Chapter 13, Problem 13.1P

(a)

Interpretation Introduction

Interpretation:

Pressure of $\text{75 atm}$ has to be expressed in terms of $\text{torr}$ and bar.

Concept Introduction:

Units often used in representation of pressure includes Pascal $\left(\text{Pa}\right)$, $\left(\text{atm}\right)$, and $\text{torr}$.

SI unit of pressure is the Pascal $\left(\text{Pa}\right)$ named Blaise Pascal. $\text{1 Pa}$ is defined as the pressure built when $1\text{N}$ force acts on ${\text{1m}}^2$. The more common unit is for chemists 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:

  $1\text{ atm}=\text{760 mm Hg}$

Answer to Problem 13.1P

Pressure of $\text{75 atm}$ in terms of $\text{torr}$ and bar is $57000\text{ torr}$ and $75.99\text{ bar}$ respectively.

Explanation of Solution

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

  $1\text{ atm}=\text{760 torr}$

Thus $\text{75 atm}$ is converted to $\text{torr}$ as follows:

  $\begin{array}{c}\text{Pressure}=\left(\text{75 atm}\right)\left(\frac{760\text{ torr}}{1\text{ atm}}\right)\\ =57000\text{ torr}\end{array}$

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

  $1\text{ atm}=1.01325\text{ bar}$

Thus $1.01325\text{ bar}$ is converted to $\text{torr}$ as follows:

  $\begin{array}{c}\text{Pressure}=\left(\text{75 atm}\right)\left(\frac{1.01325\text{ bar}}{1\text{ atm}}\right)\\ =75.99\text{ bar}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The pressure of $\text{580 torr}$ has to be expressed in terms of $\text{atm}$ and millibar.

Concept Introduction:

Refer to part (a).

Answer to Problem 13.1P

Pressure of $\text{580 torr}$ in terms of $\text{atm}$ and millibar is $0.763\text{ atm}$ and $773.109\text{ mbar}$ respectively.

Explanation of Solution

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

  $1\text{ atm}=\text{760 torr}$

Thus, $\text{580 torr}$ is converted to $\text{atm}$ as follows:

  $\begin{array}{c}\text{Pressure}=\left(\text{580 torr}\right)\left(\frac{1\text{ atm}}{760\text{ torr}}\right)\\ =0.763\text{ atm}\end{array}$

Thus, $0.763\text{ atm}$ is converted to millibaras follows:

  $\begin{array}{c}\text{Pressure}=\left(0.763\text{ atm}\right)\left(\frac{1013.25\text{mbar}}{1\text{ atm}}\right)\\ =773.109\text{ mbar}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The pressure of $\text{5}\text{.2 atm}$ has to be expressed in terms of $\text{kPa}$ and $\text{Pa}$.

Concept Introduction:

Refer to part (a).

Answer to Problem 13.1P

Pressure of $\text{5}\text{.2 atm}$ in terms of $\text{kPa}$ and $\text{Pa}$ is $526,890\text{ Pa}$ and $526.890\text{ kPa}$ respectively.

Explanation of Solution

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

  $1\text{ atm}=101325\text{ Pa}$

Thus, $\text{5}\text{.2 atm}$ is converted to $\text{Pa}$ as follows:

  $\begin{array}{c}\text{Pressure}=\left(\text{5}\text{.2 atm}\right)\left(\frac{101325\text{ Pa}}{1\text{ atm}}\right)\\ =526,890\text{ Pa}\end{array}$

The conversion factor to convert $\text{kPa}$ to $\text{Pa}$ is as follows:

  $1\text{ kPa}={10}^3\text{ Pa}$

Therefore $526,890\text{ Pa}$ is converted to $\text{kPa}$ as follows:

  $\begin{array}{c}\text{Pressure}=\left(526,890\text{ Pa}\right)\left(\frac{1\text{ kPa}}{{10}^3\text{ Pa}}\right)\\ =526.890\text{ kPa}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The pressure of $\text{920 torr}$ has to be expressed in terms of $\text{atm}$ and $\text{Pa}$.

Concept Introduction:

Refer to part (a).

Answer to Problem 13.1P

Pressure of $\text{920 torr}$ in terms of $\text{atm}$ and $\text{Pa}$ is $\text{1}\text{.2105 atm}$ and $122656.579\text{ Pa}$ respectively.

Explanation of Solution

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

  $1\text{ atm}=\text{760 torr}$

Thus $\text{920 torr}$ is converted to $\text{torr}$ as follows:

  $\begin{array}{c}\text{Pressure}=\left(\text{920 torr}\right)\left(\frac{1\text{ atm}}{760\text{ torr}}\right)\\ \text{=1}\text{.2105 atm}\end{array}$

Thus, $\text{1}\text{.2105 atm}$ is converted to $\text{Pa}$ as follows:

  $\begin{array}{c}\text{Pressure}=\left(\text{1}\text{.2105 atm}\right)\left(\frac{101325\text{ Pa}}{1\text{ atm}}\right)\\ =122656.579\text{ Pa}\end{array}$