**Problem 6:** A diver releases an air bubble of volume 2.50 cm³ from a depth of 15.5 m below the surface of a lake, where the temperature is 5.4°C. What is the volume of the bubble when it reaches the surface of the lake, where the temperature is 25°C?The solution requires applying the principles of gas laws, particularly Charles's Law, which states that the volume of a gas is directly proportional to its temperature (in Kelvin) at constant pressure. To solve this, convert the temperatures from Celsius to Kelvin, then apply the formula:\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]where:- \( V_1 = 2.50 \, \text{cm}^3 \)- \( T_1 = 5.4°C = 278.4 \, \text{K} \)- \( T_2 = 25°C = 298 \, \text{K} \) Solve for \( V_2 \), the volume at the surface.

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6. A diver releases an air bubble of volume 2.50 cm3 from a depth of 15.5 m below the surface of a lake; there the temperature is 5.4 °C. What is the volume of the bubble when it reaches the surface of the lake, where the temperature is 25 °C? cm3 f60 ss sf60

6. A diver releases an air bubble of volume 2.50 cm3 from a depth of 15.5 m below the surface of a lake; there the temperature is 5.4 °C. What is the volume of the bubble when it reaches the surface of the lake, where the temperature is 25 °C? cm3 f60 ss sf60

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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