Chapter 13, Problem 14A

(a)

Interpretation Introduction

Interpretation:

The volume of the gas and temperature at various conditions has to be calculated.

Concept Introduction:

Charles’s law: The volume of a given amount of gas is proportional to its temperature (in K) of the gas at constant pressure.

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

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

Charles’s law can also be expressed in mathematical form as-

  $\frac{{\mathrm{V}}_{\text{1}}}{{\mathrm{T}}_{\text{1}}}\mathrm{}=\mathrm{}\frac{{\mathrm{V}}_{\text{2}}}{{\mathrm{T}}_{\text{2}}}$

In the above equation, ${\mathrm{V}}_1$ is initial volume and ${\mathrm{T}}_1$ is initial temperature, while ${\mathrm{V}}_2$ is final or new volume and ${\mathrm{T}}_2$ is final or new temperature.

Answer to Problem 14A

Temperature $=167.38°\text{C}$

Explanation of Solution

Given Information:

  ${\mathrm{V}}_1=2.03\text{ L}$

  $\begin{array}{c}{\mathrm{T}}_1=24\text{°C }\\ \text{= }\left(\text{273 + 24}\right)\text{ K }\\ \text{= 297 K}\end{array}$

  ${\mathrm{V}}_2=3.01\text{ L}$

Calculation:

  $\begin{array}{l}{\mathrm{V}}_2=\frac{{\mathrm{V}}_1\times {\mathrm{T}}_2}{{\mathrm{T}}_1}\mathrm{}\\ =\frac{3.01\mathrm{L}\times 297\mathrm{K}}{2.03\mathrm{L}}\\ =440.38\mathrm{K}=167.38°\mathrm{C}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The volume of the gas and temperature at various conditions has to be calculated.

Concept Introduction:

Charles’s law: The volume of a given amount of gas is proportional to its temperature (in K) of the gas at constant pressure.

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

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

Charles’s law can also be expressed in mathematical form as-

  $\frac{{\mathrm{V}}_{\text{1}}}{{\mathrm{T}}_{\text{1}}}\mathrm{}=\mathrm{}\frac{{\mathrm{V}}_{\text{2}}}{{\mathrm{T}}_{\text{2}}}$

In the above equation, ${\mathrm{V}}_1$ is initial volume and ${\mathrm{T}}_1$ is initial temperature, while ${\mathrm{V}}_2$ is final or new volume and ${\mathrm{T}}_2$ is final or new temperature.

Answer to Problem 14A

Volume of the gas $=173.52\text{ mL}$ .

Explanation of Solution

Given Information:

  $\begin{array}{l}{\mathrm{V}}_1=127\text{ mL}\hfill \\ {\mathrm{T}}_1=273\text{ K}\hfill \\ {\mathrm{T}}_2=373\text{ K}\hfill \end{array}$

Calculation:

  $\begin{array}{l}{\mathrm{V}}_2=\frac{{\mathrm{V}}_1\times {\mathrm{T}}_2}{{\mathrm{T}}_1}\mathrm{}\\ =\frac{127\mathrm{m}\mathrm{L}\times 373\mathrm{K}}{273\mathrm{K}}=173.52\mathrm{m}\mathrm{L}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The volume of the gas and temperature at various conditions has to be calculated.

Concept Introduction:

Charles’s law: The volume of a given amount of gas is proportional to its temperature (in K) of the gas at constant pressure.

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

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

Charles’s law can also be expressed in mathematical form as-

  $\frac{{\mathrm{V}}_{\text{1}}}{{\mathrm{T}}_{\text{1}}}\mathrm{}=\mathrm{}\frac{{\mathrm{V}}_{\text{2}}}{{\mathrm{T}}_{\text{2}}}$

In the above equation, ${\mathrm{V}}_1$ is initial volume and ${\mathrm{T}}_1$ is initial temperature, while ${\mathrm{V}}_2$ is final or new volume and ${\mathrm{T}}_2$ is final or new temperature.

Answer to Problem 14A

Volume of the gas $=56.66\text{ mL}.$

Explanation of Solution

Given Information:

  $\begin{array}{l}{\mathrm{V}}_1=49.7\text{ mL}\hfill \\ \begin{array}{c}{\mathrm{T}}_1=34°\text{C }\\ =\left(273+34\right)\text{ K }\\ =307\text{ K}\end{array}\hfill \\ {\mathrm{T}}_2=350\text{ K}\hfill \end{array}$

Calculation:

  $\begin{array}{l}{\mathrm{V}}_2=\frac{{\mathrm{V}}_1\times {\mathrm{T}}_2}{{\mathrm{T}}_1}\mathrm{}\\ =\frac{49.7\mathrm{m}\mathrm{L}\times 350\mathrm{K}}{307\mathrm{K}}=56.66\mathrm{m}\mathrm{L}\end{array}$