Chapter 5, Problem 5.22QP

(a)

Interpretation Introduction

Interpretation:

The final volume of given air sample has to be calculated at $\text{6}\text{.6}\text{atm}$.

Concept Introduction:

Boyle’s law states that the volume of a sample of a gas at constant temperature is inversely proportional to pressure

$\begin{array}{l}\text{V=}{\text{k}}_{\text{1}}\frac{\text{1}}{\text{P}}\text{(or)}\\ {\text{P}}_{\text{1}}{\text{V}}_{\text{1}}{\text{= P}}_{\text{2}}{\text{V}}_{\text{2}}\left(\text{at constant temperature}\right)\end{array}$

Temperature constant.

Answer to Problem 5.22QP

The final volume of the given air sample is $\text{0}\text{.69}\text{L}$

Explanation of Solution

Record the Given data

$\begin{array}{l}{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}\text{=}{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}\\ \text{Initial}\text{Conditions}\\ {\text{P}}_{\text{1}}\text{=}1.2\text{atm}\\ {\text{V}}_{\text{1}}\text{=}3.8\text{L}\\ \\ \text{Final}\text{Conditions}\\ {\text{P}}_{\text{2}}\text{=}6.6\text{atm}\end{array}$

The final volume is given by:

$\begin{array}{l}{\text{V}}_{\text{2}}\text{=}\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}}{{\text{P}}_{\text{2}}}\\ {\text{V}}_{\text{2}}\text{=}\frac{\text{(1}\text{.2}\text{atm)}\text{(3}\text{.8}\text{L)}}{\text{(6}\text{.6}\text{atm)}}\text{=}\text{0}\text{.69}\text{L}\end{array}$

The final volume of the given air sample was calculated by plugging in the values of the given pressure $\left({\text{P}}_{\text{1}}\right)$, $\left({\text{P}}_{\text{2}}\right)$ and initial volume $\left({\text{V}}_{\text{1}}\right)$.  The volume was found to be $\text{0}\text{.69}\text{L}$.

(b)

Interpretation Introduction

Interpretation:

The required pressure in order to compress the given air sample to $\text{0}\text{.075 L}$ has to be calculated.

Concept Introduction:

Boyle’s law states that the volume of a sample of a gas at constant temperature is inversely proportional to pressure

$\begin{array}{l}\text{V=}{\text{k}}_{\text{1}}\frac{\text{1}}{\text{P}}\text{(or)}\\ {\text{P}}_{\text{1}}{\text{V}}_{\text{1}}{\text{= P}}_{\text{2}}{\text{V}}_{\text{2}}\left(\text{at constant temperature}\right)\end{array}$

Answer to Problem 5.22QP

The required pressure in order to compress the given air sample to $\text{0}\text{.075 L}$ is $\text{61}\text{atm}$.

Explanation of Solution

Record the Given data

$\begin{array}{l}{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}\text{=}{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}\\ \text{Initial}\text{Conditions}\\ {\text{P}}_{\text{1}}\text{=}\text{1}\text{.2}\text{atm}\\ {\text{V}}_{\text{1}}\text{=}\text{3}\text{.8}\text{L}\\ \\ \text{Final}\text{Conditions}\\ {\text{V}}_{\text{2}}\text{=}\text{0}\text{.075}\text{L}\end{array}$

The final pressure is given by:

$\begin{array}{l}{\text{P}}_{\text{2}}\text{=}\frac{{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}}{{\text{V}}_{\text{2}}}\\ {\text{P}}_{\text{2}}\text{=}\frac{\text{(1}\text{.2}\text{atm)}\text{(3}\text{.8}\text{L)}}{\text{(0}\text{.075}\text{L)}}\text{=}\text{61}\text{atm}\end{array}$

The required pressure was calculated by plugging in the values of the given pressure $\left({\text{P}}_{\text{1}}\right)$ Volume $\left({\text{V}}_{\text{1}}\right)$ and $\left({\text{V}}_{\text{2}}\right)$.  The required pressure was found to be $\text{61}\text{atm}$.