Question

A normal blood pressure reading is less than 120/80 where both numbers are gauge pressures measured in millimeters of mercury (mmHg). What are the absolute and gauge pressures in pascals at the base of a 0.0930 m column of mercury? (The density of mercury is rhoHg = 13.6 ✕ 103 kg/m3. Assume the system is at sea level.)

Answer 1

The pressure value is given by the equation,

$P= \rho gh$

Where,

$\rho$ represents the density of the liquid

g= gravity

h= Heigth

A) For the measurement of the guage pressure we have the data data,

$\rho_{mercury} = 13.6*10^3kg/m^3$

$h=0.0930m$

$g=9.8m/s^2$

Replacing we get,

$P_g=\rho_{mercury}gh$

$[tex]P_g = (13.6*10^3)(9.8)(0.0930)$P_g = 12395Pa[/tex]

In order to find the Absolute pressure, we perform a sum between the atmospheric pressure and that of the Gauge,

B) The atmospheric pressure at sea level is 101325Pa, assuming ideal conditions, we will take this pressure for our calculation, so

$P_T = P_a+P_g\\P_T = 101325Pa+12395Pa\\P_T = 113720Pa\\P_T = 113.7kPa$