Chapter 13, Problem 15A

(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 15A

Temperature $=-237.6°\text{C}$

Explanation of Solution

Given Information:

  $\begin{array}{l}{\mathrm{V}}_1=2.01\times {10}^2\text{L}\hfill \\ \begin{array}{c}{\mathrm{T}}_1=1150°\text{C }\\ =\left(273+1150\right)\text{ K }\\ \text{= 1423 K}\end{array}\hfill \\ {\mathrm{V}}_2=5.00\text{ L}\hfill \end{array}$

Calculation:

  $\begin{array}{l}{\mathrm{V}}_2=\frac{{\mathrm{V}}_1\times {\mathrm{T}}_2}{{\mathrm{T}}_1}\mathrm{}\mathrm{}\\ =\frac{5.00\mathrm{L}\times 1423\mathrm{K}}{2.01\times \mathrm{}10^2\mathrm{L}}\\ =35.4\mathrm{K}=-237.6°\mathrm{C}\end{array}$

(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 15A

Volume of the gas $=0\text{ mL}$

Explanation of Solution

Given Information:

  $\begin{array}{l}{\mathrm{V}}_1=44.2\text{ mL}\hfill \\ {\mathrm{T}}_1=298\text{ K}\hfill \\ {\mathrm{T}}_2=0\text{ K}\hfill \end{array}$

Calculation:

  $\begin{array}{l}{\mathrm{V}}_2=\frac{{\mathrm{V}}_1\times {\mathrm{T}}_2}{{\mathrm{T}}_1}\mathrm{}\\ =\frac{44.2\mathrm{m}\mathrm{L}\times 0\mathrm{K}}{298\mathrm{K}}=0\mathrm{m}\mathrm{L}\end{array}$

(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 15A

Volume of the gas $=40.5\text{ mL}$

Explanation of Solution

Given Information:

  $\begin{array}{l}{\mathrm{V}}_1=44.2\text{ mL}\hfill \\ {\mathrm{T}}_1=298\text{ K}\hfill \\ \begin{array}{c}{\mathrm{T}}_2=0°\text{C }\\ =273\text{ K}\end{array}\hfill \end{array}$

Calculation:

  $\begin{array}{l}{\mathrm{V}}_2=\frac{{\mathrm{V}}_1\times {\mathrm{T}}_2}{{\mathrm{T}}_1}\mathrm{}\mathrm{}\\ =\frac{44.2\mathrm{m}\mathrm{L}\times 273\mathrm{K}}{298\mathrm{K}}=40.5\mathrm{m}\mathrm{L}\end{array}$