A bubble of air is rising from the sea floor, from a depth of about 500 m. At thisdepth the pressure is 5.1*106 Pa (about 50.3 atm) and is at a temperature of about 7.0° C.It rises to the surface with the atmosphere, at a temperature of 20° C. If the bubble isobserved to have a volume of 4 cm³ at the surface, what was its volume at the sea floor?Assume the number of moles of gas in the bubble is constant.B.

The given values are, P1=5.1×106 Pa=50.3 atmT1=7.0° C=7.0+273 K=280 KP2=Patm=1 atmT2=20° C=20+273…

Answered: В. A bubble of air is rising from the…

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

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A bubble of air is rising from the sea floor, from a depth of about 500 m. At this
depth the pressure is 5.1*106 Pa (about 50.3 atm) and is at a temperature of about 7.0° C.
It rises to the surface with the atmosphere, at a temperature of 20° C. If the bubble is
observed to have a volume of 4 cm³ at the surface, what was its volume at the sea floor?
Assume the number of moles of gas in the bubble is constant.
B.

Expert Solution

Step 1

The given values are,

$P_{1} = 5.1 \times 10^{6} Pa = 50.3 atm T_{1} = 7.0 ° C = (7.0 + 273) K = 280 K P_{2} = P_{a t m} = 1 atm T_{2} = 20 ° C = (20 + 273) K = 293 K V_{2} = 4 {cm}^{3} V_{1} = ?$

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