Chemistry: Matter and Change
1st Edition
ISBN: 9780078746376
Author: Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Publisher: Glencoe/McGraw-Hill School Pub Co
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Question
Chapter 13, Problem 116A
Interpretation Introduction
Interpretation:
The effect of air temperature on a trip in hot air balloon is to be explained
Concept introduction:
It is based on the fact that hot air is lighter than cold air, so it rises upwards.
Expert Solution & Answer
Answer to Problem 116A
The hot air balloon rises upward when burner in it fires the heat. When this air gets cool down, its start descending.
Explanation of Solution
A hot air balloon trip is completely dependent on air temperature. Hot air is lighter than cold air. Therefore, hot air ascends, and cold air descends. The hot air balloon ascends in air till its burner fires the hot air. As this hot air gets cool down, hot air balloon start descending.
Conclusion
Thus, the movement of balloon trip in upward and downward direction depends on the temperature of air as the hot air causes the balloon to rises whereas the balloon starts descending when air cools down.
Chapter 13 Solutions
Chemistry: Matter and Change
Ch. 13.1 - Prob. 1PPCh. 13.1 - Prob. 2PPCh. 13.1 - Prob. 3PPCh. 13.1 - Prob. 4PPCh. 13.1 - Prob. 5PPCh. 13.1 - Prob. 6PPCh. 13.1 - Prob. 7PPCh. 13.1 - Prob. 8PPCh. 13.1 - Prob. 9PPCh. 13.1 - Prob. 10PP
Ch. 13.1 - Prob. 11PPCh. 13.1 - Prob. 12PPCh. 13.1 - Prob. 13PPCh. 13.1 - Prob. 14SSCCh. 13.1 - Prob. 15SSCCh. 13.1 - Prob. 16SSCCh. 13.1 - Prob. 17SSCCh. 13.1 - Prob. 18SSCCh. 13.1 - Prob. 19SSCCh. 13.2 - Prob. 20PPCh. 13.2 - Prob. 21PPCh. 13.2 - Prob. 22PPCh. 13.2 - Prob. 23PPCh. 13.2 - Prob. 24PPCh. 13.2 - Prob. 25PPCh. 13.2 - Prob. 26PPCh. 13.2 - Prob. 27PPCh. 13.2 - Prob. 28PPCh. 13.2 - Prob. 29PPCh. 13.2 - Prob. 30PPCh. 13.2 - Prob. 31SSCCh. 13.2 - Prob. 32SSCCh. 13.2 - Prob. 33SSCCh. 13.2 - Prob. 34SSCCh. 13.2 - Prob. 35SSCCh. 13.2 - Prob. 36SSCCh. 13.2 - Prob. 37SSCCh. 13.3 - Prob. 38PPCh. 13.3 - Prob. 39PPCh. 13.3 - Prob. 40PPCh. 13.3 - Prob. 41PPCh. 13.3 - Prob. 42PPCh. 13.3 - Prob. 43PPCh. 13.3 - Prob. 44PPCh. 13.3 - Prob. 45PPCh. 13.3 - Prob. 46SSCCh. 13.3 - Prob. 47SSCCh. 13.3 - Prob. 48SSCCh. 13.3 - Prob. 49SSCCh. 13 - Prob. 50ACh. 13 - Prob. 51ACh. 13 - Prob. 52ACh. 13 - Prob. 53ACh. 13 - Prob. 54ACh. 13 - Prob. 55ACh. 13 - Prob. 56ACh. 13 - Prob. 57ACh. 13 - Prob. 58ACh. 13 - Prob. 59ACh. 13 - Prob. 60ACh. 13 - Prob. 61ACh. 13 - Prob. 62ACh. 13 - Prob. 63ACh. 13 - Prob. 64ACh. 13 - Prob. 65ACh. 13 - Prob. 66ACh. 13 - Prob. 67ACh. 13 - Prob. 68ACh. 13 - Prob. 69ACh. 13 - Prob. 70ACh. 13 - Prob. 71ACh. 13 - Prob. 72ACh. 13 - Prob. 73ACh. 13 - Prob. 74ACh. 13 - Prob. 75ACh. 13 - Prob. 76ACh. 13 - Prob. 77ACh. 13 - Prob. 78ACh. 13 - Prob. 79ACh. 13 - Prob. 80ACh. 13 - Prob. 81ACh. 13 - Prob. 82ACh. 13 - Prob. 83ACh. 13 - Prob. 84ACh. 13 - Prob. 85ACh. 13 - Prob. 86ACh. 13 - Prob. 87ACh. 13 - Prob. 88ACh. 13 - Prob. 89ACh. 13 - Prob. 90ACh. 13 - Prob. 91ACh. 13 - Prob. 92ACh. 13 - Prob. 93ACh. 13 - Prob. 94ACh. 13 - Oxygen Consumption If 5.00 L of hydrogen...Ch. 13 - Prob. 96ACh. 13 - If 2.33 L of propane at 24°C and 67.2 kPa is...Ch. 13 - Prob. 98ACh. 13 - Prob. 99ACh. 13 - Prob. 100ACh. 13 - Prob. 101ACh. 13 - Apply Calculate the pressure of (4.671022)...Ch. 13 - Analyze When nitroglycerin (C3H5N3O9) explodes,...Ch. 13 - Prob. 104ACh. 13 - Prob. 105ACh. 13 - Prob. 106ACh. 13 - Prob. 107ACh. 13 - Prob. 108ACh. 13 - Prob. 109ACh. 13 - Prob. 110ACh. 13 - Prob. 111ACh. 13 - Prob. 112ACh. 13 - Prob. 113ACh. 13 - Prob. 114ACh. 13 - Prob. 115ACh. 13 - Prob. 116ACh. 13 - Prob. 117ACh. 13 - Prob. 118ACh. 13 - Prob. 119ACh. 13 - Prob. 1STPCh. 13 - Prob. 2STPCh. 13 - Prob. 3STPCh. 13 - Prob. 4STPCh. 13 - Prob. 5STPCh. 13 - Prob. 6STPCh. 13 - Prob. 7STPCh. 13 - Prob. 8STPCh. 13 - Prob. 9STPCh. 13 - Prob. 10STPCh. 13 - Prob. 11STPCh. 13 - Prob. 12STPCh. 13 - Prob. 13STPCh. 13 - Prob. 14STP
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