Chapter 5, Problem 5.13QP

Interpretation Introduction

Interpretation:

The unit of $\text{mmHg}$ has to be converted into atm, torr, and kPa.

Concept Introduction:

Standard atmospheric pressure: Standard atmospheric pressure is equal to the pressure that supports a column of mercury exactly $\text{760 mm }\left(\text{or}\text{76}\text{cm}\right)$ high at ${\text{0}}^{\text{o}}\text{C}$ at sea level.

Unit conversion of pressure:

$\begin{array}{l}\text{1 torr = 1 mmHg}\\ \\ \text{1 atm = 760 mmHg}\\ \text{ = 760 torr}\end{array}$

Relation between atmospheres and Pascal is,

$\begin{array}{l}\text{1 atm = 101,325}\text{Pa}\\ \text{ = 1}\text{.01325}\times {10}^5\text{Pa }\left(or\right)\text{1}\text{.01325}\times {10}^2\text{kPa}\end{array}$