--%>

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 the final mixture Can

    Can someone please help me in getting through this problem. Two solutions of a substance (that is, non electrolyte) are mixed in the given manner 480 ml of 1.5M first solution + 520 ml of 1.2M second solution. Determine the molarity of the final mixture

  • Q : Ionization Potential Second ionization

    Second ionization potential of Li, Be and B is in the order (a)Li>Be>B (b)Li>B>Be (c)Be>Li>B (d)B>Be>Li

  • Q : Problem on normality Help me to solve

    Help me to solve this problem. 0.5 M of H2AO4 is diluted from 1 lire to 10 litre, normality of resulting solution is : (a)1 N (b) 0.1 N (c)10 N (d)11 N

  • Q : Mole fraction in vapours Choose the

    Choose the right answer from following. If two substances A and B have P0A P0B= 1:2 and have mole fraction in solution 1 : 2 then mole fraction of A in vapours: (a) 0.33 (b) 0.25 (c) 0.52 (d) 0.2

  • Q : Volume of solution containing solute

    What volume of solution contains 0.1 mole of the solute: (a) 100ml (b) 125ml  (c) 500ml (d) 62.5ml Choose the right answer from above.

  • Q : What is depression in freezing point?

    Freezing point of a substance is the temperature at which solid and liquid phases of the substance coexist. It is defined as the temperature at which its solid and liquid phases have the same vapour pressure. The freezing point o

  • Q : 6. 20 gm of hydrogen is present in 5

    6. 20 gm of hydrogen is present in 5 litre vessel. The molar concentration of hydrogen is

  • Q : Explain the catalyst definition and

    Catalyst is a substance which accelerates the rate of a chemical reaction without undergoing any change in its chemical composition or mass during the reaction. The phenomenon of increasing the rate of a reaction with the help of a catalyst is known as catalysis.

  • Q : Number of electrons in the benzene

    Describe the number of electrons in the benzene? Write a short note on it?

  • Q : What type of bond does HCl encompass

    What type of bond does HCl encompass? Describe briefly?