What pressure difference is needed across the length of a 15 cm vertical drinking straw in order to drink a liquid of density 1.0 g/cm³?
The formula is $p =dgh$
$d$ is density = $1000\ \mathrm{kg/m^3}$
$g$ is gravity = $9.80665\ \mathrm{m/s^2}$
$h$ is height = $0.15\ \mathrm m$
$$p = 1.0\ \mathrm{g\ cm^{3}} \times 9.80665\ \mathrm{m\ s^{2}} \times 15\ \mathrm{cm} = 1471\ \mathrm{N/m^2}$$
This is the pressure you must overcome (about 0.21335 psi) to drink your water.
See the note by @Nicolau Saker Neto for references.
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Slugger
Updated on August 03, 2022Comments

Slugger less than a minute
What pressure difference must be generated across the length of a $15\ \mathrm{cm}$ vertical drinking straw in order to drink a waterlike liquid of density $1.0\ \mathrm{g\ cm^{−3}}$?
I am really unsure how to proceed on this one. I was thinking of using the ideal gas law for the air inside the straw initially, and then to increase the pressure difference so the volume of gas in the straw decreases. But this doesn't get me far at all. If any could help me with this one I would be very grateful!
P.S. This was probably clear but I am very new to chemistry, I am a mathematics student.

Nicolau Saker Neto over 8 years

ParaH2 about 5 yearsThis is not thermodynamics. I thought mathematics students also study physics am I wrong ?

Slugger about 5 years@HexacoordinateC Indeed in my mathematics study I did study quite some physics, but mostly in the areas of relativity theory, waves, optics and some assorted problems that give rise to (partial)differential equations. The course I did on thermodynamics was just for fun, and this is one of the questions that the professor mentioned at some point. I then (naively) assumed that that must mean it lies within the realm of thermodynamics.


ParaH2 about 5 yearsThis is ridiculous to use $g = 9.80665 m/s^2$ because the value of $g$ depends on where you are in earth. Using $9.81$ is enough. :)