Gibbs Phase Rule : Exercises

Introduction

Gibbs Phase Rule states that for a system in equilibrium the number of independent thermodynamic variables, $F$, is given by: $$ F = C - \pi + 2 $$ where $C$ is the number of chemical components we can tell apart (i.e. number of atom types/molecule types) and $\pi$ is the number of phases. We are going to examine this rule through some examples.

Example problems

Click on the problems to reveal the solution

Problem 1

The gas in the canister has only one component - Argon atoms. Furthermore, as all these atoms are in a gas there is only one phase in this system. We thus have: $$ F = C - \pi + 2 = 1 - 1 + 2 = 2 $$ It is necessary to give the values of two thermodynamic variables in order to state the thermodynamic state of the Argon gas.

Consider now the equation of state for the ideal gas: $$ PV = nRT $$ If you are given the temperature and the volume you can determine the pressure of the gas using this expression. Similarly if you are given the pressure and the volume of the gas you can determine its temperature. In other words, once you are given the values of two thermodynamic variables you have fully charaterized the gasses' thermodynamic state. The equation of state for an ideal gas is thus consistent with Gibbs Phase rule.

Problem 2

This solution to this question is very similar to the solution to the question on Argon gas. There is one component (water molecules) and a beaker of water is a one phase system. As such: $$ F = C - \pi + 2 = 1 - 1 + 2 = 2 $$ The reason I ask this question is that it is important to note here that when we have molecular systems the number of chemical components is the number of different types of molecule that are present. It is not the number of different atom types.

Problem 3

Once again there is only one chemical component in this particular system - water molecules. These water molecules can, however, be in one of two different phases - the liquid or the steam. As such Gibbs phase rule tells us that: $$ F = C - \pi + 2 = 1 - 2 + 2 = 1 $$ This result is once again in line with your intuition if you think about it a little. The critical thing to realise is that the liquid and the gas must be in equilibrium in order for Gibbs phase rule to hold. In other words, the chemical potential for the gas must be equal to that of of the liquid. This only happens when the liquid is at its boiling point. Furthermore, we know that there is an relation that allows us to calculate the temperature at which a liquid will boil when the system is at a particular pressure. As the fact that there is only one independent thermodynamic variable when we have two coexisiting phases of the same substance makes total sense.

Problem 4

In this system there are 2 components (argon and water) but there is only 1 phase (the solution). We thus have: $$ F = C - \pi + 2 = 2 - 1 + 2 = 3 $$ The thermodynamic variables we might use in this case could be the temperature, the volume and the chemical potential of the Argon. Incidentally, there are expressions for calculating the chemical potential of a solvent from the concentration of the solution. When analysing experiments it is often said that the thermodynamic variables are the temperature, the volume and the concentration. Strictly speaking, however, you need some way of calculating the chemical potential from the concentration, which is often far from trivial.

Problem 5

In this system there are 2 components (argon and water) and 2 phases. We thus have: $$ F = C - \pi + 2 = 2 - 2 + 2 = 2 $$ Once again the key point to recognise is that Gibbs Phase rule only holds when the gas and the solution are in equilibrium. The system will attain this equilibrium with the gas by having molecules of argon (or for that matter water) evapourate from the solution and enter the gas or condense from the gas so that they enter the solution. At any given temperature and pressure the concentration of argon (or chemical potential) will be determined through this equilibrium.

Contact Details

School of Mathematics and Physics,
Queen's University Belfast,
Belfast,
BT7 1NN

Email: g.tribello@qub.ac.uk
Website: mywebsite