# 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?

3,651

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, 2022

• 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 water-like 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
This is basic hydrostatics, so it's technically physics, not chemistry. The problem has nothing to do with ideal gasses, you solve it simply by a force balance. See here and here for the theory requirements needed to solve your problem. Otherwise, try posting on Physics.SE.
• • • This 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. :-)