Chapter 12, Problem 14PE

(a)

Interpretation Introduction

Interpretation:

Final volume of ${\text{CH}}_4$ gas if temperature is changed from $-25°\text{C}$ to $298\text{ K}$ 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 14PE

Final volume of ${\text{CH}}_4$ gas is $0.690\text{ L}$.

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 $-25°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{temperature}\text{in}\text{celsius}+273\\ =-25°\text{C}+273\\ =248\text{K}\end{array}$

Substitute $575\text{ mL}$ for $V_1$, $248\text{K}$ for $T_1$, and $298\text{K}$ for $T_2$ in equation (1).

  $\begin{array}{c}{V}_2=\left(\frac{\left(575\text{ mL}\right)\left(298\text{K}\right)}{248\text{K}}\right)\left(\frac{1\text{ L}}{{10}^3\text{ mL}}\right)\\ =0.690\text{ L}\end{array}$

Hence final volume of ${\text{CH}}_4$ gas is $0.690\text{ L}$.

(b)

Interpretation Introduction

Interpretation:

Final volume of ${\text{CH}}_4$ gas if temperature is changed from $-25°\text{C}$ to $32°\text{F}$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 14PE

Final volume of ${\text{CH}}_4$ gas is $0.633\text{ L}$.

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 $-25°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{temperature}\text{in}\text{celsius}+273\\ =-25°\text{C}+273\\ =248\text{K}\end{array}$

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 $32°\text{F}$ in equation (2).

  $\begin{array}{c}T\left(°\text{C}\right)=\frac{5}{9}\left(32°\text{F}-32\right)\\ =0°\text{C}\end{array}$

Conversion factor to convert $0°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{temperature}\text{in}\text{celsius}+273\\ =0°\text{C}+273\\ =273\text{K}\end{array}$

Substitute $575\text{ mL}$ for $V_1$, $248\text{K}$ for $T_1$, and $273\text{K}$ for $T_2$ in equation (1).

  $\begin{array}{c}{V}_2=\left(\frac{\left(575\text{ mL}\right)\left(273\text{K}\right)}{248\text{K}}\right)\left(\frac{1\text{ L}}{{10}^3\text{ mL}}\right)\\ =0.633\text{ L}\end{array}$

Hence final volume of ${\text{CH}}_4$ gas is $0.633\text{ L}$.

(c)

Interpretation Introduction

Interpretation:

Final volume of ${\text{CH}}_4$ gas if temperature is decreased from $-25°\text{C}$ to $45°\text{C}$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 14PE

Final volume of ${\text{CH}}_4$ gas is $0.737\text{ L}$.

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 $-25°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{temperature}\text{in}\text{celsius}+273\\ =-25°\text{C}+273\\ =248\text{K}\end{array}$

Conversion factor to convert $45°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{temperature}\text{in}\text{celsius}+273\\ =45°\text{C}+273\\ =318\text{K}\end{array}$

Substitute $575\text{ mL}$ for $V_1$, $248\text{K}$ for $T_1$, and $318\text{K}$ for $T_2$ in equation (1).

  $\begin{array}{c}{V}_2=\left(\frac{\left(575\text{ mL}\right)\left(318\text{K}\right)}{248\text{K}}\right)\left(\frac{1\text{ L}}{{10}^3\text{ mL}}\right)\\ =0.737\text{ L}\end{array}$

Hence final volume of ${\text{CH}}_4$ gas is $0.737\text{ L}$.