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Calculating air pressure in flask


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#1 chrisfris

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Posted 23 December 2016 - 03:31 PM

Hello all,

 

I have been stuck in this question for at least 3 hours, and i would like to get some tips on how to solve it. First, please check the drawing in the files to understand the water-barometer.

 

The question is as followed: 

"The water-barometer is very sensitive for temperature changes in the flask. For the increase Δh in water level per degree of increase in temperature, the following formula can roughly be applied:

 

Δh / ΔT = p / (T * ρwater * g)

 

T is the absolute temperature and p the pressure inside the flask. The atmospherical pressure (outside the flask) is equal to 10^5 Pa. Now roughly calculate the increase of the water level in cm when the temperature of the gas increases from 20° C to 23° C.

 

Useful data:

ΔT = 3 K

T = 293 K

ρwater = 10^3 kg/m^3

g = 9,81 m/s^2

patmosferical = 10^5 Pa

pflask = unknown

Δh = unknown

h = unknown

Δp = unknown

V = 1,1 dm^3 = (almost) unchanged

n = amount of gas in mole = unknown

 

How can i calculate p or Δh?

 

Thanks a lot and merry Christmas!

 

PS: I am not a native speaker, so excuse me if my language was incorrect.

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#2 swansont

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Posted 23 December 2016 - 04:04 PM

How does the pressure inside relate to the pressure outside?
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#3 chrisfris

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Posted 23 December 2016 - 07:02 PM

I does not relate in the way that you can calculate the pressure inside with the pressure outside. You can, however, calculate the height in the pipe with the difference between both pressures in the formula:

Δp = ρwater * g * h


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#4 swansont

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Posted 23 December 2016 - 07:15 PM

I does not relate in the way that you can calculate the pressure inside with the pressure outside. You can, however, calculate the height in the pipe with the difference between both pressures in the formula:
Δp = ρwater * g * h


But you have the outside pressure. So you can estimate the inside oressure. The height is most likely of order a meter, at most. So the pressure difference is ~10^4 Pa, or 0.1 of atmosphere, at biggest. IOW, if you are estimating the answer, you can ignore this difference.

Now you have an equation where you know everything but the result you want. Calculate away.
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#5 chrisfris

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Posted 23 December 2016 - 07:47 PM

Okay thanks I think this can really help me!


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