--%>

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 : Relationship between free energy and

    The free energy of a gas depends on the pressure that confines the gas. The standard free energies of formation, like those allow predictions to be made of the possibility of a reaction at 25°C for each reagent at 

  • Q : Molal elevation constant of water The

    The boiling point of 0.1 molal aqueous solution of urea is 100.18oC  at 1 atm. The molal elevation constant of water is: (a) 1.8    (b) 0.18   (c) 18    (d) 18.6Answer: (a) Kb

  • Q : Determining Mole fraction of water Can

    Can someone please help me in getting through this problem. The mole fraction of water in 20% aqueous solution of H2O2 is: (a) 77/68 (b) 68/77 (c) 20/80  (d) 80/20

  • Q : P- block why pentahalids are more

    why pentahalids are more covalent than tetrahalids

  • Q : Question on Mole fraction Mole fraction

    Mole fraction of any solution is equavalent to: (a) No. of moles of solute/ volume of solution in litter (b) no. of gram equivalent of solute/volume of solution in litters (c) no. of  moles of solute/ Mass of solvent in kg (d) no. of moles of any

  • Q : Acid Solutions Choose the right answer

    Choose the right answer from following. Volume of water needed to mix with 10 ml 10N NHO3 to get 0.1 N HNO3: (a) 1000 ml (b) 990 ml (c) 1010 ml (d) 10 ml

  • Q : Chemistry brief discription of relative

    brief discription of relative lowering of vapour pressure

  • Q : What do you mean by the term hydra What

    What do you mean by the term hydra? Briefly define it.

  • Q : Volumes of solution after concentration

    Hydrochloric acid solution A and B encompass concentration of 0.5N and 0.1N  corresspondingly. The volumes of solutions A and B needed to make 2liters of 0.2N of HCL are: (i) 0.5l of A + 1.5l of B (ii) 1.5l of A + 0.5 l of B  (iii) 1.0 l of A + 1.0l of B&nbs

  • Q : The Liver Is Responsible For Much Of

    The Liver Is Responsible For Much Of The Pentose Phosphate Activity Explain