Ideal Gas Law: Solving Air Bubble Volume at Different Depths

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An air bubble with an initial volume of 1 cm³ at a depth of 50 meters and a temperature of 17°C is analyzed using the Ideal Gas Law to determine its volume when it rises to the surface at 27°C. The pressure at 50 meters is 5 atm due to water pressure, plus 1 atm of atmospheric pressure, totaling 6 atm. The equation used is p1V1/T1 = p2V2/T2, where the initial conditions are substituted to solve for the final volume V2. The calculation initially yielded 5.2 cm³, but the discrepancy arises from not accounting for the total pressure at depth. Correcting for the total pressure leads to a different volume result.
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Homework Statement



An air bubble of volume 1cm^3 is located at a depth of 50m beneath the surface where the temperature is 17C. When the bubble goes up to the surface where the temperature is 27C, how much will its volume be? Take the atmospheric pressure to be 1atm. Assume the pressure increases 1atm for every 10 meters.


Homework Equations



Ideal Gas Law:
pV=nRT

The Attempt at a Solution



p_{1}v_{1}/T_{1}=p_{2}v_{2}/T_{2}

I plugged in:

5\times10^{5}\times1\times10^{-6}/290=1\times10^{5}\times V_{2}/300

Solving for V2, I get 5.2cm^3, but according to the solutions, that's not the answer...
I'd like to know what I'm doing wrong.
Thanks for the help!
 
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At 50 m, the pressure due to the water is 5 atm. You still have to add in the atmospheric pressure.
 
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