--%>

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 : Molarity of sodium hydroxide Can

    Can someone please help me in getting through this problem. Determine the molarity of a solution having 5g of sodium hydroxide in 250ml  solution is: (i) 0.5  (ii) 1.0  (iii) 2.0   (d) 0.1Answer: The right answer i

  • Q : Mole fraction Give me answer of

    Give me answer of following question. The sum of the mole fraction of the components of a solution is : (a) 0 (b) 1 (c) 2 (d) 4.

  • 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 : Concentration factor affected by

    Can someone please help me in getting through this problem. Which of the given concentration factor is affected by the change in temperature: (1) Molarity (2) Molality (3) Mole fraction (4) Weight fraction

  • Q : Problem based on mole concept Choose

    Choose the right answer from following. An aqueous solution of glucose is 10% in strength. The volume in which mole of it is dissolved will be : (a) 18 litre (b) 9 litre (c) 0.9 litre (d) 1.8 litre

  • Q : Production of alcoholic drinks give all

    give all physical aspects in the production of alcohol

  • Q : Osmotic Pressure The O.P. (Osmotic

    The O.P. (Osmotic Pressure) of equimolar solution of Urea, BaCl2 and AlCl3, will be in the order:(a) AlCl3 > BaCl2 > Urea  (b) BaCl2 > AlCl3 > Urea  (c) Urea > BaCl2<

  • Q : Calculation of concentration of the

    Choose the right answer from following. 200ml of a solution contains 5.85 dissolved sodium chloride. The concentration of the solution will be(Na= 23: cl = 35.5 ) (a) 1 molar (b) 2 molar (c) 0.5 molar (d) 0.25 molar

  • Q : Equimolar solutions Select the right

    Select the right answer of the question. Equimolar solutions in the same solvent have : (a)Same boiling point but different freezing point (b) Same freezing point but different boiling poin (c)Same boiling and same freezing points (d) Different boiling and differe

  • Q : Sedimentation and Velocity The first

    The first method begins with a well defined layer, or boundary, of solution near the center of rotation and tracks the movement of this layer to the outside of the cell as a function of time. Such a method is termed a sedimentary velocity experiment. A