--%>

Theory of one dimensional motion

For motion in one dimension, the distribution of the molecules over quantum states, speeds, and energies can be deduced.

Here we show that the energy of a macroscopic gas sample can be described on the basis of our knowledge of the quantum states allowed to the molecules of the gas and the distribution expressed by the Boltzmann expression. We begin by studying the translational motion in one dimension of a collection of molecules. You will see how the procedure is fascillated by the partition function.

Partition function: the molecules of a gas that move along one dimension can have, any of the energies given by

1676_one dimensional motion.png 

For gas samples we can assume a cubic container and express a as V1/3, where V is the volume of the sample.

The partition function for one-dimensional translational motion can be developed by recognizing that
    
The translational energy of the lowest-energy state is small compared with the energies of most of the populated states and can be set equal to zero.
    
The translational-energy spacing between successive energy levels is small compared with the range of energies of the populated states.
    
The degeneracy of each energy level is unity.

On this basis, the partition function summation over the translational energies can be replaced by integration, and the partition function is expressed as

83_one dimensional motion1.png 

The integral is one of the definite integrals dealt by using the general result shown there, we obtain

1990_one dimensional motion2.png 

Example: calculate the partition function for the translational motion of N2 molecules free to move along one dimension of a 1-L cubic container. The temperature is 25°C.

Solution: the translational-energy factor h2/(8ma2) can be calculated conveniently from the expression of this equation. The mass of M of 1 mol of N2 molecules is 0.02801 kg, and V = 1 L = 10-3 m3. Thus
2163_one dimensional motion3.png 

962_one dimensional motion4.png 

= 1.180 × 10-40 J

The value of kT, to which the energy spacing factor is compared, is

kT = (1.3807 × 10-23 J K-1) (298.15 K) = 4.116 × 10-21 J

The partition function is calculated as

1331_one dimensional motion5.png 

this large partition function value indicates that very many states are available to the molecules. This result, in the calculations, from the smallness of h2/(8ma2compared to kT.

Average energy: the one dimensional translational energy of 1 mol of gas molecules can now be deduced. The general thermal-energy expression is

864_one dimensional motion6.png 

The partition function for one-dimensional translational motion gives
1661_one dimensional motion7.png 

substitution of the equation expressions in the equation for U - U0 gives

U - U0 = ½ RT

We have come by this long route to the result that we obtained from the simple classical kinetic-molecular theory. The translational energy per degree of freedom is ½ RT

   Related Questions in Chemistry

  • Q : Units of Measurement Unit of

      Unit of measurement- These are also some systems for units:      (1) 

  • Q : Calculating molarity of a solution

    Select the right answer of the question .The molarity of a 0.2 N N2Co3 solution will be: (a) 0.05 M (b) 0.2 M (c) 0.1 M (d)0.4 M

  • Q : Modern periodic table and Mendeleevs

    Differentiate between the modern periodic table and Mendeleevs table?

  • Q : Explain Photoelectron Spectroscopy. The

    The energies of both the outer and inner orbitals of atoms and molecules can be determined by photoelectron spectroscopy.Energy changes of the outermost or highest energy electron of molecules were dealt with here in a different passion. The energies of ot

  • Q : State substituted hydrocarbon Elaborate

    Elaborate a substituted hydrocarbon?

  • Q : Significance of the organic chemistry

    Describe some of the significance of the organic chemistry in brief?

  • Q : Mole fraction in vapours Choose the

    Choose the right answer from following. If two substances A and B have P0A P0B= 1:2 and have mole fraction in solution 1 : 2 then mole fraction of A in vapours: (a) 0.33 (b) 0.25 (c) 0.52 (d) 0.2

  • Q : Neutralization of sodium hydroxide How

    How much of NaOH is needed to neutralise 1500 cm3 of 0.1N HCl (given = At. wt. of Na =23): (i) 4 g  (ii) 6 g (iii) 40 g  (iv) 60 g

  • Q : Vapour pressure of volatile substance

    Provide solution of this question. According to Raoult's law the relative lowering of vapour pressure of a solution of volatile substance is equal to: (a) Mole fraction of the solvent (b) Mole fraction of the solute (c) Weight percentage of a solute (d) Weight perc

  • Q : Preparation of ammonium sulphate Select

    Select the right answer of the question. Essential quantity of ammonium sulphate taken for preparation of 1 molar solution in 2 litres is: (a)132gm (b)264gm (c) 198gm (d) 212gm