--%>

Molecular energies and speeds

The average translational kinetic energies and speeds of the molecules of a gas can be calculated.

The result that the kinetic energy of 1 mol of the molecules of a gas is equal to 3/2 RT can be used to obtain numerical values for the average energies and speeds of these molecules. Notice, first, the remarkable generality of the relation KE = 3/2 RT. The translational kinetic energy of 1 mol of molecules, and therefore the average translational energy of the individual molecules, and therefore the average translational energy of the individual molecules, depends on only the temperature of the gas. None of the properties of the molecules not the atomic makeup, not the mass, not the shape-need is considered. The average kinetic energy of gas molecules depends on only the temperature.

Molecular translational energies: the value of R was obtained as 8.3143 J K-1 mol-1. The translational kinetic energy of 1 mol of gas molecules at 25°C (298.15 K) is

3/2 RT = 3/2 (8.3143 J K-1 mol-1) (298.15 K)

= 3718 J mol-1 = 3.718 kJ mol-1

This quantity, about 4 kJ/mol, will be a useful reference energy amount. It is a measure of the readily available, or "loose-change, " energy.

The average energy of a single molecule is given by

ke? = KE/ 639_molecular energy.png = (3/2 RT)/ 639_molecular energy.png 

For dealing with the energies of individual atoms or molecules, it is convenient to introduce a constant k, called the Boltzmann constant, as

K = R/ 639_molecular energy.png = 1.3806 × 10-23 J K-1

Notice that the Boltzmann constant is the gas constant per molecule. With this new constant we can express the average translational kinetic energy of a molecule of a gas as

ke? = 3/2 kT 

This energy, at 25°C, is

ke? = 3/2 (1.3806 × 10-23 J K-1) (298.15 K)

= 6.174 × 10-23 J


Speeds of molecules: energies have broader application in chemistry than do speeds. But at first it is easier to appreciate speeds.

Consider a gas that contains molecules of a particular mass. Molecular speed values can be obtained by writing the kinetic energy of 1 mol of these molecules as

KE = 639_molecular energy.png (1/2 mv2?) = ½( 639_molecular energy.png m)v2? = ½ Mv2?

Where M is the mass of 1 mol of molecules. This kinetic energy is given, according to our kinetic-molecular theory deviation, by

KE = 3/2 RT

Equating these expressions and rearranging give

√v2 = √3RT/M

The cumbersome term √v2 is known as the root mean square (rms) speed. It is the value that would be obtained if each molecular speed were squared, the average value of the squared terms was calculated, and finally the square root of this average is obtained. The rms value is only slightly different from a simple average if the individual contributions are bunched closely together. The rms value is typically about 10 percent higher than the simple average. We can, for the moment, take the rms value as being indicative of the average molecular speed.

Average speeds of gas molecules (equal to 0.921 √v2) at 25°C (298 K) and 1000°C (1273 K)

357_molecular energy1.png

   Related Questions in Chemistry

  • Q : Describe the properties of the

    Briefly describe the properties of the carbohydrates?

  • Q : Question based on relative lowering of

    Give me answer of this question. When a non-volatile solute is dissolved in a solvent, the relative lowering of vapour pressure is equal to: (a) Mole fraction of solute (b) Mole fraction of solvent (c) Concentration of the solute in grams per litre (d) Concentratio

  • Q : What is Distillation Separation by

    Separation by distillation can be described with a boiling point diagram. The important process of distillation can now be investigated. From the boiling point diagram one can see that if a small amount of vapour were removed from a liquid of composit

  • Q : Problem on distribution law The

    The distribution law is exerted for the distribution of basic acid among: (i) Water and ethyl alcohol (ii) Water and amyl alcohol (iii) Water and sulphuric acid (iv) Water and liquor ammonia What is the right answer.

  • Q : Question on colligative property Choose

    Choose the right answer from following. Which of the following is a colligative property: (a) Osmotic pressure (b) Boiling point (c) Vapour pressure (d) Freezing point

  • Q : Finding Active mass of urea Can someone

    Can someone please help me in getting through this problem. 10 litre solution of urea comprises of 240 gm urea. The active mass of urea is: (i) 0.04 (ii) 0.02 (iii) 0.4 (iv) 0.2

  • Q : Reason for medications contain hcl What

    What is the reason behind this that some medications contain hcl?

  • 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 : What is Ideal Mixtures Ideal mixing

    Ideal mixing properties can be recognized in the formation of an ideal gas mixture from ideal gases. Consider the formation of a mixture of gases i.e. a gaseous solution, from two mixtures of pure gases. A useful characterization of an ideal mixture, or soluti

  • Q : Solubility are halides are halogens

    are halides are halogens more soluble? why?