--%>

Theory of three dimensional motion

Partition function; that the translational energy of 1 mol of molecules is 3/2 RT will come as no surprise. But the calculation of this result further illustrates the use of quantized states and the partition function to obtain macroscopic properties. The partition function is:

 
qtrans = Σ exp [- (n2x + n2y + n2z) h2/ (8ma2)/kT]  

= Σ exp [- n2x h2/ (8ma2)/kT] Σ exp [- n2y h2/ (8ma2)/kT] × Σexp [- n2z h2/ (8ma2)/kT]

= Σ exp [-n2x h2/(8ma2)/kT] Σ exp [-n2y h2/(8ma2)/kT] × Σexp [-n2z h2/(8ma2)/kT]

= qx qy qz

Each of the three partition function terms is like the one-dimensional term. We therefore can use:

qx = qy = qz = √∏/2 [kT/h2/(8ma2)] ½ 

to obtain, with V = a3,

qtrans = qx qy qz = (2∏mkT/h2)3/2 V

The Three dimensional translation energy: the three dimensional translation energy is derivative with respect to temperature can be used to reach an expression for the normal energy of three dimensional translational motions. Although qtrans depends on the particles and the volume of the container, the thermal energy (U - U0)trans has, for 1 mol of any gas in any volume the value 3/2 RT.

Distribution over quantum states: the distribution expressions for three dimensional motions can be derived by following the same procedure as we do for one dimensional motion before. First, however, we see that we can use one "effective" quantum number n in place of the three dimensional quantum numbers are nx, ny, and nz.

It is enough for us to deal with a quantity that shows the sum of the square of the equation of quantum numbers rather than with the individual values. We introduce the variable n defined by n2 = n2x + n2y + n2z.

Then the allowed energies are given instead of the more detailed manner than the previous one which we have done above. In using the effective quantum number n, we must recognize that there are number of states all with the same value of the energy. The display of states as point shows that, for large n, the additional number of states included when n increases by 1 is equal to 1/2πn2. Thus, if we use n as an effective quantum number, we must use gn = 1/2πn2.

Distribution over Quantum states: the distribution expressions for dimensional motion can be derived by following the same procedure as we did for one dimensional motion. First, however, we see that we can use one 'effective" quantum number n in place of the three quantum numbers nx, ny and nz.

(n2x + n2y + n2z) (h2/8ma2)

It is enough for us to deal with a quantity that shows the sum of the squares of the quantum numbers rather than with the individual values. We introduces the variable n defined by n2 = n2x + n2y + n2z. then the allowed energies are given by n2h2/(8ma2) instead of the more detailed, but no more useful, expression involving nx, ny and nz.

In using the effective quantum number n, we must recognize that there are a number of states all with the same value of n, or of energy εn. The number of states at this energy is the degeneracy gn. The display of states as points shows that, for large n, the additional number of states included when n increases by 1 is equal to ½ ∏n2. Thus if we use n as an effective quantum number we must use gn, ½ ∏n2 as the degeneracy.

   Related Questions in Chemistry

  • Q : Problem on thermodynamic equilibrium In

    In the manufacture of sulphuric acid by the contact process, S02 is oxidized to SO3 over a vanadium catalyst: The reactor is adiabatic and operates at atmospheric pressure. The gases enter the reactor at 410&d

  • Q : Thermodynamics 1 Lab Report I already

    I already did Materials and Methods section. I uploaded it with the instructions. Also, make sure to see Concept Questions and Thinking Ahead in the instructions that I uploaded. deadline is tomorow at 8 am here is the link to download all instructions because I couldn't attach all of t

  • 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 : Haloalkene with the help of polarity of

    with the help of polarity of c-x bond show that aryl halides are less reactive than alkyl halides

  • 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 : Unit of mole fraction Provide solution

    Provide solution of this question. Unit of mole fraction is: (a) Moles/litre (b) Moles/litre2 (c) Moles-litre (d) Dimensionless

  • 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 : Decision about dipole moment is present

    How can you decide if there is a dipole moment or not?

  • Q : Problem on preparing of a solution Give

    Give me answer of this question. How many grams of CH3OH should be added to water to prepare 150 solution of@M CH3 OH: (a) 9.6 (b) 2.4 (c) 9.6x 103 (d) 2.4 x103

  • Q : Colligative properties give atleast two

    give atleast two application of following colligative properties