Chapter 13, Problem 7A

(a)

Interpretation Introduction

Interpretation:

The new volume of the gas sample has to be calculated.

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 7A

Volume of the gas $=121\text{ mL}$

Explanation of Solution

Given Information:

  $\begin{array}{l}{\mathrm{P}}_1=755\text{ mm Hg}\hfill \\ {\mathrm{P}}_2=780\text{ mm Hg}\hfill \\ {\mathrm{V}}_1=125\text{ mL}\hfill \end{array}$

Calculation:

  $\begin{array}{l}{\mathrm{V}}_{\text{2}}=\frac{{\mathrm{P}}_{\text{1}}{\mathrm{V}}_{\text{1}}}{{\mathrm{P}}_{\text{2}}}\\ =\frac{755\mathrm{m}\mathrm{m}\mathrm{H}\mathrm{g}\times 125\mathrm{m}\mathrm{L}}{780\mathrm{m}\mathrm{m}\mathrm{H}\mathrm{g}}\\ =121\mathrm{m}\mathrm{L}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The new volume of the gas sample has to be calculated.

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 7A

Volume of the gas $=253.25\text{ mL}.$

Explanation of Solution

Given Information:

  $\begin{array}{l}{\mathrm{P}}_1=1.08\text{ atm}\hfill \\ {\mathrm{P}}_2=0.951\text{ atm}\hfill \\ {\mathrm{V}}_1=223\text{ mL}\hfill \end{array}$

Calculation:

  $\begin{array}{l}{\mathrm{V}}_{\text{2}}=\frac{{\mathrm{P}}_{\text{1}}{\mathrm{V}}_{\text{1}}}{{\mathrm{P}}_{\text{2}}}\\ =\frac{1.08\mathrm{a}\mathrm{t}\mathrm{m}\times 223\mathrm{m}\mathrm{L}}{0.951\mathrm{a}\mathrm{t}\mathrm{m}}\\ =253.25\mathrm{m}\mathrm{L}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The new volume of the gas sample has to be calculated.

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 7A

Volume of the gas $=2.571\text{ L}$

Explanation of Solution

Given Information:

  $\begin{array}{l}{\mathrm{P}}_1=103\text{ kPa}\hfill \\ {\mathrm{P}}_2=121\text{ kPa}\hfill \\ {\mathrm{V}}_1=3.02\text{ L}\hfill \end{array}$

Calculation:

  $\begin{array}{l}{\mathrm{V}}_{\text{2}}=\frac{{\mathrm{P}}_{\text{1}}{\mathrm{V}}_{\text{1}}}{{\mathrm{P}}_{\text{2}}}\\ =\frac{103\mathrm{k}\mathrm{P}\mathrm{a}\times 3.02\mathrm{L}}{121\mathrm{k}\mathrm{P}\mathrm{a}}\\ =2.571\mathrm{L}\end{array}$