--%>

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 : Meaning of molality of a solution The

    The molality of a solution will be: (i) Number of moles of solute per 1000 ml of solvent (ii) Number of moles of solute per 1000 gm of solvent (iii) Number of moles of solute per 1000 ml of solution (iv) Number of gram equivalents of solute per 1000 m

  • Q : Problem on molecular weight of solid

    The vapor pressure of pure benzene at a certain temperature is 200 mm Hg. At the same temperature the vapor pressure of a solution containing 2g of non-volatile non-electrolyte solid in 78g of benzene is 195 mm Hg. What is the molecular weight of solid:

  • Q : Dipole moment of chloro-octane Describe

    Describe the dipole moment of chloro-octane in brief?

  • Q : DNA Organic Explain DNA organic in

    Explain DNA organic in brief?

  • Q : Problem on equilibrium composition The

    The catalytic dehydrogenation of 1-butene to 1,3-butadiene, C4H8(g) = C4H6(g)+H2(g) is carried out at 900 K and 1 atm.

    Q : Coordination compounds discuss

    discuss practical uses of coordination compounds

  • Q : Solution and colligative properties

    what is molarity of a solution of hcl which contains 49% by weight of solute and whose specific gravity is 1.41

  • Q : Finding strength of HCL solution Can

    Can someone please help me in getting through this problem. 1.0 gm of pure calcium carbonate was found to require 50 ml of dilute  HCL for complete reaction. The strength of the HCL  solution is given by: (a) 4 N  (b) 2 N  (c) 0.4 N  (d) 0.2 N

  • Q : Problem on volumetric flow rate Methane

    Methane containing 4 mol% N2 is flowing through a pipeline at 105.1 kpa and 22 °C. To check this flow rate, N2 at the same temperature and pressure are introduced to the pipeline at the rate of 2.83 m3/min. At the end of the pipe (

  • Q : Molecular energies and speeds The

    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