Question

As a hurricane passes through a region, there is an associated drop in atmospheric pressure. If the height of a mercury barometer drops by 21.4 mm from the normal height, what is the atmospheric pressure (in Pa)? Normal atmospheric pressure is 1.013 ✕ 105 Pa and the density of mercury is 13.6 g/cm3 (NOTE: This is g/cm^3, not SI units of kg/m^3).

Answer 1

The atmospheric pressure is
`9.845 * 10^4\ Pa`

Step-by-step explanation:

There are two ways of solving this exercise:

1)

In physics you can find that mmHg is a unit of pressure.

Pressure = Force/area.

If you consider the weight of mercury as your force (mass* acceleration of gravity = density*volume* acceleration of gravity ), then

`P =(\rho Vg)/(A) = (\rho Ahg)/(A) =\rho g h`

where h is the height of the mercury column and rho its density.

`\Delta P= P_(atm) - P_(hur) = 21.4\ mm Hg`.

if normal atmospheric pressure is
`1.013 * 10^5\ Pa = 759.81\ mmHg`

then the pressure in the presence of the hurricane is

`P_(hur) = 759.81 - 21.4 = 738.41\ mmHg = 9.845 * 10^4\ Pa`

2)

Considering the definition of pressure

`\Delta P = \rho g h`

where
`\rho = 13.6\ g/cm^3`,
`g = 9.8\ m/s^2 =980\ cm/s^2` and
`h = 21.4\ mm = 2.14\ cm`.

`\Delta P = P_(atm) - P_(hur) = 28521,92\ g/cms^2 = 2852,192\ kg/ms^2`, where
`kg/ms^2 = Pa`.

if

`P_(atm) = 1.013 * 10^5\ Pa`,

then

`P_(hur) = 1.013 * 10^5 - 2852,192 =9.845 * 10^4\ Pa`.