--%>

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 : What are different mechanisms for

    Nucleophilic substitution reactions in halides containing  - X bond may take place through either of the two different mechanisms,S<

  • Q : Calculating Formulae Superphosphate has

    Superphosphate has the formula CaH4(PO4)2 H2O, what is the calculation to get the percentage of Phosphorus, I need to show the calculation. I know it is 30.9737622 u in weight and 2 atoms of the formula, but not sure how to work the calculation backwards.

  • Q : Laws of Chemical Combination Laws of

    Laws of Chemical Combination- In order to understand the composition of the compounds, it is necessary to have a theory which accounts for both qualitative and quantitative observations during chem

  • Q : Mass percent Help me to go through this

    Help me to go through this problem. 10 grams of a solute is dissolved in 90 grams of a solvent. Its mass percent in solution is : (a) 0.01 (b) 11.1 (c)10 (d) 9

  • Q : Anti-aromatic and the non-aromatic

    What is main difference among anti-aromatic and the non-aromatic compounds?

  • 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 : Oxoacids of halogens Why oxidising

    Why oxidising character of oxoacids of halogens decreases as oxidation number increases?

  • Q : Sugar solution The solution of sugar in

    The solution of sugar in water comprises: (i) Free atoms (ii) Free ions (iii) Free molecules (iv) Free atom and molecules. Choose the right answer from the above.

  • Q : Precipitation Addition of conc. HCl to

    Addition of conc. HCl to saturated Bacl2 solution precipitates Bacl2 ; because of the following reason : (a) It follows from Le Chatelier's principle (b) Of common-ion effect (c) Ionic product (Ba++)(cl) remains constant in a saturated sol

  • Q : Mole 2.0gram of dolomite is heated to a

    2.0gram of dolomite is heated to a constant weight of 1.0g. Calculate the total volume of CO2 produced at STP by this reation