Chapter 3, Problem 3A.1AST

Interpretation Introduction

Interpretation:

Atmospheric pressure in $\text{kPa}$ when height of the mercury column in a barometer is $756\text{mm}$ has to be determined.

Concept Introduction:

Force exerted by gas on unit area is pressure. It is can be calculated as follows:

  $P=\frac{F}{A}$

Here,

$\text{P}$ is pressure, $\text{F}$ is force and $\text{A}$ is area of an object.

Pressure of an atmosphere can be measured through barometer by the relation as follows:

  $P=dhg$

Here,

$d$ is density

$h$ is height of mercury.

$g$ is acceleration of free fall.

Answer to Problem 3A.1AST

Atmospheric pressure when $756\text{mm}$ is height of the mercury column in a barometer is $100.784\text{kPa}$.

Explanation of Solution

Pressure of an atmosphere can be measured through barometer by employing the relation as follows:

  $P=dhg$        (1)

Substitute $756\text{mm}$ for $h$, $13595\text{kg}\cdot {\text{m}}^{-3}$ for $d$ and $9.806\text{m}\cdot {\text{s}}^{-2}$ for $g$ in equation (1).

  $\begin{array}{c}P=\left(13595\text{kg}\cdot {\text{m}}^{-3}\right)\left(756\text{mm}\right)\left(\frac{0.001\text{m}}{\text{1}\text{mm}}\right)\left(9.806\text{m}\cdot {\text{s}}^{-2}\right)\\ =100784.303\text{kg}\cdot {\text{m}}^{-\text{1}}\cdot {\text{s}}^{-2}\left(\frac{\text{1}\text{Pa}}{\text{1}\text{kg}\cdot {\text{m}}^{-\text{1}}\cdot {\text{s}}^{-2}}\right)\\ =100784.303\text{Pa}\left(\frac{{10}^{-3}\text{kPa}}{\text{1}\text{Pa}}\right)\\ =100.784\text{kPa}\end{array}$

Therefore, atmospheric pressure when $756\text{mm}$ is height of the mercury column is $100.784\text{kPa}$.