--%>

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/(8ma2) compared 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 : Problem on vapour pressure and mole

    Provide solution of this question. The vapour pressure of a solvent decreased by 10 mm of mercury, when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is 0.2. What should be the mole fraction of the solvent, if decrea

  • Q : What is Elevation in boiling point? The

    The boiling of a liquid may be defused by the temperature at which its vapour pressure which is equal to atmospheric pressure. The effect of addition in a non-volatile solute on the boiling point shown and its solution containing non-volatile solute with tempe

  • Q : Molecular crystals Among the below

    Among the below shown which crystal will be soft and have low melting point: (a) Covalent  (b) Ionic  (c) Metallic  (d) MolecularAnswer: (d) Molecular crystals are soft and have low melting point.

  • Q : Number of moles of potassium chloride

    Choose the right answer from following. The number of moles of KCL in 1000ml of 3 molar solution is: (a)1 (b)2 (c)3 (d)1.5

  • Q : Concentration of Calcium carbonate Help

    Help me to go through this problem. 1000 gms aqueous solution of CaCO3 contains 10 gms of carbonate. Concentration of the solution is : (a)10 ppm (b)100 ppm (c)1000 ppm (d)10000 ppm

  • Q : Volumes of solution after concentration

    Hydrochloric acid solution A and B encompass concentration of 0.5N and 0.1N  corresspondingly. The volumes of solutions A and B needed to make 2liters of 0.2N of HCL are: (i) 0.5l of A + 1.5l of B (ii) 1.5l of A + 0.5 l of B  (iii) 1.0 l of A + 1.0l of B&nbs

  • Q : Define Bond Energies - Bond Charges

    Energy changes in some chemical reactions can be used to deduce the energies of chemical bonds. Our understanding of the molecular basis of thermodynamic properties is extended when we ask why the enthalpy change for a reaction is what it is. We deduce,

  • Q : Molality of a glucose solution What

    What will be the molality of a solution containing 18g of glucose (having mol. wt. = 180) dissolved in 500g of water: (i) 1m  (ii) 0.5m  (iii) 0.2m  (iv) 2m

  • 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 : Describe Enzyme Catalyzed reactions

    Many enzyme catalyzed reactions obeys a complex rate equation that can be written as the total quantity of enzyme and the whole amount of substrate in the reaction system. Many rate equations that are more complex than first and se