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Q&A for How to Demonstrate Charles's Law
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QuestionWhy does the balloon shrink inside the freezer?Bess Ruff is a Geography PhD student at Florida State University. She received her MA in Environmental Science and Management from the University of California, Santa Barbara in 2016. She has conducted survey work for marine spatial planning projects in the Caribbean and provided research support as a graduate fellow for the Sustainable Fisheries Group.The balloon shrinks inside the freezer because, as Charles's Law explains, air contracts in lower temperatures. The cooler temperature slows the speed at which the air molecules are moving within the balloon, which means there isn't as much space between molecules, causing the air, and therefore the balloon, to contract.
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QuestionWhat is happening to the balloon in these experiments?Community AnswerAs you heat the air inside the balloon, it expands to take up more volume. This pushes the walls of the balloon out, making it bigger. When you cool the air inside the balloon, the opposite happens. The air contracts to take up less volume, which allows the walls of the balloon to contract as well. This makes the balloon smaller.
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QuestionIf given 277V/147.5 = 1, how do I solve the equation for V?Community AnswerThe first step is to get your variable, V, on one side of the equation and all of your known values on the other side. In this case, you would start by multiplying both sides of the equation by 147.5. This gives you 277V = 147.5. Next, divide both sides of the equation by 277 to get V = 147.5/277. Doing the math leaves you with V = 0.53. As you can see, this equation doesn't contain any units, which is a problem. 0.53 L is very different from 0.53 mL. Always be sure to include units for all values when solving the equation.
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QuestionBased on the result of the activity, what can you infer about the relationship about volume and temperature and constant pressure?Community AnswerWe can infer that volume is proportional to temperature, given a constant pressure. This is the case for ideal gases. However, for real gases, this is not exactly correct.
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