--%>

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 : 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 : Cons of eating organic foods Illustrate

    Illustrate the cons of eating organic foods?

  • Q : Polymers comparison of biodegradable

    comparison of biodegradable and non-biodegradable polymers

  • Q : Vapour pressure of water Give me answer

    Give me answer of this question. 5cm3 of acetone is added to 100cm3 of water, the vapour pressure of water over the solution: (a) It will be equal to the vapour pressure of pure water (b) It will be less than the vapour pressure of pure water

  • Q : Quastion of finding vapour pressure

    Vapour pressure of CCl425Degree C at is 143mm of Hg0.5gm of a non-volatile solute (mol. wt. = 65) is dissolved in 100ml CCl4 .Find the vapour pressure of the solution (Density of CCl4 = = 1.58g /cm2): (a)141.43mm (b)

  • Q : Hybridization Atomic orbitals can be

    Atomic orbitals can be combined, in a process called hybridization, to describe the bonding in polyatomic molecules. Descriptions of the bonding in CH4 can be used to illustrate the valence bond procedure. We must arrive a

  • Q : Dissolving Group IV Carbonate Explain

    Explain how dissolving the Group IV carbonate precipitate with 6M CH3COOH, followed by the addition of extra acetic acid.

  • 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 : Real vapour pressure Choose the right

    Choose the right answer from following. The pressure under which liquid and vapour can coexist at equilibrium is called the : (a) Limiting vapour pressure (b) Real vapour pressure (c) Normal vapour pressure (d) Saturated vapour pressure

  • Q : What are Vander Waal's Radii? Vander

    Vander Waal's radii can be assigned to the atoms of molecules on the basis of the closeness of approach of these atoms in crystalline substances. Diffraction studies of crystals give information about hoe molecules can approach each other and can pack