A system releases 622 kJ of heat and does 105 kJ of work

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

Problem 41E 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 41E

A system releases 622 kJ of heat and does 105 kJ of work on the surroundings. What is the change in internal energy of the system?

Step-by-Step Solution:
Step 1 of 3

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(final orbit) may be given as: f --------(1) Step1: In the Balmer series, for n = 3, the possible values of n = 4, 5, 6,.... Considering n = 3 and n = f i f i 4, the amount of energy emitted is given by: -18 1 1 E = (-2.18 x 10 J) ( 32 42 ) -18 1 1 = (-2.18 x 10 J) ( 9 16) = (-2.18 x 10 -18J)(7 / 144) = (-2.18 x 10 -18J) x (0.049) -18 = -0.107 x 10 J = -1.07 x 10 -19J Here, (-)ve sign indicates emission of energy. Step2: 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 Substituting the value in the above expression, we get, = [(6.626 x 10 J.s) x (3 x 10 m/sec)] / (1.07 x 10 -19J) = 18.58 x 10 m -7 -6 = 1.858 x 10 m = 1858 x 10 m -9 Thus, wavelength of the radiation is 1858 nm. Step3: 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 emitted lies in the infrared region and hence are not seen in the visible region of the spectrum. Hence, the correct option is (ii). Step4: b) In the Balmer series, for n = 2f nd n = 3,ihe amount of energy emitted is given by: E = (-2.18 x 10 -18J) 12 2 ( 2 3 ) = (-2.18 x 10 -18J) 1 1 ( 4 9) = (-2.18 x 10 -18J)(5 / 36) -18 = (-2.18 x 10 J) x (0.14) = -0.3052 x 10 -18J -19 = -3.05 x 10 J Here, (-)ve sign indicates emission of energy. Step5: Wavelength of the radiation, is given by- = hc / E, where h is the Planck’s constant and c...

Step 2 of 3

Chapter 6, Problem 41E is Solved
Step 3 of 3

Textbook: Chemistry: A Molecular Approach
Edition: 3rd
Author: Nivaldo J. Tro
ISBN: 9780321809247

The full step-by-step solution to problem: 41E from chapter: 6 was answered by , our top Chemistry solution expert on 02/22/17, 04:35PM. Since the solution to 41E from 6 chapter was answered, more than 658 students have viewed the full step-by-step answer. The answer to “A system releases 622 kJ of heat and does 105 kJ of work on the surroundings. What is the change in internal energy of the system?” is broken down into a number of easy to follow steps, and 26 words. This textbook survival guide was created for the textbook: Chemistry: A Molecular Approach, edition: 3rd. Chemistry: A Molecular Approach was written by and is associated to the ISBN: 9780321809247. This full solution covers the following key subjects: Lines, transitions, spectrum, Photons, portion. This expansive textbook survival guide covers 82 chapters, and 9464 solutions.

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