Chapter 13, Problem 8A

(a)

Interpretation Introduction

Interpretation:

The missing quantity has to be found.

Concept Introduction:

Boyle’s law: The pressure of a gas is inversely proportional to the volume of the gas at constant temperature.

The relationship between pressure and volume of a gas is expressed as-

  $\begin{array}{l}\mathrm{P}\propto \frac{1}{\mathrm{V}}\\ \mathrm{or},\mathrm{P}\mathrm{V}\mathrm{}=\mathrm{}\mathrm{C}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}\end{array}$ When T is constant.

The equation of Boyle’s law at constant temperature and amount of gas is

  ${\mathrm{P}}_{\text{1}}{\mathrm{V}}_{\text{1}}={\mathrm{P}}_{\text{2}}{\mathrm{V}}_{\text{2}}$

Initial pressure = ${\mathrm{P}}_1$   Final pressure $={\mathrm{P}}_2$

Initial volume $={\mathrm{V}}_1$  Final volume $={\mathrm{V}}_2$

Answer to Problem 8A

Volume of the gas $=145.5\text{ mL}.$

Explanation of Solution

Given Information:

  $\begin{array}{l}{\mathrm{P}}_1=1.07\text{ atm}\hfill \\ {\mathrm{P}}_2=2.14\text{ atm}\hfill \\ {\mathrm{V}}_1=291\text{ mL}\hfill \end{array}$

Calculation:

  $\begin{array}{l}{\mathrm{V}}_{\text{2}}=\frac{{\mathrm{P}}_{\text{1}}{\mathrm{V}}_{\text{1}}}{{\mathrm{P}}_{\text{2}}}\\ =\frac{1.07\mathrm{a}\mathrm{t}\mathrm{m}\times 291\mathrm{m}\mathrm{L}}{2.14\mathrm{a}\mathrm{t}\mathrm{m}}\\ =145.5\mathrm{m}\mathrm{L}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The missing quantity has to be found.

Concept Introduction:

Boyle’s law: The pressure of a gas is inversely proportional to the volume of the gas at constant temperature.

The relationship between pressure and volume of a gas is expressed as-

  $\begin{array}{l}\mathrm{P}\propto \frac{1}{\mathrm{V}}\\ \mathrm{or},\mathrm{P}\mathrm{V}\mathrm{}=\mathrm{}\mathrm{C}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}\end{array}$ When T is constant.

The equation of Boyle’s law at constant temperature and amount of gas is

  ${\mathrm{P}}_{\text{1}}{\mathrm{V}}_{\text{1}}={\mathrm{P}}_{\text{2}}{\mathrm{V}}_{\text{2}}$

Initial pressure = ${\mathrm{P}}_1$   Final pressure $={\mathrm{P}}_2$

Initial volume $={\mathrm{V}}_1$  Final volume $={\mathrm{V}}_2$

Answer to Problem 8A

Volume of the gas $=354\text{ mL}.$

Explanation of Solution

Given Information:

  $\begin{array}{l}\begin{array}{c}{\mathrm{P}}_1=755\text{ mm Hg }\\ \text{= 0}\text{.9934 atm}\end{array}\hfill \\ {\mathrm{P}}_2=3.51\text{ atm}\hfill \\ {\mathrm{V}}_1=1.25\text{ L}\hfill \end{array}$

Calculation:

  $\begin{array}{l}{\mathrm{V}}_{\text{2}}=\frac{{\mathrm{P}}_{\text{1}}{\mathrm{V}}_{\text{1}}}{{\mathrm{P}}_{\text{2}}}\\ =\frac{0.9934\mathrm{a}\mathrm{t}\mathrm{m}\times 1.25\mathrm{L}}{3.51\mathrm{a}\mathrm{t}\mathrm{m}}\\ =0.354\mathrm{L}=354\mathrm{m}\mathrm{L}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The missing quantity has to be found.

Concept Introduction:

Boyle’s law: The pressure of a gas is inversely proportional to the volume of the gas at constant temperature.

The relationship between pressure and volume of a gas is expressed as-

  $\begin{array}{l}\mathrm{P}\propto \frac{1}{\mathrm{V}}\\ \mathrm{or},\mathrm{P}\mathrm{V}\mathrm{}=\mathrm{}\mathrm{C}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}\end{array}$ When T is constant.

The equation of Boyle’s law at constant temperature and amount of gas is

  ${\mathrm{P}}_{\text{1}}{\mathrm{V}}_{\text{1}}={\mathrm{P}}_{\text{2}}{\mathrm{V}}_{\text{2}}$

Initial pressure = ${\mathrm{P}}_1$   Final pressure $={\mathrm{P}}_2$

Initial volume $={\mathrm{V}}_1$  Final volume $={\mathrm{V}}_2$

Answer to Problem 8A

Pressure $=687.04\text{ mm Hg}.$

Explanation of Solution

Given information:

  $\begin{array}{l}\begin{array}{c}{\mathrm{P}}_1=101.4\text{ kPa }\\ \text{= 760}\text{.56 mm Hg}\end{array}\hfill \\ {\mathrm{V}}_1=2.71\text{ L}\hfill \\ {\mathrm{V}}_2=3.00\text{ L}\hfill \end{array}$

Calculation:

  $\begin{array}{l}{\mathrm{V}}_{\text{2}}=\frac{{\mathrm{P}}_{\text{1}}{\mathrm{V}}_{\text{1}}}{{\mathrm{P}}_{\text{2}}}\\ =\frac{760.56\mathrm{m}\mathrm{m}\mathrm{H}\mathrm{g}\times 2.71\mathrm{L}}{3.00\mathrm{L}}\\ =687.04\mathrm{m}\mathrm{m}\mathrm{H}\mathrm{g}\end{array}$