--%>

Molecular basis of third law.

The molecular, or statistical, basis of the third law can be seen by investigating S = k in W.

The molecular deductions of the preceding sections have led to the same conclusions as that stated in the third law of thermodynamics, namely, that a value can be assigned to the entropy of any substance. When the entropy values calculated from the details of the molecular energies are compared with those obtained calorimetric third law measurements, arrangements with in experimental error in usually found, but there are some exceptions. It is the perfectly ordered state of the crystal, with all the molecules in the same lowest energy level that is the molecular basis of the third law that the entropy is zero at absolute zero.

The third law value obtained for the entropy of CO at 1 bar and 298.15 K is 193.3 JK -1 mol-1. This value is obtained lower than the statistical result of 197.6 J K mol-1 obtained by the methods of the preceding section. Similar descriptions are found for NO and N2O. The third law result forH2O vapour is lower than the statistically calculated value by 3.3 J K-1 mol-1  there discrepancies can now be attributed to the failure of these materials to form the perfect crystalline state required at absolute zero for the third law to be applied. It is the perfectly ordered state of the crystal, with all the molecules in the same lowest energy level that is the molecular basis of the third law that the entropy is zero at absolute zero. (The positive value at the entropies of all compounds at temperature above absolute zero result from the fact as the temperature is raised, more and more energy levels become accessible to the molecules. The entropy at such temperature is, of course very characteristic of the individual molecule, since each molecule has its own particular energy level pattern.)

The discrepancies between calculated and third law entropies can now be attributed to a nonzero value of the entropy at absolute zero. Thus we must explain absolute zero entropy of, for example, about 4.3 J K-1 mol-1 for CO.

A disorder to be expected for such a material is that in which the molecular alignment in the crystal is not CO CO CO CO .... But rather a disorder pattern in the crystal like CO CO OC CO.... a crystal formed initially in this way could have the disorder "frozen" in as the temperature is lowered, there being too little thermal energy for the molecules to rearrange to the ordered structure. Thus, instead of each molecule having a single state to occupy, the randomness makes two states available to each molecule. The entropy of such a crystal can then be expected to be greater by k In 2N = R In 2 = 5.8 JK-1 mol-1 than it would be for a perfect crystal. This is, in fact, the approximate discrepancy found for CO.

Other types of disorder can now be expected to persist at absolute zero and to lead to apparent discrepancies in the third law. For example, a glassy material at entropy of zero will not have the necessary molecular order to guarantee as entropy of zero at absolute zero. In view of such difficulties, the third law statement must include the restriction that only perfectly ordered crystalline materials have zero entropy at absolute zero.

   Related Questions in Chemistry

  • Q : What is synthetic rubber and how it

    To meet human needs, scientists have started preparing synthetic rubbers. Besides having similar properties as natural rubbers they are tougher, more flexible and more durable than natural rubber. They are capable of getting stretched to twice its length. Though, it reverts to its original shape

  • Q : Problem based on lowering in vapour

    Help me to solve this problem. An aqueous solution of glucose was prepared by dissolving 18 g of glucose in 90 g of water. The relative lowering in vapour pressure is: (a) 0.02 (b)1 (c) 20 (d)180

  • Q : Molecular weight of substance The

    The boiling point of a solution of 0.11 gm of a substance in 15 gm of ether was found to be 0.1oC higher than that of the pure ether. The molecular weight of the substance will be (Kb = 2.16)       (a) 148 &nbs

  • Q : Relationship between free energy and

    The free energy of a gas depends on the pressure that confines the gas. The standard free energies of formation, like those allow predictions to be made of the possibility of a reaction at 25°C for each reagent at 

  • Q : Molecular basis of third law. The

    The molecular, or statistical, basis of the third law can be seen by investigating S = k in W.The molecular deductions of the preceding sections have led to the same conclusions as that stated in the third law of thermodynamics, namely, that a value can be

  • Q : Mole 2.0gram of dolomite is heated to a

    2.0gram of dolomite is heated to a constant weight of 1.0g. Calculate the total volume of CO2 produced at STP by this reation

  • Q : Vapour pressure related question Help

    Help me to solve this question. Which of the following is incorrect: (a) Relative lowering of vapour pressure is independent (b)The vapour pressure is a colligative property (c)Vapour pressure of a solution is lower than the vapour pressure of the solvent (d)The

  • Q : Vapour pressure of the pure hydrocarbons

    Give me answer of this question. A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at 20°C are 440 mmHg for pentane and 120 mmHg for hexane. The mole fraction of pentane in the vapour phase would be: (a) 0.549 (b)

  • Q : Molarity of solution Help me to go

    Help me to go through this problem. When 7.1gm Na2SO4 (molecular mass 142) dissolves in 100ml H2O , the molarity of the solution is: (a) 2.0 M (b) 1.0 M (c) 0.5 M (d) 0.05 M

  • Q : Neutralisation of phosphorous acids

    Provide solution of this question. To neutralise completely 20 mL of 0.1 M aqueous solution of phosphorous acid (H3 PO3) the volume of 0.1 M aqueous KOH solution required is: (a) 40 mL (b) 20 mL (c) 10 mL (d) 60 mL