TBR, G. Chem, Gases, Passage 3, #17

[Sammy1024]

17. If fresh water were used rather than salt water, how would the results be affected?

C. The change in volume of the ball would be greater when submerged in salt water than fresh water, because the salt water is denser than fresh water.

I knew that salt water would be more dense, but I don't understand the first part of that answer.

The answer said: Fresh water is less dense than salt water, which causes the ball to be less buoyant in fresh water, although that's not the focus of this question. Because salt water is more dense than fresh water, choices A and B are eliminated. A greater volume of the ball would initially be submerged in fresh water. However, as the ball is submerged below the surface, the pressure exerted by the less massive fresh water is less than the pressure exerted by the more massive salt water. The greater the pressure, the more the volume decreases so salt water reduces the volume of the ball more than fresh water. This means that the change in volume of the ball is greater when submerged in salt water than in fresh water.

From this answer, I feel like they're saying that the ball drops into the salt water, and goes below the surface and since the salt water is more dense, it applies a greater force on the ball, leading to a greater change in volume. BUT IF i'm correct, then how was I supposed to know that the ball does underwater? From the question it kind of just sounds like they put the ball on the water or something.

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[inasensegone]

17. If fresh water were used rather than salt water, how would the results be affected?
C. The change in volume of the ball would be greater when submerged in salt water than fresh water, because the salt water is denser than fresh water.

I knew that salt water would be more dense, but I don't understand the first part of that answer.

The answer said: Fresh water is less dense than salt water, which causes the ball to be less buoyant in fresh water, although that's not the focus of this question. Because salt water is more dense than fresh water, choices A and B are eliminated. A greater volume of the ball would initially be submerged in fresh water. However, as the ball is submerged below the surface, the pressure exerted by the less massive fresh water is less than the pressure exerted by the more massive salt water. The greater the pressure, the more the volume decreases so salt water reduces the volume of the ball more than fresh water. This means that the change in volume of the ball is greater when submerged in salt water than in fresh water.

From this answer, I feel like they're saying that the ball drops into the salt water, and goes below the surface and since the salt water is more dense, it applies a greater force on the ball, leading to a greater change in volume. BUT IF i'm correct, then how was I supposed to know that the ball does underwater? From the question it kind of just sounds like they put the ball on the water or something.

To recap: The passage basically describes an experiment where a ball is initially floating on water. This ball has a metal ring around it that allows it to be attracted by a magnet. A magnet at the bottom of the container of water causes the ball to be pulled downward deeper into the water. Basically, the ball would float normally (because it is less dense than the water), but because of the magnet, it is being pulled downward. As it is being pulled downward, the pressure at that depth causes the ball to decrease in volume, similar to the way a simple submarine could be crushed at deep depths. This pressure is given as:

P= ρgh

ρ or "rho" is the density of the fluid (water)
g is gravity
h is depth

Therefore as the ball is dragged to greater depths, h increases, so the pressure increases causing volume to decrease.

If saltwater is used, ρ (rho) increases leading to a greater pressure (than pure water) at equivalent depths.

Their answer explanation is a roundabout way of explaining the concept: If saltwater is more dense, then when no magnet is present at the bottom of the tank, the ball will float with less of the ball submerged under the surface (This is because buoyancy force = V * g * ρ ..... volume = volume of fluid displaced, and since the density, ρ, is greater in saltwater, less volume is displaced in order to provide a buoyancy force great enough for the ball).

 

[Sammy1024]

Ooooo! When it says more volume is displaced, they mean like compared to the displacement in water, the salt water causes more buoyancy which leads to a greater volume displacement!

 

[inasensegone]

Ooooo! When it says more volume is displaced, they mean like compared to the displacement in water, the salt water causes more buoyancy which leads to a greater volume displacement!

Not quite... because salt water is more dense, the ball needs to displace a smaller volume of water in order to be provided with a sufficient buoyancy force to float. Maybe try re-writing the equations a couple times and play with some numbers to see how the relationships change.

But again, this question isn't really about buoyancy... it's about pressure under water. When the ball is submerged, the saltwater is more dense so it exerts a greater pressure on the ball, and causes the volume of the ball to decrease more than the fresh water does, at equivalent depths.

 

[Sammy1024]

oooo! I get it now. Wow. I guess I was fixated on the "more" part of the volume and even decreasing something a lot more than another thing is still "more". Thank you so much!

 

[inasensegone]

oooo! I get it now. Wow. I guess I was fixated on the "more" part of the volume and even decreasing something a lot more than another thing is still "more". Thank you so much!

You're welcome!