Chapter 12, Problem 1PE

Interpretation Introduction

Interpretation:

Blank values in table below have to be determined.

$\begin{array}{llll}\hfill & \text{torr}\hfill & \text{in}\text{. Hg}\hfill & \text{kilopascals }\left(\text{kPa}\right)\hfill \\ \left(\text{a}\right)\hfill & \hfill & 30.2\hfill & \hfill \\ \left(\text{b}\right)\hfill & 752\hfill & \hfill & \hfill \\ \left(\text{c}\right)\hfill & \hfill & \hfill & 99.3\hfill \end{array}$

Concept Introduction:

Barometer is used for measurement of pressure of gas in atmospheric units $\left(\text{atm}\right)$. Pressure can also be represented in many units such as $\text{torr}$, $\text{mm Hg}$, and pascal $\left(\text{Pa}\right)$. These units can be related to one another as follows:

  $\begin{array}{c}1\text{ atm}=760\text{ torr}\\ =760\text{ mm Hg}\\ =\text{101325 Pa}\\ =\text{101}\text{.325 kPa}\end{array}$

Explanation of Solution

Conversion factor to convert $\text{mm Hg}$ to $\text{torr}$ is as follows:

  $\left(\frac{1\text{ torr}}{1\text{ mm Hg}}\right)$

Pressure $30.2\text{ in}\text{. Hg}$ can be converted to $\text{torr}$ as follows:

  $\begin{array}{c}\text{Pressure}\left(\text{torr}\right)=\left(30.2\text{ in}\text{. Hg}\right)\left(\frac{1\text{ torr}}{1\text{ mm Hg}}\right)\left(\frac{\text{25}\text{.4 mm Hg}}{1\text{ in}\text{. Hg}}\right)\\ =767\text{ torr}\end{array}$

Conversion factor to convert $\text{torr}$ to $\text{mm Hg}$ is as follows:

  $\left(\frac{1\text{ mm Hg}}{1\text{ torr}}\right)$

Pressure $\text{752 torr}$ can be converted to $\text{in}\text{. Hg}$ as follows:

  $\begin{array}{c}\text{Pressure}\left(\text{in}\text{. Hg}\right)=\left(\text{752 torr}\right)\left(\frac{1\text{ mm Hg}}{1\text{ torr}}\right)\left(\frac{1\text{ in}\text{. Hg}}{\text{25}\text{.4 mm Hg}}\right)\\ =29.6\text{ in}\text{. Hg}\end{array}$

Conversion factor to convert $\text{mm Hg}$ to $\text{kPa}$ is as follows:

  $\left(\frac{101.325\text{ kPa}}{760\text{ mm Hg}}\right)$

Pressure $30.2\text{ in}\text{. Hg}$ can be converted to $\text{kPa}$ as follows:

  $\begin{array}{c}\text{Pressure}\left(\text{kPa}\right)=\left(30.2\text{ in}\text{. Hg}\right)\left(\frac{101.325\text{ kPa}}{760\text{ mm Hg}}\right)\left(\frac{\text{25}\text{.4 mm Hg}}{1\text{ in}\text{. Hg}}\right)\\ =102\text{ kPa}\end{array}$

Pressure $29.6\text{ in}\text{. Hg}$ can be converted to $\text{kPa}$ as follows:

  $\begin{array}{c}\text{Pressure}\left(\text{kPa}\right)=\left(29.6\text{ in}\text{. Hg}\right)\left(\frac{101.325\text{ kPa}}{760\text{ mm Hg}}\right)\left(\frac{\text{25}\text{.4 mm Hg}}{1\text{ in}\text{. Hg}}\right)\\ =100\text{ kPa}\end{array}$

Conversion factor to convert $\text{kPa}$ to $\text{torr}$ is as follows:

  $\left(\frac{760\text{ torr}}{101.325\text{ kPa}}\right)$

Pressure $99.3\text{ kPa}$ can be converted to $\text{kPa}$ as follows:

  $\begin{array}{c}\text{Pressure}\left(\text{torr}\right)=\left(99.3\text{ kPa}\right)\left(\frac{760\text{ torr}}{101.325\text{ kPa}}\right)\\ =745\text{ torr}\end{array}$

Pressure $\text{745 torr}$ can be converted to $\text{in}\text{. Hg}$ as follows:

  $\begin{array}{c}\text{Pressure}\left(\text{in}\text{. Hg}\right)=\left(\text{745 torr}\right)\left(\frac{1\text{ mm Hg}}{1\text{ torr}}\right)\left(\frac{1\text{ in}\text{. Hg}}{\text{25}\text{.4 mm Hg}}\right)\\ =29.3\text{ in}\text{. Hg}\end{array}$

Therefore, table can be filled as follows:

$\begin{array}{llll}\hfill & \text{torr}\hfill & \text{in}\text{. Hg}\hfill & \text{kilopascals }\left(\text{kPa}\right)\hfill \\ \left(\text{a}\right)\hfill & \underset{\_}{767}\hfill & 30.2\hfill & \underset{\_}{102}\hfill \\ \left(\text{b}\right)\hfill & 752\hfill & \underset{\_}{29.6}\hfill & \underset{\_}{100}\hfill \\ \left(\text{c}\right)\hfill & \underset{\_}{745}\hfill & \underset{\_}{29.3}\hfill & 99.3\hfill \end{array}$