--%>

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 : Haloalkane how haloalkane can be

    how haloalkane can be prepared by refluxing alcohol with hydrohalic acids

  • Q : Value of molar solution Select the

    Select the right answer of the question. Molar solution contains: (a)1000g of solute (b)1000g of solvent (c)1 litre of solvent (d)1 litre of solution

  • Q : Question of vapour pressure Choose the

    Choose the right answer from following. Vapour pressure of a solution is: (a) Directly proportional to the mole fraction of the solvent (b) Inversely proportional to the mole fraction of the solute (c) Inversely proportional to the mole fraction of the solvent (d

  • Q : Means of molality Give me answer of

    Give me answer of this question. The number of moles of solute per kg of a solvent is called its: (a) Molarity (b) Normality (c) Molar fraction (d) Molality

  • Q : Concentration of urea Help me to go

    Help me to go through this problem. 6.02x 1020 molecules of urea are present in 100 ml of its solution. The concentration of urea solution is: (a) 0.02 M (b) 0.01 M (c) 0.001 M (d) 0.1 M (Avogadro constant, N4= 6.02x 1023mol -1)<

  • Q : Concentration of an aqueous solution

    Give me answer of this question. The concentration of an aqueous solution of 0.01M CH3OH solution is very nearly equal to which of the following : (a) 0.01%CH3OH (b) 0.1%CH3OH (c) xCH3OH= 0.01 (d) 0.99MH2O (

  • Q : Carnot cycle show how a mathematical

    show how a mathematical definition of entropy can be obtauined from a consideration of carnot cycle?

  • Q : Rotational energy and entropy due to

    The entropy due to the rotational motion of the molecules of a gas can be calculated. Linear molecules: as was pointed out, any rotating molecule has a set of allowed rotational energies. For a linear molecule the

  • Q : Explain methods for industrial

    The important methods for the preparation of alcohol on large-scale are given below:    

  • Q : What are heterogenous catalysis? Give

    When the catalyst exists in a different phase than that of reactants, it is said to be heterogeneous catalyst, and the catalysis is called heterogeneous catalysis. For example, SO2 can be oxidized to SO3