Entropy and Probability Background · ‘Classical approach’ to thermodynamics .

1. Entropy and Probability Background · ‘Classical approach’ to thermodynamics. ·What entropy ‘means’ ·What is happening in terms of the atoms and molecules when we change the state of the system ·S>0, there is always an increase in ‘randomness’ or ‘disorder’. ·

2. 1. Expansion of a gas at constant T e.g. 10 moles ideal gas expanded from 1 to 10 m3 2. Compression of gas at constant T

3. 1. Heat up a solid, liquid or gas at constant P If Cp=const e.g. Heat 2 moles of Al (~54g) from 298K to 373K, Cp(Al,s)=24.36JK-1mol-1.

4. 1. Mixing gases e.g. mix 1 mole of O2 with 3 moles of N2

5. Melting solid . ==Enthalpy of Fusion e.g. melt 10 moles of Pb at its melting point 600K, =4812 J.mole-1

6. 1. Oxidise a metal to its oxide e.g. oxidise 2 moles of Al(s) to Al2O3(s) at 298K =50.99-2(28.33)-3/2(205.0)=-313.2JK-1.

7. Now examine these in terms of changes in degree of ‘disorder’. 1. Gas molecules occupy large volume. Each molecule has more space. Increase in disorder. 2. Gas molecules occupy less volume. Each molecule as less space. Decrease in disorder, ie. Increase in order. 3. For crystals, atoms vibrate about fixed lattice points. When the crystal is heated up, amplitude of vibrations has increased, ie. More disordered. 4. When gases have formed a mixture, there is an increase in disorder. 5. When a crystal is melted, ordered array of atoms on crystal lattice become free movement of atoms within liquid phase. 6. When gas reacts with solid to form another solid, the net result is removal of gas phase. Therefore, it leads to decrease in disorder, ie. Increase in order.

8. What do we means by ‘disorder’ and how do we quantify this? By ‘probability’ A ‘disordered’ state is one where there are a large number of possible equally probable arrangements, ie a high probability state. The more disordered state has a higher probability f (larger number of different arrangements). The less disordered state has a lower probability i (smaller number of different arrangements). Example: two colour balls mixing

9. ·Thus entropy changes can be regarded as a measure of the increase in the ‘degree of mixing’ such that for Less disordered statemore disordered state S=Sfinal -Sinitial >0 The more disordered state has a higher probability f (larger number of different arrangements). The less disordered state has a lower probability i (smaller number of different arrangements).

10. These are related by the Boltzman Equation

11. When we use the term ‘disorder’, we are referring to a situation which can be realised in many different ways whereas ‘ordered’ states refer to those which can only be obtained in a few ways.

12. The equation proposed by Boltzman is as follows: Where final = Number of ways of achieving the final state (final probability) initial = number of ways of achieving the initial state (initial probability). final>initial, then S>0 then process will occur spontaneously in an isolated system.

13. Example: expansion of gas into a vaccum We consider the situation after opening the valve and before the gas has expanded. What is the probability of finding a particular molecule in V1 as opposed to finding it in V2. Probability must be related to volumes of V1 and V2, ie. Chance of finding a molecule in V1 is V1/(V1+V2). Now add a second molecule, chance of finding this in V1 is also V1/(V1+V2). Probability of finding them both in V1 =

14. For N molecules, chance of all N molecules being in V1=, which is the probability of initial state  initial. Probability of final state=1 since all of the molecules must be somewhere inside the volume (V1+V2) i.e.  final =1

15. Apply Boltzman Equation to obtain S for 1 moles i.e. N=6.0231023.  Thus we get the same result as before.

16. Conclusion: There are two ways to calculate entropy changes

17. What is the reversible process? [5 marks] • Give an example of the reversible process. [5 marks] • Explain why a gas expansion to vacuum is not reversible process? [5 marks] • For the reaction Calculate the heat of reaction at 1000 K if the heat capacities in J.mol-1K-1 are given by Cp= 25.7 for CO(g), Cp=24.8 for CO2(g) Cp=24.6 for O2(g) Enthalpies of formation for CO(g) –105.6 kJ.mol-1, for CO2(g) -376 kJ mol-1 at 298 K.