Calculate the average translational kinetic energy, Ex, for one mole of gas at 669 K. Translational kinetic energy is sometimes called average kinetic energy. Ex = Calculate the average translational kinetic energy for a single gas molecule at 669 K. J/mol J/molecule

Calculate the average translational kinetic energy, Ex, for one mole of gas at 669 K. Translational kinetic energy is sometimes called average kinetic energy. Ex = Calculate the average translational kinetic energy for a single gas molecule at 669 K. J/mol J/molecule

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter5: Gases

Section: Chapter Questions

Problem 116E

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Concept explainers

Mole Concept

Mole concept is a method to determine the amount of any substance that consists of some elemental particles using a grouping unit called ‘mole’. The mole concept enables one to handle the large amount of small elementary particles. The mole concept can be used in many calculations related to mole calculation, molar amount, mole ratio, and so on, which are significant in chemistry.

Question

Calculate the average translational kinetic energy, Ex, for one mole of gas at 669 K. Translational kinetic energy is sometimes
called average kinetic energy.
Ex =
Calculate the average translational kinetic energy for a single gas molecule at 669 K.
J/mol
J/molecule

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In both the questions we have to calculate the average kinetic energy of gas particles at 669K in different units.

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12
This relationship is known as Graham’s Law of Effusion. Since both gases are at the same temperature, they must have the same average kinetic energy (½ mv2), where m is mass and v is velocity (like speed). Since both gases have the same average kinetic energy, you can state that ½ mLvL2 = ½ mHvH2. Multiplying both sides by 2 gives you mLvL2 = mHvH2. Rearranging the equation to get both masses on the same side of the equation will give you mL/mH = VH2/VL2. In 3a and 3b, you probably noticed that the heavy gas particles took twice as long to diffuse as the light gas particles. This means that the light gas particles are moving twice as fast, VH/VL = ½. Therefore, VH2/VL2 = ¼. If the light gas was Ne, what would be a reasonable identity for the heavy gas? Explain.