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 : Benzoic acid is weaker than paranitro

    Briefly state that Benzoic acid is weaker than paranitro benzoic acid?

  • Q : Calculation of concentration of the

    Choose the right answer from following. 200ml of a solution contains 5.85 dissolved sodium chloride. The concentration of the solution will be(Na= 23: cl = 35.5 ) (a) 1 molar (b) 2 molar (c) 0.5 molar (d) 0.25 molar

  • Q : Decision about dipole moment is present

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

  • Q : Atmospheric pressure Give me answer of

    Give me answer of this question. The atmospheric pressure is sum of the: (a) Pressure of the biomolecules (b) Vapour pressure of atmospheric constituents (c) Vapour pressure of chemicals and vapour pressure of volatile (d) Pressure created on to atmospheric molecules

  • Q : Haloalkanes define primary secondary

    define primary secondary and tertiary alkyl halides with examples

  • 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 : 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 : 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 : Problem on solutions The 2N aqueous

    The 2N aqueous solution of H2S04 contains: (a) 49 gm of H2S04 per litre of solution (b) 4.9 gm of H2S04 per litre of solution (c) 98 gm of H2S04

  • Q : Problem on making solutions The weight

    The weight of pure NaOH needed to made 250cm3 of 0.1 N solution is: (a) 4g  (b) 1g  (c) 2g  (d) 10g Choose the right answer from above.

©TutorsGlobe All rights reserved 2022-2023.