Chapter 10, Problem 6PE

Interpretation Introduction

Interpretation:

The pressure of the gas in a manometer that consists of mercury liquid at one end, which rises higher than the arm and towards the open air, is to be calculated.

Concept Introduction:

Atmospheric pressure, also called barometric pressure, is the pressure exerted per unit of area of an atmospheric column.

Manometer is a scientific device used to measure the difference in pressure between the pressure of two gases, or that of one gas against the atmospheric pressure.

To convert the unit from $\text{cm Hg}$ to $\text{mm Hg}$, first, convert centimeters into millimeters.

$\begin{array}{l}1\text{ cm}=\text{10 mm}\\ \text{The conversion factor for cm to mm is as:}\\ \frac{10\text{ mm}}{1\text{ cm}}\end{array}$

The required relationship for the conversion of the given atmospheric pressure $770\text{ torr}$ in terms $\text{mm Hg}$ is:

$\begin{array}{l}1\text{ torr}=\text{1 mm Hg}\\ \text{The coversion factor for torr to mm Hg}\\ \frac{\text{1 mm Hg}}{1\text{ torr}}\end{array}$