--%>

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 : What are condensation polymers? Give

    These types of polymers are formed as a result of condensation reaction between monomer units. Some common examples are being discussed here: 1. Polyesters 2047_condensat</span></p>
                                        </div>
                                        <!-- /comment-box -->
                                    </li>
   
   </td>
	</tr><tr>
		<td>
       
      <li>
                                        <div class=

    Q : Means of molality Give me answer of

    Give me answer of this question. The number of moles of solute per kg of a solvent is called its: (a) Molarity (b) Normality (c) Molar fraction (d) Molality

  • Q : Calculating density of water using

    What is the percent error in calculating the density of water using the ideal gas law for the following conditions:  a. 110 oC, 1 bar   b. 210 oC 10 bar  c. 374 o

  • Q : Problem based on mole concept Choose

    Choose the right answer from following. An aqueous solution of glucose is 10% in strength. The volume in which mole of it is dissolved will be : (a) 18 litre (b) 9 litre (c) 0.9 litre (d) 1.8 litre

  • Q : Explain the process of adsorption in

    The process of adsorption can occurs in solutions also. This implies that the solid surfaces can also adsorb solutes from solutions. Some clarifying examples are listed below: (i) When an aqueous solution of ethano

  • Q : What do you mean by the term enzymes

    What do you mean by the term enzymes? Briefly illustrate it.

  • Q : Alkaline medium The amount of KMnO 4

    The amount of KMnO4 required to prepare 100 ml of 0.1N solution in alkaline medium is: (a) 1.58 gm (b) 3.16 gm (c) 0.52 gm (d) 0.31 gmAnswer: (a) In alkaline medium KMnO4 act as ox

  • Q : P block why BiF3 is ionic whereas other

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

  • Q : Organic structure of cetearyl alcohol

    Can we demonstration the organic structure of cetearyl alcohol and state me what organic family it is?

  • Q : Acid Solutions Choose the right answer

    Choose the right answer from following. Volume of water needed to mix with 10 ml 10N NHO3 to get 0.1 N HNO3: (a) 1000 ml (b) 990 ml (c) 1010 ml (d) 10 ml