What is the relationship between pressure and depth in a scuba diver's lungs?

[notsam]

Homework Statement

If a scuba diver fills his lungs to full capasity of 5.5 L when 8.0 m below the surface, to what volume would his lungs expand if he quickly rose to the surface?

Homework Equations

P=pg(H)

The Attempt at a Solution

Heyguys! :) I know that P=pgh but I have volume...

 

[gneill]

Below the surface of the Sun?

 

[ideasrule]

From the ideal gas law, which says PV=nRT, you can see that PV is constant if T is constant. So P1*V1=P2*V2, and you already know how to calculate the pressures. Just don't forget to add in the contribution from atmospheric pressure.

If T were constant, that would be all you need to do. However, a more realistic model is to assume an adiabatic process, since there's no heat exchange if the diver rises quickly. If you haven't learned about adiabatic processes yet, don't worry about it. If you have, do you know the equation relating pressure and volume?

 

[supratim1]

what is the link with Power radiated by the Sun?

 

[notsam]

Sorry about the power radiated by the Sun I posted this late last night. The Power radiated by the Sun went to a diffrent problem and I guess I just tagged it as this one.

OK! So I think I get it So the expanded equation looks like pgh underwater*V=pgh surface* V?! YES? V at the surface is my only unknown and my relative height of the diver will be 0?

 

[gneill]

The pressure at the surface is 1 atmosphere. The pressure increases with depth according to ρgh. The ideal gas law allows you to relate the volumes with the pressures.

 

[notsam]

Do I need to use pa or atm?

 

[gneill]

Since your ρgh formula will yield Pascals, you might want to head in that direction.

1 atm = 101,325 pascals

 

[notsam]

Thanks :) Y'all are awesome.

 

[notsam]

And I'll have to change my Volume to M^3 so that it can cancel out with the density of water. Right?

 

[gneill]

And I'll have to change my Volume to M^3 so that it can cancel out with the density of water. Right?

If your equations are set up as ratios (the usual case with ideal gas law type questions), then the conversion factors would cancel. You should be able to leave the volume units as-is.

p1*v1 = p2*v2

v2 = (p1/p2)*v1

The "trick" is to have the pressures in the same units, since you have to assume that the pressure at the surface is one atmosphere, and you calculate the pressure with depth by means which yield Pascals.

 

[supratim1]

its better you convert everything in SI units before you calculate, and at the end change the units of the final answer as per requirement.