Pressure and Fluids question

[dray5150]

This is quest 7.2a from Berkeley physics review:
At a specific depth in a swimming pool, a barometer measures the total pressure to be twice that of atmospheric pressure. If the barometer is now submerged to a depth that is twice its inital depth, how much does the total pressure increase?

a. the pressure increses by 50%
b. the pressure increses by 100%
c. the pressure increases by 200%
d. the pressure increases by 300%

The answer is a 50%

I really don't understand the explanation to this at all in the book. Can someone help?

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[MBHockey]

To answer this question you have to know where the total pressure is coming from.

When something is submerged under water, there are two pressures acting on it. One from the water above the object pushing down on it (hydrostatic pressure), and then the normal atmospheric pressure pushing down on to the top of the water (and everything else that is above the water surface).

We know that hydrostatic pressure is simply: p = rho*g*h

They tell us that at the first position under the water the barometer reads twice atmospheric pressure. Since atmospheric pressure is 1 atm, we know the total pressure here is 2 atm.

We know from the 2nd paragraph of this post that Ptotal = Phydrostatic + Patm. Since we know what Ptotal and Patm are, we can deduce that the hydrostatic pressure is 1 atm.

Now let's move down to point 2, where the depth (h) is twice its initial depth. Atmospheric pressure has not changed, so we know that the total pressure here will be equal to 1 atm (Patm) + our new (larger) hydrostatic pressure. Since hydrostatic pressure increases linearly with depth (p = rho*g*h) we know that the hydrostatic pressure at point 2 is double what it was at point 1. Since it was 1 atm at point 1, it must be 2 atm at point 2.

Therefore, our new total pressure at point 2 is: Ptotal = 1 atm + 2 atm = 3 atm.

3 atm is 50% more than 2 atm.

I'm sorry if this is the same explanation as the book (I don't know what the book says).

 

[FutureInternist]

This is quest 7.2a from Berkeley physics review:
At a specific depth in a swimming pool, a barometer measures the total pressure to be twice that of atmospheric pressure. If the barometer is now submerged to a depth that is twice its inital depth, how much does the total pressure increase?

a. the pressure increses by 50%
b. the pressure increses by 100%
c. the pressure increases by 200%
d. the pressure increases by 300%

The answer is a 50%

I really don't understand the explanation to this at all in the book. Can someone help?

For every 10 ft of water you have approx 1 atm of pressure. This is in addition to the pressure of the actual atmosphere. So total pressure at 10 ft = 2 atm.
Now if you go down a further 10 ft (total depth now 20ft) you have an additional atm added to the previous pressure, hence 3 atm.

% change = (final - original) / original = (3 - 2) / 2 = 0.5 = 50%


Edit - Oops.........1 minute too late 🙂

 

[dray5150]

Thanks guys... both of those explanations were great.

Right when it made sense the next example kind of through me off.

A swimmer is at the bottom of a 5m pool, where the gauge pressure is .5Patm. If the swimmer rises to the surface, by how much does the total pressure change?

a. increase by 33%
b. increase by 66%
c. decrease by 33%
d. decrease by 66%

Obviously I know it increases but how can you be 5m below the surface and have a pressure less than atomospheric pressure at the bottom. What am I missing?

 

[FutureInternist]

It's so much easier when we have the answer & can reverse engineer it 🙂

Is the answer 33%??

I'm not sure why you think the pressure would increase...I would say decrease since like you said at the bottom of the pool you have atm + water pressure so the atm has to be less than the total

I think, you could do it this way:-

For every 10 ft (i.e. 3 m) you go up by 1 atm...although as this example shows, that 1 atm does NOT have to equal 760mm Hg (he could be near Everest)
So if 5m = .5atm then 3m = .3atm (which would equal the atm pressure)
So % change = (.5 - .3) / .5 = 40%

 

[dray5150]

I meant to say pressure decreases but the answer in the back says decreases by 33%.

 

[FutureInternist]

I meant to say pressure decreases but the answer in the back says decreases by 33%.

40 is close enough 🙂....although I may have edited this after you already read it.

 

[MBHockey]

Thanks guys... both of those explanations were great.

Right when it made sense the next example kind of through me off.

A swimmer is at the bottom of a 5m pool, where the gauge pressure is .5Patm. If the swimmer rises to the surface, by how much does the total pressure change?

a. increase by 33%
b. increase by 66%
c. decrease by 33%
d. decrease by 66%

Obviously I know it increases but how can you be 5m below the surface and have a pressure less than atomospheric pressure at the bottom. What am I missing?

This is a bad question. There is no way that the gage pressure could be 0.5 atm at 5 meters below the surface of water (do you see why?). But you can still use general relations to arrive at the exact answer of a 33% decrease in total pressure from bottom to top:

To answer this one you have to understand what gage pressure is. It's a relative pressure measurement (relative to ambient conditions), so in this case, it's the pressure at that point minus the atmospheric pressure. This is why gage pressure, in general, can be less than 1 atm.

Going with that, since we know from the previous problem that the only two pressures acting on a submerged object are atmospheric pressure and hydrostatic pressure and that gage pressure is a pressure reading relative to (not including) atmospheric, then we can deduce that the hydrostatic pressure is 0.5 atm (it has to be, it's the only other pressure).

So they are telling you that Phydrostatic is 0.5 atm. With atmospheric pressure being 1 atm, this gives you 1.5 atm of total pressure at the bottom.

When the swimmer moves to the surface, he "loses" all hydrostatic pressure. So at the top he is only experiencing 1 atm of pressure (all from the atmosphere). He "lost" 0.5 atm of pressure out of the total 1.5 atm acting on him at the bottom.

0.5 atm/1.5 atm = 33% decrease in pressure from h = 5 m to h = 0 m.

 

[FutureInternist]

This is a bad question. There is no way that the gage pressure could be 0.5 atm at 5 meters below the surface of water (do you see why?). But you can still use general relations to arrive at the exact answer of a 33% decrease in total pressure from bottom to top:

To answer this one you have to understand what gage pressure is. It's a relative pressure measurement (relative to ambient conditions), so in this case, it's the pressure at that point minus the atmospheric pressure. This is why gage pressure, in general, can be less than 1 atm.

Going with that, since we know from the previous problem that the only two pressures acting on a submerged object are atmospheric pressure and hydrostatic pressure and that gage pressure is a pressure reading relative to (not including) atmospheric, then we can deduce that the hydrostatic pressure is 0.5 atm (it has to be, it's the only other pressure).

So they are telling you that Phydrostatic is 0.5 atm. With atmospheric pressure being 1 atm, this gives you 1.5 atm of total pressure at the bottom.

When the swimmer moves to the surface, he "loses" all hydrostatic pressure. So at the top he is only experiencing 1 atm of pressure (all from the atmosphere). He "lost" 0.5 atm of pressure out of the total 1.5 atm acting on him at the bottom.

0.5 atm/1.5 atm = 33% decrease in pressure from h = 5 m to h = 0 m.

That does make more sense....I stand corrected. Although situations like these show how someone who knows jack (me) & someone who knows the material (MB Hockey) can still end up w/ the same grade on a multiple choice 😀

 

[BerkReviewTeach]

This is a bad question. There is no way that the gage pressure could be 0.5 atm at 5 meters below the surface of water (do you see why?).

I would say that at 5 meters below the surface the gauge pressure has to be 0.50 atm. If you happen to scuba dive, then you know that the gauge pressure at the surface (where it is actual pressure is 1.00 atm.) is said to be 0. A 10 m column of water exerts approximately 1.00 atm of pressure, so at a depth of 5.0 m below the surface, the pressure reading on your gauge should be 0.5 atm. In essence, they are using the term gauge pressure in lieu of hydrostatic pressure.

But you can still use general relations to arrive at the exact answer of a 33% decrease in total pressure from bottom to top:

To answer this one you have to understand what gage pressure is. It's a relative pressure measurement (relative to ambient conditions), so in this case, it's the pressure at that point minus the atmospheric pressure. This is why gage pressure, in general, can be less than 1 atm.

Going with that, since we know from the previous problem that the only two pressures acting on a submerged object are atmospheric pressure and hydrostatic pressure and that gage pressure is a pressure reading relative to (not including) atmospheric, then we can deduce that the hydrostatic pressure is 0.5 atm (it has to be, it's the only other pressure).

So they are telling you that Phydrostatic is 0.5 atm. With atmospheric pressure being 1 atm, this gives you 1.5 atm of total pressure at the bottom.

When the swimmer moves to the surface, he "loses" all hydrostatic pressure. So at the top he is only experiencing 1 atm of pressure (all from the atmosphere). He "lost" 0.5 atm of pressure out of the total 1.5 atm acting on him at the bottom.

0.5 atm/1.5 atm = 33% decrease in pressure from h = 5 m to h = 0 m.

Perfect explanation, so there's nothing to add.

BTW MBHockey, I know I still owe you a PM. It's been crazy at work (works I should say). I do an occassional drive by at this site between readings and when I bored logging numbers into my lab notebook. I will reply soon, I swear.

 

[MBHockey]

I would say that at 5 meters below the surface the gauge pressure has to be 0.50 atm. If you happen to scuba dive, then you know that the gauge pressure at the surface (where it is actual pressure is 1.00 atm.) is said to be 0. A 10 m column of water exerts approximately 1.00 atm of pressure, so at a depth of 5.0 m below the surface, the pressure reading on your gauge should be 0.5 atm. In essence, they are using the term gauge pressure in lieu of hydrostatic pressure.

Berk, you're absolutely right. I actually realized this while I was feverishly typing up the response and then forgot to go back and change it. 🙂

No prob about the PM...whenever you get a chance.

 

[TRN1983]

As an aside, a good thing to remember is that 1ATM increases in pressure increase with every 10m of depth in water.

The answer is definitely a 50% increase.

You can think of it this way. You have three total parts contributing to the total pressure. (1) atmosphere, (2) depth of water with pressure equal to one atmosphere, (3) double the depth of water. This gives ATM + ATM + ATM = 3ATM of pressure. If the second depth is 3ATM and the first is 2ATM then 3ATM/2ATM x 100% = 150%, or a 50% increase.

edit: Just saw that MBHockey pretty much said the same thing as me ,so my post was not needed.

 

[TRN1983]

For every 10 ft of water you have approx 1 atm of pressure.

Just wanted to clarify, this is for every 10m 🙂

I.e. P=(density)gy = 1000kg/m^3x9.8m/s^2*10m = 98,000 Pa ~ 101,000Pa = ATM.

 

[mR LaZy]

nm

 

[ryblake]

MBhockey, even though this was many years ago, it helped me today, so thanks. So much clearer than the book's explanation.