--%>

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 : The three facts on the evaporation

    Describe briefly the three facts on the evaporation?

  • Q : Mass percent Help me to go through this

    Help me to go through this problem. 10 grams of a solute is dissolved in 90 grams of a solvent. Its mass percent in solution is : (a) 0.01 (b) 11.1 (c)10 (d) 9

  • Q : Problem on Molar solution Can someone

    Can someone please help me in getting through this problem. 2.0 molar solution is acquired, when 0.5 mole solute is dissolved in: (i) 250 ml solvent (ii) 250 g solvent (iii) 250 ml solution (iv) 1000 ml solvent

  • Q : Calculation of molecular weight Provide

    Provide solution of this question. In an experiment, 1 g of a non-volatile solute was dissolved in 100 g of acetone (mol. mass = 58) at 298K. The vapour pressure of the solution was found to be 192.5 mm Hg. The molecular weight of the solute is (vapour pressure of ace

  • Q : Question of vapour pressure Choose the

    Choose the right answer from following. Vapour pressure of a solution is: (a) Directly proportional to the mole fraction of the solvent (b) Inversely proportional to the mole fraction of the solute (c) Inversely proportional to the mole fraction of the solvent (d

  • Q : Means of molal solution Choose the

    Choose the right answer from following. A molal solution is one that contains one mole of a solute in: (a) 1000 gm of the solvent (b) One litre of the solvent (c) One litre of the solution (d) 22.4 litres of the solution

  • Q : P block why BiF3 is ionic whereas other

    why BiF3 is ionic whereas other trihalides are covalent in nature?

  • Q : What are the various types of drugs

    Drugs are broadly classified into following types depending on the purpose for which they are used. 1. Antipyretics

  • Q : Problem on Osmotic Pressure of solution

    The osmotic pressure of a 5% solution of cane sugar at 150oC  is (mol. wt. of cane sugar = 342)(a) 4 atm (b) 3.4 atm (c) 5.07 atm (d) 2.45 atmAnswer: (c) Π = (5 x 0.0821 x 1000 x 423)/(342 x 100) = 5.07 atm

  • Q : Define thermal energy The thermal part

    The thermal part of the internal energy and the enthalpy of an ideal gas can be given a molecular level explanation. All the earlier development of internal energy and enthalpy has been "thermodynamic". We have made no use o