Air bubble volume at surface using ideal gas law

[Jimmy87]

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]

 

[mfb]

I get the same answer as the book.
Did you convert the temperature values to kelvin?

 

[Jimmy87]

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.