Calculating Volume of Escaped Air Bubble Using Ideal Gas Law | Homework Question

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SUMMARY

The discussion centers on calculating the volume of an air bubble using the Ideal Gas Law, specifically under varying conditions of pressure and temperature. The initial volume of the bubble is 3.0x10-5 m3 at a depth of 45m with a temperature of 5°C. The correct final volume at the surface, where the temperature is 12°C, is derived using the equation P1V1/T1 = P2V2/T2, leading to a final volume of 1.7x10-4 m3 after converting temperatures to Kelvin. The error in the initial calculation stemmed from not converting temperature values, which was crucial for obtaining the correct answer.

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  • Ability to convert temperature from Celsius to Kelvin
  • Familiarity with basic algebra for solving equations
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Jimmy87
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Homework Statement



An air bubble of volume 3.0x10-5m3 escapes from a divers equipment at a depth of 45m where the water temperatures is 5 degrees C. What is its volume as it reaches the surface, where the temperature is 12 degrees C? Atmospheric pressure = 101kPa, density of sea water = 1020kgm-3.

I know how to answer question, I'm just no sure on how to find the pressure of the air bubble at a depth of 45m.

Homework Equations


PV = nRT

The Attempt at a Solution


I said P1V1/T1 = P2V2/T2 and then solved for V2 giving V2 = P1V1T2/T1P2. Where P1, V1 and T1 are the initial values and P2, V2 and T2 are the final values. I said that P2 = 101,000Pa and P1 = 101,000 + (1020 x 9.81 x 45) Pa. I can't think of anything I have done wrong but the answer in the back of the book is 1.7 x 10-4 m3 whereas I get 3.92 x 10-4 m3. Is the back of the book wrong?

Thanks for any help offered. [/B]
 
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I get the same answer as the book.
Did you convert the temperature values to kelvin?
 
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mfb said:
I get the same answer as the book.
Did you convert the temperature values to kelvin?

Of course, how stupid of me - the ideal gas equation must be in Kelvin. Thanks! I now get the right answer.
 

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