Chapter 6, Problem 6.20QE

(a)

Interpretation Introduction

Interpretation:

The pressure of $\text{3}\text{.2 atm}$ has to be expressed in terms of $\text{torr}$.

Concept Introduction:

Units used to express pressure include Pascal $\left(\text{Pa}\right)$, atmospheric pressure $\left(\text{atm}\right)$, and $\text{torr}$.

SI unit of pressure is Pascal $\left(\text{Pa}\right)$ that is named after the French scientist Blaise Pascal.  It is defined as the pressure that results when $1\text{N}$ force acts on ${\text{1m}}^2$. The unit is more useful for chemist is bar.

The relation between $\text{Pa}$ and ${\text{N/m}}^2$ is as follows:

  $\text{1 Pa}=1{\text{N/1m}}^2$

The relation between $\text{bar}$ and $\text{Pa}$ is as follows:

  $\text{1 bar}={10}^5\text{ Pa}$

The conversion factor to convert $\text{atm}$ to $\text{mm Hg}$ is as follows:

  $1\text{ atm}=\text{760 mm Hg}$

The conversion factor to convert $\text{atm}$ to $\text{mm Hg}$ is as follows:

  $1\text{ atm}=\text{760 mm Hg}$

(b)

Interpretation Introduction

Interpretation:

The pressure of $54.9\text{ atm}$ has to be expressed in terms of $\text{kPa}$.

Concept Introduction:

Refer to part (a).

(c)

Interpretation Introduction

Interpretation:

The pressure of $356\text{ torr}$ has to be expressed in terms of $\text{atm}$.

Concept Introduction:

Refer to part (a).