(N_i\) is the number of species i
Temperature and Entropy
From the Fundamental Thermodynamic Relation, we have:
$$(\frac{\partial S}{\partial E})_{N,V}=\frac{1}{T}$$
If \(T_{A}>T_{B}\), then \(\frac{1}{T_{A}}<\frac{1}{T_{B}}\)
$$(\frac{\partial S_{A}}{\partial E_{A}})_{N,V}<(\frac{\partial S_{B}}{\partial E_{B}})_{N,V}$$
by substituting in the thermodynamic definition for temperature, we see that the lower temperature system, B, would have a greater change in entropy for the same change in energy. This means that there would be a net entropy increase if energy were to flow from A to B.
So when would this heat flow stop? When entropy is maximized/equilibrium. The termination condition would be when there is no net increase in entropy due to heat flow or in mathematical terms:
$$(\frac{\partial S_{A}}{\partial E_{A}})_{N,V}=(\frac{\partial S_{B}}{\partial E_{B}})_{N,V}$$
which is equivalent to saying
$$\frac{1}{T_{A}}=\frac{1}{T_{B}}$$
or alternatively
$$T_{A}=T_{B}$$
In plain English to summarize: For the same amount of energy added to a system, the one with the lower temperature would increase in entropy more.
Conversely, for the same amount of energy taken out of a system, the one with a higher entropy would decrease in energy less. Therefore, if you take energy from the higher temperature system and put it into the lower temperature system, there will be a net increase in entropy.
This will change the temperatures of the systems, bringing them closer together. This exchange in energy will stop when the temperatures are the same, or when there is no longer a possible increase in energy from taking energy from one system and putting it in another.
Enthalpy and the Thermodynamic Definition of Entropy
Deriving the Thermodynamic Definition of Entropy
By the differential form of the first law of Thermodynamics, we have
$$dE=(\frac{\partial E}{\partial S})_{V,N}dS+(\frac{\partial E}{\partial V})_{S,N}dV+\sum(\frac{\partial E}{\partial N_i})dN_i$$
$$=TdS-p_{ext}dV+\sum\mu_idN_i$$
*Negative sign for pressure since our dV is pointing in opposite direction of \(p_{ext}\).
Let us define a quantity called Enthalpy, denoted by H, as follows:
$$H\equiv E+PV$$
Through the
product rule for differentiation:
$$dH=dE+d(PV)=dE+(pdV+Vdp+dVdp)$$
Substituting in the total differential form of the first law from above for dU:
$$=(TdS-pdV+\sum\mu_idN_i)+pdV+Vdp+dVdp$$
The dVdp term goes to 0 in the limit. The pdV term from work cancels out with one of the terms from the product rule for d(PV).
$$=TdS+Vdp\sum\mu_idN_i$$
If we are working in a constant temperature environment and no chemical reactions are taking place, we can let dp=0 and \(dN_i=0\forall i\), yielding the formula
$$dH=TdS=\delta q$$
$$dS=\frac{\delta q_{rev}}{T}$$
This is the so-called "Thermodynamic definition of Entropy" which holds for a reversible process at constant pressure
Irreversible Processes
However, what about for irreversible processes? The first law still holds such that
$$U=q+w$$
but
\(q_{rev}\leq q_{irr}\)
and
\(w_{rev}\geq w_{irr}\).
A more general statement which holds for irreversible processes is
$$dS\geq \frac{\delta Q}{T}$$
with equality when the process is reversible. This is known as the Clausius Inequality, which is another way of formulating the 2nd law. In the basic Thermodynamics chapter, we saw such a process in adiabatic free expansion: where gas was expanding into a vacuum.
An Aside on Legendre Transforms
In the first law, we had
$$dU=TdS-pdV+\sum\mu_idN_i$$
and when we defined Enthalpy as \(H\equiv U+PV\), we ended up getting
$$dH=TdS+Vdp+\sum\mu_idN_i$$
By adding PV, we were able to switch the positions of the Volume and Pressure variables (along with a sign change).
By doing this, we can now have a quantity that is more convenient to work with when we assume different conditions (e.g. constant pressure instead of constant volume).
This is from which we got heat being the change in energy at a fixed volume and the change in enthalpy at fixed pressure equivalencies (set dV=0 or dP=0 in the above equations).
This is a theme that will recur for the Helmholtz and Gibbs free energies where instead we will switch temperature and entropy, as dT=0 is more feasible than dS=0.
These paired variables are known as "conjugate variables," and the process of switching the differential between the extensive and intensive variable is known as a Legendre transformation.
Helmholtz Free Energy and Gibbs Free Energy
Let us define a quantity Helmholtz Free Energy, denoted by A (The letter A was chosen for the German word for work as H was already being used for Enthalpy; some books use F for free energy):
$$A\equiv U-TS$$
Doing a similar trick as we did for Enthalpy - taking the derivative of both sides and using the linearity of the derivative operator:
$$dA=dU-d(TS)$$
Substituting in for dU using the first law; product rule for d(TS)
$$dA=(TdS-pdV+\sum\mu_idN_i)-(TdS+SdT+dSdT)$$
Canceling out TdS terms and dSdT goes to 0
$$dA=-pdV-SdT+\sum\mu_idN_i$$
If we are working in a constant temperature system (dT=0) and constant Volume, then we find that the change in Helmholtz free energy is the amount of chemical work being done.
It is also possible to do what we did for Enthalpy and Helmholtz for a single quantity.
Let us define a quantity Gibbs Free Energy, denoted by G:
$$G\equiv U+PV-TS=A+PV=H-TS$$
In general chemistry, Gibbs free energy is often introduced as
$$G=H-TS$$
so the derivation will go from that point.
$$dG=dH-d(TS)$$
Substituting in our above result for dH; product rule for d(TS)
$$dG=(TdS+Vdp+\sum\mu_idN_i)-(TdS+SdT+dSdT)$$
Cancelling out TdS terms; dSdT goes to 0
$$dG=Vdp-SdT+\sum\mu_idN_i$$
If we are working in a constant temperature system and constant pressure, then the change in Gibbs free energy is the amount of chemical work.
Maxwell Relations
Under mild conditions, the order of partial differentiation does not matter (Schwarz's Theorem).
We can apply this principle to the differential form of the first law to get relationships between state variables.
Restating the First Law
$$dE=(\frac{\partial E}{\partial S})_{V,N}dS+(\frac{\partial E}{\partial V})_{S,N}dV+\sum(\frac{\partial E}{\partial N_i})dN_i$$
$$=TdS-p_{ext}dV+\sum\mu_idN_i$$
4 Main Maxwell Relations
$$(\frac{\partial T}{\partial V})_S=-(\frac{\partial P}{\partial S})_V$$
We can further apply this technique to Enthalpy, and free energy to get more relationships.
$$(\frac{\partial T}{\partial P})_S=(\frac{\partial V}{\partial S})_P$$
$$(\frac{\partial S}{\partial P})_T=(\frac{\partial P}{\partial T})_V$$
$$-(\frac{\partial S}{\partial P})_T=(\frac{\partial V}{\partial T})_P$$