Question: Suppose a person can reduce the pressure in his lungs to \({\bf{ - 75\;mm}}\)- Hg gauge pressure. How high can water then be "sucked" up a straw?

Whenever a specific solid body is immersed in a liquid, the pressure exerted by the vertical height of the liquid above the measurement level refers as hydrostatic pressure. Similarly, suppose a vacuum generates in a closed volume and connects to the liquid. In that case, the liquid rises in the closed volume so that the pressure difference between the two points becomes equal.

The pressure in the person lungs is \(\Delta P = \left( {\left( {75\;{\rm{mm}} - Hg} \right) \times \frac{{\left( {133.32\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}} \right)}}{{\left( {1\;{\rm{mm}} - Hg} \right)}}} \right) = 9999\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}\).

The expression of water height in straw is given by,

\(\begin{array}{c}\Delta P = \rho gh\\h = \frac{{\Delta P}}{{\rho g}}\end{array}\)

Here, \(\rho \) is the density of water \(\left( {\rho = {{10}^3}\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}} \right)\), \(g\) is the acceleration due to gravity\(\left( {g = 9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\).

Substitute all the known values in the above expression,

\(\begin{array}{c}h = \frac{{\left( {9999\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}} \right)}}{{\left( {{{10}^3}\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)}}\\ \approx 1.02\;{\rm{m}}\end{array}\)

Therefore, the height of water column in the straw is \(1.02\;\;{\rm{m}}\).