Lecture 13

10/16

Extensive and Intensive Properties

Intensive: state variable that is independent of system size. These are the variables that determine equilibrium between two bodies:

Temperature T - two systems share a temperature when they are in thermal equilibrium, regardless of their size and composition
Pressure P - two systems are in mechanical equilibrium (forces are balanced) when they have same pressure.
Chemical Potential $μ$ - determines diffusive equilibrium

Extensive: state variable that grows with system size, keeping intensive variables fixed. Examples include:

Internal Energy U
Entropy S
Volume V
Number of Particles N

Natural Variables

Just recalling the thermodynamic identity:

d U = T d S - P d V + μ d N

This relation implies that the natural coordinates for the energy are $S, V, N$ , i.e. changes in internal energy $d U$ are related to changes in entropy, volume, and number. All of the arguments are extensive. So we may write $U = U (S, V, N)$ , i.e. the internal energy as a function of $S$ , $V$ and $N$ .

Euler Relation

A particularly interesting identity comes from considering how extensive properties change with system size.

Let us consider changing $S \to λ S$ , $V \to λ V$ and $N \to λ N$ , $U \to λ U$ . As we observed from the thermodynamic, the internal energy is a natural function of these extensive variables, so it depends on $S$ , $V$ , and $N$ . However, since $U$ is also extensive, as a result of this transformation of all extensive variables, $U$ will also transform like all of the other extensive variables, $U \to λ U$ . Writing this as an equation, we get:

$U (λ S, λ V, λ N) = λ U (S, V, N)$

Differentiating both sides then setting $λ = 1$ , we get

U = {(\frac{\partial U}{\partial S})}_{V, N} S + {(\frac{\partial U}{\partial V})}_{S, N} V + {(\frac{\partial U}{\partial N})}_{S, V} N

From the thermodynamic identity, we get the partial derivative relations

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

Therefore, we arrive at the very handy and remarkable Euler relation for Thermodynamics:

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

Thermodynamic Potentials

Legendre Transformation

It is sometimes not convenient to work with purely extensive dependent variables. $U$ depends naturally on extensive variables. This motivates the question: is there a thermodynamic state function that depends naturally on different variables? How do we even change variables?

This is what the Legendre transformation is for.

Take a function $f (x)$ that depends on $x$ . It's differential is given by

d f = (\frac{\partial f}{\partial x}) d x

We can define $p = \partial_{x} f$ , and define a new function $g = x p - f$ . Then differentials of $g$ are

$d g = x d p + p d x - p d x = x d p$
So $d g$ is naturally a variable of $p$ .

(N.B. This comes up in classical mechanics, when we want to go from lagrangian to hamiltonian: $H (x, p) = \dot{x} p - L (x, \dot{x})$ , with $p = \partial_{\dot{x}} L$ )

So it is possible to change variables by redefining the function. But we cannot do it willy-nilly. Each extensive variables as a natural partner (conjugate) intensive variable:

$S \to T$
$V \to P$
$N \to μ$

So we can take a function of $S$ and convert it to a function of $T$ . Similarly with $V$ and $P$ , and $N$ and $μ$ .

Energy

We start with the energy, and the conventional thermodynamic identity

$U = U (S, V, N)$ , $d U = T d S - P d V + μ d N$

I will use the notation $Φ (S, V, N)$

Enthalpy: $H = U + P V$

If we replace volume with pressure, we get the enthalpy:

$H (S, P, N) = U + P V$

The differential of enthalpy then satisfies a new thermodynamic identity:

$d H = d U + P d V + V d P = T d S + V d P + μ d N$

To get the second equality, just substitute the conventional thermodynamic identity to cancel $P d V$ .

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

Replacing $S$ with $T$ gives the Helmholtz Free energy:

$F (T, V, N) = U - T S$ , which satisfies the identity

$d F = d U - T d S - S d T = - P d V + μ d N - S d T$

Partial derivative relations imply

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

The first relation implies we can determine the internal energy from $F$ using only derivatives wrt temperature:

$U = F - T \partial_{T} F = - T^{2} \partial_{T} {(F / T)}_{V}$

If you know $F (T, V, N)$ , you can extract the internal energy. In this sense, F is called a thermodynamic potential (analogous to electric or gravitational potential)

Gibbs Free Energy: $G = U - T S + P V$

Replacing both $S$ and $V$ with their intensive conjugate variables ends up giving the Gibbs Free energy:

$G = Φ (T, P, N) = U - T S + P V$ which satisfies the differential identity:

$d G = d U - T d S - S d T + P d V + V d P + μ d N = - S d T + V d P + μ d N$

A few partial derivative relations that will be useful later on 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: $Φ = U - T S - μ N$

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: $Φ = - P V$

See problem 5.23 in Schroeder for more on this thermodynamic potential.

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