Equations of state

In which we consider two equations of state:

  • The ideal gas law, $Pv=RT$.
  • The van der Waals equation of state.

Extensive and intensive variables

Consider one thermodynamic system at equilibrium at a particuler $P$, $V$, and $T$, and then divide it in half...

The volume $V$ of each subsequent half-system is half of its original value. $\Rightarrow V$ is an extensive parameter.

But the pressure and temperature of each half-system is the same as the original. $\Rightarrow P$ and $T$ are intensive variables.

"Specific" variables

  • You can convert an extensive variable to an intensive one by dividing by the quantity of material $n$ (number of kilomoles).
  • You prepend the word "specific" onto such quantities, indicating that they are "per (kilo)mole".

Specific volume: $$v = V/n = \text{volume per kilomole}.$$

Ideally, we would like to write all extensive quantities in upper-case, e.g. $V$ and...

all intensive variables in lower-case.

Unfortunately, temperature as $t$ looks too much like time, and pressure as $p$ looks too much like momentum. So, we'll continue to use $P$ and $T$ (even though they're intensive).

Equations of state

The central assumption of thermodynamics is that there is a simple state function: $$f(P, V, T, .... ) = 0$$ ...which defines a surface in state space.

The most familiar equation of state that relates $P$, $V$, and $T$ is the...

Ideal gas law

Over a century or so, laws relating $P$, $V$, and $T$ where put forth by.. Avogadro, Boyle, Charles, Gay-Lussac.

In 1834 Emile Clapeyron synthesized these together as... $$PV = nRT,$$

where...

  • $P$ - pressure in Pa,
  • $V$ - volume in ${\text m}^3$,
  • $n$ - quantity of matter in kilomoles,
  • $T$ - absolute temperature in degrees K,

Then, the gas constant $ 8.314472 \times 10^3 {\rm J} \cdot {\rm K}^{−1}\cdot {\rm kmol}^{−1}$. (* warning chemists...)

Or, using $n=m/M$ where

  • $m$ - mass in kilograms,
  • $M$ - molecular weight: *kg per kmol. E.g. for Carbon: $M =$ 12.0107 grams / mole = 12.0107 kg / kilomole.

Here is the ideal gas law in terms of intensive variables only: $$Pv=RT.$$

How ideal?

The ideal gas law will eventually be shown to be an exact result for an idealized gas made up of...

  • point particles: volume of each particle $\to 0$),
  • noninteracting: interparticle force $\to 0$.,

So it's a good approximation for real gases...

  • at low pressures (relatively small volume per particle), and...
  • at high temperatures (attractive forces negligible compared to impact forces),
  • or for noble gases (smaller attractive forces).

Ideal gas diagrams

You can examine the equation of state and plot one variable as a function of another while holding the third constant. That way of thinking of the equation of state leads us to consider

Isochore plots

Solve the ideal gas law for $P(v,T)$. Now hold the volume constant (isochore) and plot $\left(P(t)\right)_v$ .. $$P = \( \frac{R}{v} \) T \propto T.$$

- $0.7 v_0$
- $v_0$
- $1.2 v_0$

Isobar plots

Solve the ideal gas law for $v$. Now hold the pressure constant (isobar) and sketch $v(T)$...

$$v = \( \frac{R}{P} \)T \propto T.$$

- $0.7 P_0$
- $P_0$
- $1.2 P_0$

Isotherm plots

Solve the ideal gas law for pressure $P$. Now holding the temperature constant (isotherm) and plot $P(v)$...

$$P = \(RT\) \frac{1}{v} \propto \frac{1}{v}.$$

- $1.2 T_0$
- $T_0$
- $0.7 T_0$

State-space surfaces

Here, we've written the temperature $T(P,v)$ as a function of pressure and volume, namely $$T(P,v)=\frac{Pv}{R}.$$

Using, Mathematica, we can picture the state surface (a 2-dimensional surface embedded in 3-dimensions) directly.

Ideal gas law surface

What features of real materials does the ideal gas law fail to capture??

Van der Waals

The van der Waals equation of state results in a phase transition as temperature drops... $$(P+a/v^2)(v-b) = RT.$$ What happens to this equation as $a\to 0$ and $b\to 0$??

Notice that we're using $v$ (intensive) not $V$ (extensive).

Solving for $P$, $$P=\frac{RT}{v-b}-\frac{a}{v^2}.$$

Below a certain critical temperature (in purple) an unstable region appears, and phase separation occurs.

Parameters:

a=20
b=1
=>
v_c=3b=3
T_c = 8a / 27Rb =>  RT_c = 160/27
p_c = a / 27b^2 = 20/27

- $T \gt T_c$
- $T_c$
- $T \lt T_c$

The critical temperature occurs as a point with slope=0 appears on the $P-v$ plot: $$\(\frac{\partial P}{\partial v}\) = 0 = -\frac{RT}{(v-b)^2}+\frac{2a}{v^3},$$

and a point with zero curvature appears: $$\(\frac{\partial^2 P}{\partial v^2}\) = 0 = \frac{2RT}{(v-b)^3}+\frac{6a}{v^4},$$

What happens if you cool a system with the critical volume $v_c$ to a temperature below $T_c$ in a container with a fixed volume?

  • The system can reduce its pressure by reducing its volume. So, what happens?
  • If $v$ changes, is the gas still occupying the whole container volume??
  • phase separation is the idea that some of the material exists at one $v=v_A$ and another portion exists at a different molecular volume, $v_B$. How would this get you out of the quandary above?
  • Now, these two portions of the material are in the same container. What thermodynamic variables must be the same for each portion if they are in equilibrium?

 

Image credits

Tim Patterson