Exam 2 Review

Practice Exam 2 Problems

Chapters 4.1 - 4.2

The Carnot Cycle represents the maximal efficiency of a heat engine operating between two reservoirs at fixed temperature:

η_{C} = 1 - \frac{T_{l}}{T_{h}}

where $T_{l}$ is the low temperature reservoir, and $T_{h}$ is the high temperature reservoir. The
Carnot efficiency is realized with the cycle is reversible.

Two very useful representations of a cycle: the P-V diagram, and the T-S plane. The area under the curve of a process in the P-V diagram is the work done by the gas. The area under the curve during a T-S process is the total heat absorbed by the system.

Chapter 5.1 - 5.2

In Lecture 13, I discussed extensive and intensive variables. An especially interesting relation that comes from extensivity is the Euler relation:

\begin{matrix} (1) & U = T S - P V + μ N \end{matrix}

These sections also extensively discussed free energies, and from them derived some useful partial derivative relations. The starting point is always the thermodynamic identity:

d U = T d S - P d V + μ d N

A relation such as this implies the internal energy $U$ has natural variables $S, V, N$ , i.e. $U (S, V, N)$ It also implies that there is a way to transform variables between these extensive variables, and their intensive counterparts. The following are so-called "conjugate" variables, in this sense:

Entropy $S$ and Temperature $T$
Volume $V$ and Pressure $P$
Number $N$ and chemical potential $μ$

There are many possible ways to change variables, but there are a few ways in particular that are especially useful and relevant in practice:

Enthalpy:

Definition: $H = U + P V$

Exchanges $V$ for $P$ , so the natural variables of H are $S, P, N$ .

Using the thermodynamic identity, one can show: $d H = T d S + V d P + μ d N$

Helmholtz Free Energy: $F = U - T S$

Definition: $F (T, V, N) = U - T S$

Exchanges $S$ for $T$ , so the natural variables are $T, V, N$

Differential identity is: $d F = - P d V + μ d N - S d T$

This implies the partial derivative relations

{(\frac{\partial F}{\partial T})}_{V, N} = - S, {(\frac{\partial F}{\partial V})}_{T, N} = - P

Gibbs Free Energy:

Definition: $G = U - T S + P V$

Exchanges $S$ for $T$ , and exchanges $V$ for $P$ .

Differential identity is: $d G = - S d T + V d P + μ d N$

A few partial derivative relations that are useful come directly from this:

S = - {(\frac{\partial G}{\partial T})}_{P, N}, V = {(\frac{\partial G}{\partial P})}_{T, N}

Notice that if we use the Euler relation Eq. (1), it implies the Gibbs free energy is directly proportional to the number of particles:

G = μ N

Grand Potential:

Definition: $Φ = U - T S - μ N$

We now consider replacing $S \to T$ , $N \to μ$ . This produces the Grand potential:

$Φ (T, V, μ) = U - T S - μ N$

which satisfies

$d Φ = - S d T - N d μ - P d V$

Notice again that using the Euler relation Eq. (1) implies: $Φ = - P V$

Gibbs-Duhem Equation

There is an interesting relation which follows when we replace all the extensive variables with their conjugate intensive variables. Following the logic above, we should define a new thermodynamic potential

$ϕ (T, P, μ) = U - T S + P V - μ N$

However, by the Euler relation Eq. (1), $ϕ = 0$ , and furthermore the differentials must be exactly zero. However, this implies a nontrivial relation connecting variations between intensive variables, called the Gibbs-Duhem equation:

- S d T + V d P - N d μ = 0

Chapter 6.1 -6.2, 6.5

The Boltzmann distribution describes the so-called canonical ensemble in which the temperature is fixed. The probability for a particular microstate in this ensemble is related to the energy of that microstate $E (s)$ , and is given by

P (s) = \frac{1}{Z} e^{- E (s) / k T}

where the partition function is

Z = \sum_{s} e^{- E (s) / k T}

Is the sum over all configurations. From this probability distribution, you can compute averages, such as the internal energy:

U = ⟨ E ⟩ = \frac{1}{Z} \sum_{s} E (s) e^{- E (s) / k T} = k T^{2} \partial_{T} \ln Z

In the last equality I showed a powerful relation, which we can use to connect the partition to the Helmholtz Free energy (from Sec. 6.5):

F = - k T \ln Z