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A system absorbs 196 kJ of heat and the surroundings do

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

Solution for problem 42E 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 42E

Problem 42E

A system absorbs 196 kJ of heat and the surroundings do 117 kJ of work on the system. 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 nfinal orbit) may be given as: --------(1) Step1: a) The lines of the Lyman series are observed in the ultraviolet region of the electromagnetic spectrum. Step2: b) In the Lyman series, for n = 1,f d n = 2, ti amount of energy emitted is given by: -18 1 1 E = (-2.18 x 10 J) ( 12 2 2) -18 1 1 = (-2.18 x 10 J)( 1 4) = (-2.18 x 10 -18J)(3 / 4) -18 = (-2.18 x 10 J) x (0.75) = -1.635 x 10 -18J Here, (-)ve sign indicates emission of energy. 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 Substituting the value in the above expression, we get, = [(6.626 x 10 J.s) x (3 x 10 m/sec)] / (1.635 x 10 -18J) = 12.16 x 10 m-8 -9 = 121.6 x 10 m Thus, wavelength of the radiation is 121.6 nm. Step4: In the Lyman series, for n = 1f nd n = 3, ti amount of energy emitted is given by: -18 1 1 E = (-2.18 x 10 J) ( 12 3 2) 1 1 = (-2.18 x 10 -18J) (1 9) -18 = (-2.18 x 10 J)(8 / 9) = (-2.18 x 10 -18J) x (0.89) -18 = -1.94 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 is the velocity of light Here, Planck's constant, h = 6.626 x 10 J.s -34 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)] / (1.94 x 10 -18J) -8 = 10.25 x 10 m = 102.5 x 10 m -9 Thus, wavelength of the radiation is 102.5 nm. Step6: In the Lyman series, for n = 1,f d n = 4, thiamount of energy emitted is given by: -18 1 1 E = (-2.18 x 10 J) ( 12 42) -18 1 1 = (-2.18 x 10 J) (1 16 ) = (-2.18 x 10 -18J)(15 / 16) -18 = (-2.18 x 10 J) x (0.9375) = -2.04 x 10 -18J Here, (-)ve sign indicates emission of energy. Step7: 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, -34 8 -18 = [(6.626 x 10 J.s) x (3 x 10 m/sec)] / (2.04 x 10 J) = 9.74 x 10 m-8 -9 = 97.4 x 10 m Thus, wavelength of the radiation is 97.4 nm. ---------------------

Step 2 of 3

Chapter 6, Problem 42E is Solved
Step 3 of 3

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

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A system absorbs 196 kJ of heat and the surroundings do