Chapter 8.1, Problem 8.1E

(a)

Interpretation Introduction

Interpretation:

The given pressures has to be converted from $inHg$ to $torr.$

Concept Introduction:

Pressure of gas is the force applied on a unit area.

  $\text{Pressure}\text{=}\frac{\text{Force}}{\text{Area}}$

Pressure can be expressed in different units

General unit conversions:

  $\begin{array}{l}\text{1}\text{Pa}\text{=}{\text{1Nm}}^{-2}{\text{ = 1 kgm}}^{-1}{s}^{-2};{\text{ 1 kPa = 10}}^{\text{3}}\text{ Pa}\\ \\ \text{1}\text{bar}\text{=}{\text{10}}^{\text{5}}{\text{Pa = 10}}^{\text{3}}{\text{ kPa; mbar = 10}}^{\text{2}}\text{ Pa}\\ \\ \text{1}\text{atm}\text{=}1.01325\times {10}^5\text{ Pa = 101}\text{.325 kPa}\\ \\ 1\text{ atm = 760 Torr = 760}\text{mm}\text{Hg}\\ \\ 1\text{ atm = 14}{\text{.7 lb/in}}^{\text{2}}\left(psi\right)\text{ = 1}\text{.01325 bar}\end{array}$

Unit conversion:

  $1\text{ in = 25}\text{.4 mm = 25}\text{.4}\times {10}^{-3}\text{ m}$

(b)

Interpretation Introduction

Interpretation:

The given pressures has to be converted from $inHg$ to $mbar.$

Concept Introduction:

Refer to part (a).

(c)

Interpretation Introduction

Interpretation:

The given pressures has to be converted from $inHg$ to $kPa.$

Concept Introduction:

Refer to part (a).

(d)

Interpretation Introduction

Interpretation:

The given pressures has to be converted from $inHg$ to $atm.$

Concept Introduction:

Refer to part (a).