TBR Gen Chem Chapter 6-Gases Questions

[Jay2910]

---

Members do not see this ad.

Hey everyone,

Below are two questions that I don't seem to understand. Help would be great!

18) For a ball that is filled with an ideal gas and immersed in freshwater that is initially 25% submerged before the magnetic field is applied, at what depth will the ball no longer be buoyant?

A) At depths between 0 and 32 feet below the surface
B)At depths between 32 and 64 feet below the surface
c)At depths between 64 and 96 feet below the surface
d)At depths greater than 96 feet below the surface

All I get for this one, is that the density of the object is 1/4 of that of water. So the volume has to increase? 4x in order for it to equal the density of water. How do I go from there?

19) For a ball that is initially 60% submerged, what is observed when compared to the ball in figure 1 under same experimental conditions?

I'm going to go ahead and assume that since many have TBR, that you may know the answer choices to this question.

For #19, is the approach very similar to that of #18?

Thanks in advance!

 

[Patassa]

For the first one keep in mind that the Fb depends in the fluid and the volume of that fluid displaced. So as you drag the ball deeper and deeper what happens to its volume as the pressure around increases? It goes down, so that means Fb goes down. PV relationship for the gas and P=rho g dh for the depth in the fluid.

 

[Jay2910]

Patassa,

I'm not too sure on what you mean?
So far I got
1) As you go deeper, the buoyant force decreases because the volume decreases. Ok and then what do you do?
2) How are you using buoyant force and PV relationship together?

 

[Patassa]

Patassa,

I'm not too sure on what you mean?
So far I got
1) As you go deeper, the buoyant force decreases because the volume decreases. Ok and then what do you do?
2) How are you using buoyant force and PV relationship together?

Well they gave you the volume change now that I'm looking at it so it's even easier, that volume change should come out to V=nRT/P, but whatever they made it easy. So you can solve for the weight of the ball from the initial conditions Fb=W with only 50% of the ball submerged. Then when it's submerged you can solve for the depth at which the Fb (with the new volume of the ball) = the weight of the ball.

 

[mehc012]

Well they gave you the volume change now that I'm looking at it so it's even easier, that volume change should come out to V=nRT/P, but whatever they made it easy. So you can solve for the weight of the ball from the initial conditions Fb=W with only 50% of the ball submerged. Then when it's submerged you can solve for the depth at which the Fb (with the new volume of the ball) = the weight of the ball.

Just remember, ~33ft for 1atm (in seawater, but freshwater is close enough), so if you need to decrease V by a factor of 4, then you need to increase the pressure to 4x the original (1atm at the surface). Therefore, you need a final pressure of 4atm, which occurs at 3*33ft (remember that there's still 1atm of pressure from the air.) So at ~99ft it should be neutral, and anything deeper than that should be negatively buoyant. In their answer scale, that's d).

Furthermore, you know that there should be a transition into negative buoyancy, not a range...so it can't be any of the answers which say "at depths between", because those imply that once it gets deep enough, it suddenly becomes positively buoyant again, which makes no sense.