Chapter 6, Problem 6.18E

Interpretation Introduction

(a)

Interpretation:

The pressure of $\text{28}\text{.6}\text{in}\text{.Hg}$ is to be converted to $\text{atm}$.

Concept introduction:

The mathematical expressions used to demonstrate the behavior of gases are known as gas laws. These expressions are used under the several changes like temperature, pressure etc. The use of these expressions helps to understand the behavior of gases changing under a particular condition.

Answer to Problem 6.18E

The pressure of $\text{28}\text{.6}\text{in}\text{.Hg}$ is equal to $0.957\text{atm}$.

Explanation of Solution

The relation between $\text{inch}\text{Hg}$ and $\text{atm}$ is as follows.

$1\text{atm}=\text{29}\text{.9}\text{in}\text{.Hg}$

The given pressure of $28.6\text{in}\text{.Hg}$ can be converted to $\text{atm}$ as follows.

$\begin{array}{rll}1\text{atm}& =& \text{29}\text{.9}\text{in}\text{.Hg}\\ 1\text{in}\text{.Hg}& =& \frac{1\text{atm}}{\text{29}\text{.9}\text{in}\text{.Hg}}\\ 28.6\text{in}\text{.Hg}& =& \frac{1\text{atm}}{\text{29}\text{.9}\text{in}\text{.Hg}}\times 28.6\text{in}\text{.Hg}\\ & =& 0.957\text{atm}\end{array}$

Conclusion

The pressure of $\text{28}\text{.6}\text{in}\text{.Hg}$$28.6$ is equal to $0.957\text{atm}$.

Interpretation Introduction

(b)

Interpretation:

The pressure of $\text{28}\text{.6}\text{in}\text{.Hg}$ is to be converted to $\text{torr}$.

Concept introduction:

The mathematical expressions used to demonstrate the behavior of gases are known as gas laws. These expressions are used under the several changes like temperature, pressure etc. The use of these expressions helps to understand the behavior of gases changing under a particular condition.

Answer to Problem 6.18E

The pressure of $\text{28}\text{.6}\text{in}\text{.Hg}$ is equal to $727\text{torr}$.

Explanation of Solution

The relation between $\text{inch}\text{Hg}$ and $\text{atm}$ is as follows.

$1\text{atm}=\text{29}\text{.9}\text{in}\text{.Hg}$

The relation between $\text{torr}$ and $\text{atm}$ is as follows.

$1\text{atm}=\text{760}\text{torr}$

Use above two relations and calculate the given pressure in $\text{torr}$ as follows.

$\begin{array}{rll}1\text{atm}& =& \text{29}\text{.9}\text{in}\text{.Hg}\\ 1\text{in}\text{.Hg}& =& \frac{1\text{atm}}{\text{29}\text{.9}\text{in}\text{.Hg}}\times \frac{\text{760}\text{torr}}{\text{1}\text{atm}}\\ 28.6\text{in}\text{.Hg}& =& \frac{1\text{atm}}{\text{29}\text{.9}\text{in}\text{.Hg}}\times \frac{\text{760}\text{torr}}{\text{1}\text{atm}}\times 28.6\text{in}\text{.Hg}\\ & =& 727\text{torr}\end{array}$

Conclusion

The pressure of $\text{28}\text{.6}\text{in}\text{.Hg}$ is equal to $727\text{torr}$.

Interpretation Introduction

(c)

Interpretation:

The pressure of $\text{28}\text{.6}\text{in}\text{.Hg}$ is to be converted to $\text{psi}$.

Concept introduction:

The mathematical expressions used to demonstrate the behavior of gases are known as gas laws. These expressions are used under the several changes like temperature, pressure etc. The use of these expressions helps to understand the behavior of gases changing under a particular condition.

Answer to Problem 6.18E

The pressure of $\text{28}\text{.6}\text{in}\text{.Hg}$ is equal to $\text{14}\text{.1}\text{psi}$.

Explanation of Solution

The relation between $\text{inch}\text{Hg}$ and $\text{atm}$ is as follows.

$1\text{atm}=\text{29}\text{.9}\text{in}\text{.Hg}$

The relation between $\text{psi}$ and $\text{atm}$ is as follows.

$1\text{atm}=\text{14}\text{.7}\text{psi}$

Use above two relations and calculate the given pressure in $\text{psi}$ as follows.

$\begin{array}{rll}1\text{atm}& =& \text{29}\text{.9}\text{in}\text{.Hg}\\ 1\text{in}\text{.Hg}& =& \frac{1\text{atm}}{\text{29}\text{.9}\text{in}\text{.Hg}}\times \frac{\text{14}\text{.7}\text{psi}}{\text{1}\text{atm}}\\ 28.6\text{in}\text{.Hg}& =& \frac{1\text{atm}}{\text{29}\text{.9}\text{in}\text{.Hg}}\times \frac{\text{14}\text{.7}\text{psi}}{\text{1}\text{atm}}\times 28.6\text{in}\text{.Hg}\\ & =& \text{14}\text{.1}\text{psi}\end{array}$

Conclusion

The pressure of $\text{28}\text{.6}\text{in}\text{.Hg}$ is equal to $\text{14}\text{.1}\text{psi}$.

Interpretation Introduction

(d)

Interpretation:

The pressure of $\text{28}\text{.6}\text{in}\text{.Hg}$ is to be converted to $\text{bars}$.

Concept introduction:

The mathematical expressions used to demonstrate the behavior of gases are known as gas laws. These expressions are used under the several changes like temperature, pressure etc. The use of these expressions helps to understand the behavior of gases changing under a particular condition.

Answer to Problem 6.18E

The pressure of $\text{28}\text{.6}\text{in}\text{.Hg}$ is equal to $\text{0}\text{.966}\text{bar}$.

Explanation of Solution

The relation between $\text{inch}\text{Hg}$ and $\text{atm}$ is as follows.

$1\text{atm}=\text{29}\text{.9}\text{in}\text{.Hg}$

The relation between $\text{bars}$ and $\text{atm}$ is as follows.

$1\text{atm}=1.01\text{bar}$

Use above two relations and calculate the given pressure in $\text{bars}$ as follows.

$\begin{array}{rll}1\text{atm}& =& \text{29}\text{.9}\text{in}\text{.Hg}\\ 1\text{in}\text{.Hg}& =& \frac{1\text{atm}}{\text{29}\text{.9}\text{in}\text{.Hg}}\times \frac{1.01\text{bar}}{\text{1}\text{atm}}\\ 28.6\text{in}\text{.Hg}& =& \frac{1\text{atm}}{\text{29}\text{.9}\text{in}\text{.Hg}}\times \frac{1.01\text{bar}}{\text{1}\text{atm}}\times 28.6\text{in}\text{.Hg}\\ & =& \text{0}\text{.966}\text{bar}\end{array}$.

Conclusion

The pressure of $\text{28}\text{.6}\text{in}\text{.Hg}$ is equal to $\text{0}\text{.966}\text{bar}$.