--%>

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 : Molar and Volumetric flow rate problem

    Cyclohexane (C6H12) is produced by mixing Benzene and hydrogen. A process including a reactor, separator, and recycle stream is used to produce Cyclohexane. The fresh feed contains 260L/min C6H6 with 950 L/min of H2

  • Q : What is adsorption and its examples. In

    In a liquid a solid substance a molecule present within the bulk of the substance is being attracted infirmly from all sides by the neighbouring molecules. Hence there is no bet force acting on the molecule or there are no unbalanced forces of the molecule. On the oth

  • Q : Moles of HCl present in .70 L of a .33

    Detail the moles of HCl which are present in .70 L of a .33 M HCl solution?

  • Q : P block why BiF3 is ionic whereas other

    why BiF3 is ionic whereas other trihalides are covalent in nature?

  • Q : Problem on preparing of a solution Give

    Give me answer of this question. How many grams of CH3OH should be added to water to prepare 150 solution of@M CH3 OH: (a) 9.6 (b) 2.4 (c) 9.6x 103 (d) 2.4 x103

  • Q : Question 6 A student was analyzing an

    A student was analyzing an unknown containing only Group IV cations. When the unknown was treated with 3M (NH4)2CO3 solution, a white precipitate formed. Because the acetic acid bottle was empty, the student used 6M HCl to dissolve the precipitate. Following the procedure of this experiment, the stu

  • Q : Dipole moment of chlorooctane

    Illustrate the dipole moment of chlorooctane?

  • Q : Chemical formula of detergent Describe

    Describe the chemical formula of detergent?

  • Q : Number of electrons in the benzene

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

  • Q : Explain polyhalogen compounds with

    Carbon compounds containing more than one halogen atom are called polyhalogen compounds. Most of these compounds are valuable in industry and agriculture. Some important polyhalogen compounds are described as follows:

    Discover Q & A

    Leading Solution Library
    Avail More Than 1416942 Solved problems, classrooms assignments, textbook's solutions, for quick Downloads
    No hassle, Instant Access
    Start Discovering

    18,76,764

    1934684
    Asked

    3,689

    Active Tutors

    1416942

    Questions
    Answered

    Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!

    Submit Assignment

    ©TutorsGlobe All rights reserved 2022-2023.