--%>

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 : Normality of solution containing

    Can someone please help me in getting through this problem. Determine the normality of a solution having 4.9 gm H3PO4 dissolved in 500 ml water: (a) 0.3  (b) 1.0  (c) 3.0   (d) 0.1

  • Q : Whether HCl is a base or an acid

    Whether HCl is a base or an acid? Briefly state your comments?

  • 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 : Henry law question Answer the following

    Answer the following qustion. The definition “The mass of a gas dissolved in a particular mass of a solvent at any temperature is proportional to the pressure of gas over the solvent” is: (i) Dalton’s Law of Parti

  • Q : Problem on decomposition reaction

    Nitrogen tetroxide (melting point: -11.2°C, normal boiling point 21.15°C) decomposes into nitrogen dioxide according to the following reaction: N2O4(g) ↔ 2 NO2(g)<

  • Q : Short note on the function of

    Write down a short note on the function of mitochondria?

  • Q : Importance of organic chemistry

    Describe the importance of organic chemistry?

  • Q : Chemists have not created a periodic

    Explain the reason behind that the chemists have not created a periodic table of compounds?

  • Q : Explain Second Order Rate Equations.

    Integration of the second order rate equations also produces convenient expressions for dealing with concentration time results.A reaction is classified as second order if the rate of the reaction is proportional to the square of the concentration of one o

  • Q : Problem on decinormal strength Can

    Can someone please help me in getting through this problem. How many grams of dibasic acid (having mol. wt. 200) must be present in 100ml  of its aqueous solution to provide decinormal strength: (i) 1g  (ii)2g  (iii) 10g  (iv) 20g<