--%>

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 : Problem on mole fraction of glucose

    Provide solution of this question. While 1.80gm glucose dissolve in 90 of H2O , the mole fraction of glucose is: (a) 0.00399 (b) 0.00199 (c) 0.0199 (d) 0.998

  • Q : Solution problem What is the correct

    What is the correct answer. To made a solution of concentration of 0.03 g/ml of AgNO3, what quantity of AgNO3 must be added in 60 ml of solution: (a) 1.8  (b) 0.8  (c) 0.18  (d) None of these

  • Q : Describe various systems for

    Common system According to this system, the individual members are named according to alkyl groups att

  • Q : Question on Mole fraction Mole fraction

    Mole fraction of any solution is equavalent to: (a) No. of moles of solute/ volume of solution in litter (b) no. of gram equivalent of solute/volume of solution in litters (c) no. of  moles of solute/ Mass of solvent in kg (d) no. of moles of any

  • 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 : Eutectic Formation In some two

    In some two component, solid liquid systems, a eutectic mixture forms.Consider, now a two component system at some fixed pressure, where the temperature range treated is such as to include formation of one or more solid phases. A simple behavior is shown b

  • Q : Thermodynamics I) Sulphur dioxide (SO2)

    I) Sulphur dioxide (SO2) with a volumetric flow rate 5000cm3/s at 1 bar and 1000C is mixed with a second SO2 stream flowing at 2500cm3/s at 2 bar and 200C. The process occurs at steady state. You may assume ideal gas behaviour. For SO2 take the heat capacity at constant pressure to be CP/R = 3.267

  • Q : Dipole attractions-London dispersion

    Describe how dipole attractions, London dispersion forces and the hydrogen bonding identical?

  • Q : Molecular weight of solute Select right

    Select right answer of the question. A dry air is passed through the solution, containing the 10 gm of solute and 90 gm of water and then it pass through pure water. There is the depression in weight of solution wt by 2.5 gm and in weight of pure solvent by 0.05 gm. C

  • Q : Colligative property associated question

    Give me answer of this question. Which of the following is not a colligative property : (a)Optical activity (b)Elevation in boiling point (c)Osmotic pressure (d)Lowering of vapour pressure