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Identify each energy exchange as primarily heat or work

Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro ISBN: 9780321809247 1

Solution for problem 40E Chapter 6

Chemistry: A Molecular Approach | 3rd Edition

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Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro

Chemistry: A Molecular Approach | 3rd Edition

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Problem 40E

Problem 40E

Identify each energy exchange as primarily heat or work and determine whether the sign of ΔEis positive or negative for the system.

a. A rolling billiard ball collides with another billiard ball. The first billiard ball (defined as the system) stops rolling after the collision.

b. A book is dropped to the floor. (The book is the system).

c. A father pushes his daughter on a swing. (The daughter and the swing are the system).

Step-by-Step Solution:
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Solution: The line spectrum observed in case of hydrogen atom can be quantitatively explained using Bohr’s model. When an electron jumps from a lower orbit to a higher orbit, it absorbs energy. When the electron jumps from higher orbit to lower orbit, it emits energy. The amount of energy absorbed or emitted when an electron jumps from orbit n(initial orbit) to i orbit n(fnal orbit) may be given as: --------(1) In case of absorption spectrum, i.e., n > n, the term [(1/n ) - (1/n )] becomes negative so that f i f i E is positive and hence energy is absorbed. 2 2 In case of emission spectrum, i.e., n > n, the ti [f/n ) - (1/n )] becofs positiviso that E is negative and hence energy is emitted. Step1: a) For the transition, n = 4 tin = 9, nf n andi ncefE is positive. Since, the transition is from lower energy level to higher energy level, therefore, light is absorbed. Step2: b) For the transition, n = 4 tin = 9, thf verage energy is given by, -18 1 1 E = (-2.18 x 10 J) ( 92 42 ) -18 1 1 = (-2.18 x 10 J) ( 81 16 ) -18 = (-2.18 x 10 J)(-65 / 1296) = (-2.18 x 10 -18J) x (-0.05) -18 = 0.109 x 10 J = 1.09 x 10 -19J Step3: Wavelength of the radiation, is given by- = hc / E, where h is the Planck’s constant and c is the velocity of light -34 Here, Planck's constant, h = 6.626 x 10 J.s Velocity of light, c = 3 x 10 m/sec 8 Substituting the value in the above expression, we get, -34 8 -19 = [(6.626 x 10 J.s) x (3 x 10 m/sec)] / (1.09 x 10 J) = 18.24 x 10 m -7 -6 = 1.824 x 10 m -6 Thus, wavelength of the radiation is 1.824 x 10 m. Step4: Since the value of E is positive, therefore, light is absorbed. Step5: c) Out of all the radiations in the electromagnetic spectrum, the wavelength of the infrared radiation ranges from 7 x 10 m to 1 x 10 m. Thus, the electromagnetic radiation in the above problem lies in the infrared region. -------------------------

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Chapter 6, Problem 40E is Solved
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Textbook: Chemistry: A Molecular Approach
Edition: 3
Author: Nivaldo J. Tro
ISBN: 9780321809247

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Identify each energy exchange as primarily heat or work