How can you anticipate when to add the atmospheric pressure?

[hbk69]

Homework Statement

There are certain problems involving pressure where you are required to add the atmospheric pressure to solve the problems, i have done some problems where i work everything out but get the answer wrong because i did not add the atmospheric pressure. How can i overcome this problem? and how can i anticipate when i need to add the atmospheric pressure?, i will give some example questions where i made the mistake:

1)The depth of the ocean at a particular place is d=3.00km. The density of seawater Ps=1.03*10^3kgm^-3. What is the pressure P at the bottom of the ocean?

2)Two tubes are connected via a closed channel (forming an L-shaped figure). One tube is open-ended and has the length L1=100cm anthe other tube is close-ended and has the length L2=40cm. The L shaped structure is filled completely with water. What is the pressure at the top of the closed tube?

3) An ocean driver takes a deep breath at the surface, filling his lungs with air V=4L. The diver then descends to a depth of h=5m. The density of salt water=1030kgm^-3. The pressure of air at the surface is p=1.01x10^5 Pa. Assume that the body temperature T of the diver is the same at depth h as it is on the surface. Fine the volume of air in the lungs at this depth.

The Attempt at a Solution

In the questions above why was the atmospheric pressure added to each answer?

I calculated the answer for the questions above but failed to add the atmospheric pressure and although they stated the atmospheric pressure in each question (1.013*10^5 Pa) one would think that you'd automatically would know that you add it to your answer but what if that is not the case, i want to understand why we add the atmospheric pressure and how can i generally anticipate if i need to or not .

Thanks for any help or guidance

 

[BvU]

Hello hbk,

I would say that you generally have to take the atmospheric pressure into consideration, unless it is clearly not applicable.

In your first example, the pressure difference between the surface and the depth provided can be calculated from $\rho\, g\, \Delta h$. Since they ask for "the" pressure (i.e. an absolute pressure), you have to add the pressure at the surface. A small addition in this case, but nevertheless.

Example 2: same thing! The difference you calculate, "the" pressure you get by adding the atmospheric pressure. If L2 were 12 m, you would at first "calculate" a difference of 11 bar and end up with -1 bar. That is unphysical, so you realize that the difference cannot be more than the atmosperic pressure and the answer is (almost) zero. Almost, because the water vapour pressure is not 0.

Number 3 clearly mentions the pressure at the surface. No way you can then ignore that.
(I don't like the exercise, because you have to wonder if the diver's body is completely compressible).

Having said all that, your question is quite justified: I remember a post at the beginning of this year where there was a lot of confusion about tyre pressure, because the jargon is rather sloppy.

Physicists don't know anything except absolute pressure; technicians have a choice: they can use Bar for absolute and Barg for gauge pressure (look it up). But they often don't. Everybody is sloppy.

 

[haruspex]

As BvU says, you should include atmospheric pressure unless there's a reason not to.
It might be easier to clarify things for you if you provide an example where you did not need to consider it.

 

[hbk69]

Hello hbk,

I would say that you generally have to take the atmospheric pressure into consideration, unless it is clearly not applicable.

In your first example, the pressure difference between the surface and the depth provided can be calculated from $\rho\, g\, \Delta h$. Since they ask for "the" pressure (i.e. an absolute pressure), you have to add the pressure at the surface. A small addition in this case, but nevertheless.

Example 2: same thing! The difference you calculate, "the" pressure you get by adding the atmospheric pressure. If L2 were 12 m, you would at first "calculate" a difference of 11 bar and end up with -1 bar. That is unphysical, so you realize that the difference cannot be more than the atmosperic pressure and the answer is (almost) zero. Almost, because the water vapour pressure is not 0.

Number 3 clearly mentions the pressure at the surface. No way you can then ignore that.
(I don't like the exercise, because you have to wonder if the diver's body is completely compressible).

Having said all that, your question is quite justified: I remember a post at the beginning of this year where there was a lot of confusion about tyre pressure, because the jargon is rather sloppy.

Physicists don't know anything except absolute pressure; technicians have a choice: they can use Bar for absolute and Barg for gauge pressure (look it up). But they often don't. Everybody is sloppy.

thanks for this post! as a guideline what are the general scenarios where you have to add the atmospheric pressure and cases where you do not? going by your post i'd have to add it in most of my physics problems. Above tubes and above the surface of an ocean i'd have to add it? but what about underwater that would be an example where i do not require to? the reason i did not add it in the questions in the OP was because i contemplated if i should and i did not understand the importance of the atmospheric pressure or the impact it would have in the given scenario so i chose not to add it

 

[hbk69]

As BvU says, you should include atmospheric pressure unless there's a reason not to.
It might be easier to clarify things for you if you provide an example where you did not need to consider it.

Most of the problems i have done so far are ones where i had to add the atmospheric pressure but i did not, i have not done any problems where i was not required to add the atmospheric pressure i guess if i was given some problems where i did not have to add the atmospheric pressure my understanding would improve when i compare the questions where i need to add the atmospheric pressure and the questions where i do not

 

[haruspex]

Most of the problems i have done so far are ones where i had to add the atmospheric pressure but i did not, i have not done any problems where i was not required to add the atmospheric pressure i guess if i was given some problems where i did not have to add the atmospheric pressure my understanding would improve when i compare the questions where i need to add the atmospheric pressure and the questions where i do not

I can think of a few cases, but they're all sort of obvious...
- where it's already included
- where it's optional because it will cancel out
- where there is no atmosphere

 

[dauto]

In all three problems described at the OP one is calculating the pressure in a liquid which has a surface somewhere exposed to the atmosphere. By Pascal principle the atmospheric pressure affects the whole liquid.

 

[hbk69]

In all three problems described at the OP one is calculating the pressure in a liquid which has a surface somewhere exposed to the atmosphere. By Pascal principle the atmospheric pressure affects the whole liquid.

And this is the explantion for why we add the atompsheric pressure in problems where we have an object under the surface of a liquid? also how does the atmospheric pressure affect the liquid as a whole?

 

[BvU]

The atmosphere is pushing down on the surface of the liquid to the tune of 100000 N per square meter. That's quite something!

You could rephrase that: the surface of the ocean is carrying the weight of the atmosphere above it. If you realize that a cubic kilometer of air weighs some 1 000 000 000 kg (at sea level ), again: pretty hefty!

 

[hbk69]

The atmosphere is pushing down on the surface of the liquid to the tune of 100000 N per square meter. That's quite something!

You could rephrase that: the surface of the ocean is carrying the weight of the atmosphere above it. If you realize that a cubic kilometer of air weighs some 1 000 000 000 kg (at sea level ), again: pretty hefty!

Thanks for the help!

I found a question, it says:

Calculate the pressure at the bottom of the mercurly column: density of mercury=13600Pa. Pressure(h)=Pressure(Outside)=atmospheric pressure.
At standard atmospheric pressure the height of the mercury h=0.760m

In this particular problem is the atmospheric pressure included in the problem? which means we don't need to add it in the end solution?

 

[BvU]

If it says "Pressure(Outside)=atmospheric pressure" then that's a definite Yes!

I do hope the full text is a bit clearer, though ...
http://www.chem.wisc.edu/deptfiles/genchem/sstutorial/Text9/Tx94/tx94p1.GIF

 

[hbk69]

If it says "Pressure(Outside)=atmospheric pressure" then that's a definite Yes!

I do hope the full text is a bit clearer, though ...
http://www.chem.wisc.edu/deptfiles/genchem/sstutorial/Text9/Tx94/tx94p1.GIF

thanks a million bro! much clearer.

 

[Under_Pressure]

Here's a question I just came across in my textbook that led me to this forum and does not include atmospheric pressure. "A swimming pool has dimensions 30.0m x 10.0m and a flat bottom. When the pool is filled to a depth of 2.00 m with fresh water, what is the force caused by the water on the bottom? on each end? on each side?

 

[haruspex]

Here's a question I just came across in my textbook that led me to this forum and does not include atmospheric pressure. "A swimming pool has dimensions 30.0m x 10.0m and a flat bottom. When the pool is filled to a depth of 2.00 m with fresh water, what is the force caused by the water on the bottom? on each end? on each side?

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