--%>

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 based on molarity Select the

    Select the right answer of the question. If 18 gm of glucose (C6H12O6) is present in 1000 gm of an aqueous solution of glucose, it is said to be: (a)1 molal (b)1.1 molal (c)0.5 molal (d)0.1 molal

  • Q : Problem on bubble point The following

    The following mixture of hydrocarbons is obtained as one stream in a petroleum refinery.

    Q : Vitamines 7 enzyme cofactor what is the

    what is the relationship between vitamins and enzyme cofactors

  • Q : Production of alcoholic drinks give all

    give all physical aspects in the production of alcohol

  • Q : Degree of dissociation The degree of

    The degree of dissociation of Ca(No3)2 in a dilute aqueous solution containing 14g of the salt per 200g of water 100oc is 70 percent. If the vapor pressure of water at 100oc is 760 cm. Calculate the vapor pr

  • Q : Freezing point of equimolal aqueous

    The freezing point of equi-molal aqueous solution will be maximum for:            (a) C6H5NH3+Cl-(aniline hydrochloride)  (b) Ca(NO3

  • Q : Problem based on molality of glucose

    Select the right answer of the question. If 18 gm of glucose (C6H12O6) is present in 1000 gm of an aqueous solution of glucose, it is said to be: (a)1 molal (b)1.1 molal (c)0.5 molal (d)0.1 molal

  • Q : Quantum Mechanical Operators The

    The quantum mechanical methods, illustrated previously by the Schrödinger equation, are extended by the use of operators. Or, w

  • Q : Precipitation problem On passing H 2 S 

    On passing H2S  gas through a solution of Cu+ and Zn+2 ions, CuS is precipitated first because: (i) Solubility product of CuS is equal to the ionic product of ZnS (ii) Solubility product of CuS is equal to the solubility product o

  • Q : Molar mass of compound The freezing

    The freezing point of a solution having 4.8 g of a compound in 60 g of benzene is 4.48. Determine the molar mass of the compound (Kf = 5.1 Km-1) , (freezing point of  benzene = 5.5oC)          &n