You have a helium balloon at 1.00 atm and 25°C. You want to make a hot-air balloon with the same volume and same lift as the helium balloon. Assume air is 79.0% nitrogen and 21.0% oxygen by volume. The “lift” of a balloon is given by the difference between the mass of air displaced by the balloon and the mass of gas inside the balloon. a. Will the temperature in the hot-air balloon have to be higher or lower than 25°C? Explain. b. Calculate the temperature of the air required for the hot- air balloon to provide the same lift as the helium balloon at 1.00 atm and 25°C. Assume atmospheric conditions are 1.00 atm and 25°C.
You have a helium balloon at 1.00 atm and 25°C. You want to make a hot-air balloon with the same volume and same lift as the helium balloon. Assume air is 79.0% nitrogen and 21.0% oxygen by volume. The “lift” of a balloon is given by the difference between the mass of air displaced by the balloon and the mass of gas inside the balloon. a. Will the temperature in the hot-air balloon have to be higher or lower than 25°C? Explain. b. Calculate the temperature of the air required for the hot- air balloon to provide the same lift as the helium balloon at 1.00 atm and 25°C. Assume atmospheric conditions are 1.00 atm and 25°C.
Solution Summary: The author explains that volume of a gas container is directly proportional to the temperature of gas according to Charles's law at constant pressure and number of moles.
You have a helium balloon at 1.00 atm and 25°C. You want to make a hot-air balloon with the same volume and same lift as the helium balloon. Assume air is 79.0% nitrogen and 21.0% oxygen by volume. The “lift” of a balloon is given by the difference between the mass of air displaced by the balloon and the mass of gas inside the balloon.
a. Will the temperature in the hot-air balloon have to be higher or lower than 25°C? Explain.
b. Calculate the temperature of the air required for the hot- air balloon to provide the same lift as the helium balloon at 1.00 atm and 25°C. Assume atmospheric conditions are 1.00 atm and 25°C.
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Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
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Step by Step Stoichiometry Practice Problems | How to Pass ChemistryMole Conversions Made Easy: How to Convert Between Grams and Moles; Author: Ketzbook;https://www.youtube.com/watch?v=b2raanVWU6c;License: Standard YouTube License, CC-BY