Chapter 12, Problem 13PE

(a)

Interpretation Introduction

Interpretation:

Final volume of ${\text{CO}}_2$ gas if temperature is decreased from $21.0°\text{C}$ to $-5°\text{C}$ has to be determined.

Concept Introduction:

Charles’ law represents direct relation of volume of fixed mass of ideal gas with its temperature at fixed pressure.

Mathematically relation for Charles’ law is as follows:

  $V\propto T$

Or,

  $\frac{V_1}{T_1}=\frac{V_2}{T_2}$

Here,

$T_1$ is initial temperature.

$T_2$ is final temperature.

$V_1$ is initial volume.

$V_2$ is final volume.

Answer to Problem 13PE

Final volume of ${\text{CO}}_2$ gas is $0.114\text{ L}$.

Explanation of Solution

Expression to calculate final volume of gas is as follows:

  $V_2=\frac{V_1{T}_2}{T_1}$        (1)

Conversion factor to convert $21.0°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{temperature}\text{in}\text{celsius}+273\\ =21.0°\text{C}+273\\ =294\text{K}\end{array}$

Conversion factor to convert $-5°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{temperature}\text{in}\text{celsius}+273\\ =-5°\text{C}+273\\ =268\text{K}\end{array}$

Substitute $125\text{ mL}$ for $V_1$, $294\text{K}$ for $T_1$, and $268\text{K}$ for $T_2$ in equation (1).

  $\begin{array}{c}{V}_2=\left(\frac{\left(125\text{ mL}\right)\left(268\text{K}\right)}{294\text{K}}\right)\left(\frac{1\text{ L}}{{10}^3\text{ mL}}\right)\\ =0.114\text{ L}\end{array}$

Hence final volume of ${\text{CO}}_2$ gas is $0.114\text{ L}$.

(b)

Interpretation Introduction

Interpretation:

Final volume of ${\text{CO}}_2$ gas if temperature is changed from $21.0°\text{C}$ to $95°\text{F}$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 13PE

Final volume of ${\text{CO}}_2$ gas is $0.131\text{ L}$.

Explanation of Solution

Expression to calculate final volume of gas is as follows:

  $V_2=\frac{V_1{T}_2}{T_1}$        (1)

Conversion factor to convert $21.0°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{temperature}\text{in}\text{celsius}+273\\ =21.0°\text{C}+273\\ =294\text{K}\end{array}$

Conversion factor to convert degree Fahrenheit to Celsius is as follows:

  $T\left(°\text{C}\right)=\frac{5}{9}\left(T\left(\text{°F}\right)-32\right)$        (2)

Substitute $95°\text{F}$ in equation (2).

  $\begin{array}{c}T\left(°\text{C}\right)=\frac{5}{9}\left(95°\text{F}-32\right)\\ =35°\text{C}\end{array}$

Conversion factor to convert $35°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{temperature}\text{in}\text{celsius}+273\\ =35°\text{C}+273\\ =308\text{K}\end{array}$

Substitute $125\text{ mL}$ for $V_1$, $294\text{K}$ for $T_1$, and $268\text{K}$ for $T_2$ in equation (1).

  $\begin{array}{c}{V}_2=\left(\frac{\left(125\text{ mL}\right)\left(308\text{K}\right)}{294\text{K}}\right)\left(\frac{1\text{ L}}{{10}^3\text{ mL}}\right)\\ =0.131\text{ L}\end{array}$

Hence final volume of ${\text{CO}}_2$ gas is $0.131\text{ L}$.

(c)

Interpretation Introduction

Interpretation:

Final volume of ${\text{CO}}_2$ gas if temperature is changed from $21.0°\text{C}$ to $1095\text{ K}$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 13PE

Final volume of ${\text{CO}}_2$ gas is $0.466\text{ L}$.

Explanation of Solution

Expression to calculate final volume of gas is as follows:

  $V_2=\frac{V_1{T}_2}{T_1}$        (1)

Conversion factor to convert $21.0°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{temperature}\text{in}\text{celsius}+273\\ =21.0°\text{C}+273\\ =294\text{K}\end{array}$

Substitute $125\text{ mL}$ for $V_1$, $294\text{K}$ for $T_1$, and $1095\text{ K}$ for $T_2$ in equation (1).

  $\begin{array}{c}{V}_2=\left(\frac{\left(125\text{ mL}\right)\left(1095\text{ K}\right)}{294\text{K}}\right)\left(\frac{1\text{ L}}{{10}^3\text{ mL}}\right)\\ =0.466\text{ L}\end{array}$

Hence final volume of ${\text{CO}}_2$ gas is $0.466\text{ L}$.