Chapter 5.1, Problem 5.2PP

(a)

Interpretation Introduction

Interpretation:

Volume that will be occupied by nitrogen gas at $\text{100}\text{°C}$ has to be calculated if the sample of nitrogen at $\text{25}\text{°C}$ has a volume of $\text{3}\text{.00}\text{L}$.

Concept Introduction:

Charles’s law states that the volume of gas is directly proportional to the absolute temperature if the pressure and number of moles of the gas are kept as constant.  This can be expressed as,

    $\frac{V}{T}=k_c$

During temperature change, the initial and final can be related as,

    $\frac{V_i}{T_i}=\frac{V_f}{T_f}$

Explanation of Solution

Given initial volume ($V_i$) of nitrogen gas is $\text{3}\text{.00}\text{L}$ and initial temperature ($T_i$) is $\text{25}\text{°C}$.  Final temperature ($T_f$) is given as $\text{100}\text{°C}$.

Initial and final temperature conversion into Kelvin scale:

    $\begin{array}{c}{T}_i=\left(25+273.15\right)\text{K}\\ \text{=298}\text{.15}\text{K}\\ T_f=\left(100+273.15\right)\text{K}\\ \text{=373}\text{.15}\text{K}\end{array}$

Charles’s law relating the volume and temperature can be given as,

    $\frac{V_i}{T_i}=\frac{V_f}{T_f}$

Rearranging the above equation to calculate the final volume,

    $V_f=\frac{V_i{T}_f}{T_i}$        (1)

Substituting the values in equation (1),

    $\begin{array}{c}{V}_f=\frac{\left(\text{3}\text{.00}\text{L}\right)\left(\text{373}\text{.15}\text{K}\right)}{\left(\text{298}\text{.15}\text{K}\right)}\\ =3.754\text{L}\end{array}$

Therefore, the volume occupied by nitrogen gas at $\text{100}\text{°C}$ is $\text{3}\text{.754}\text{L}$.

(b)

Interpretation Introduction

Interpretation:

Volume that will be occupied by nitrogen gas at $\text{150}\text{°F}$ has to be calculated if the sample of nitrogen at $\text{25}\text{°C}$ has a volume of $\text{3}\text{.00}\text{L}$.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given initial volume ($V_i$) of nitrogen gas is $\text{3}\text{.00}\text{L}$ and initial temperature ($T_i$) is $\text{25}\text{°C}$.  Final temperature ($T_f$) is given as $\text{150}\text{°F}$.

Initial temperature conversion into Kelvin scale:

    $\begin{array}{c}{T}_i=\left(25+273.15\right)\text{K}\\ \text{=298}\text{.15}\text{K}\end{array}$

Final temperature conversion to Kelvin scale:

    $\begin{array}{c}{T}_f=\left(150-32\right)\times \frac{5}{9}+273.15\\ \text{=338}\text{.70}\text{K}\end{array}$

Charles’s law relating the volume and temperature can be given as,

    $\frac{V_i}{T_i}=\frac{V_f}{T_f}$

Rearranging the above equation to calculate the final volume,

    $V_f=\frac{V_i{T}_f}{T_i}$        (1)

Substituting the values in equation (1),

    $\begin{array}{c}{V}_f=\frac{\left(\text{3}\text{.00}\text{L}\right)\left(\text{338}\text{.70}\text{K}\right)}{\left(\text{298}\text{.15}\text{K}\right)}\\ =3.408\text{L}\end{array}$

Therefore, the volume occupied by nitrogen gas at $\text{150}\text{°F}$ is $\text{3}\text{.408}\text{L}$.

(c)

Interpretation Introduction

Interpretation:

Volume that will be occupied by nitrogen gas at $\text{273}\text{K}$ has to be calculated if the sample of nitrogen at $\text{25}\text{°C}$ has a volume of $\text{3}\text{.00}\text{L}$.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given initial volume ($V_i$) of nitrogen gas is $\text{3}\text{.00}\text{L}$ and initial temperature ($T_i$) is $\text{25}\text{°C}$.  Final temperature ($T_f$) is given as $\text{273}\text{K}$.

Initial temperature conversion into Kelvin scale:

    $\begin{array}{c}{T}_i=\left(25+273.15\right)\text{K}\\ \text{=298}\text{.15}\text{K}\end{array}$

Charles’s law relating the volume and temperature can be given as,

    $\frac{V_i}{T_i}=\frac{V_f}{T_f}$

Rearranging the above equation to calculate the final volume,

    $V_f=\frac{V_i{T}_f}{T_i}$        (1)

Substituting the values in equation (1),

    $\begin{array}{c}{V}_f=\frac{\left(\text{3}\text{.00}\text{L}\right)\left(\text{273}\text{K}\right)}{\left(\text{298}\text{.15}\text{K}\right)}\\ =2.746\text{L}\end{array}$

Therefore, the volume occupied by nitrogen gas at $\text{273}\text{K}$ is $\text{2}\text{.746}\text{L}$.

(d)

Interpretation Introduction

Interpretation:

Volume that will be occupied by nitrogen gas at $\text{546}\text{K}$ has to be calculated if the sample of nitrogen at $\text{25}\text{°C}$ has a volume of $\text{3}\text{.00}\text{L}$.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given initial volume ($V_i$) of nitrogen gas is $\text{3}\text{.00}\text{L}$ and initial temperature ($T_i$) is $\text{25}\text{°C}$.  Final temperature ($T_f$) is given as $\text{546}\text{K}$.

Initial temperature conversion into Kelvin scale:

    $\begin{array}{c}{T}_i=\left(25+273.15\right)\text{K}\\ \text{=298}\text{.15}\text{K}\end{array}$

Charles’s law relating the volume and temperature can be given as,

    $\frac{V_i}{T_i}=\frac{V_f}{T_f}$

Rearranging the above equation to calculate the final volume,

    $V_f=\frac{V_i{T}_f}{T_i}$        (1)

Substituting the values in equation (1),

    $\begin{array}{c}{V}_f=\frac{\left(\text{3}\text{.00}\text{L}\right)\left(\text{546}\text{K}\right)}{\left(\text{298}\text{.15}\text{K}\right)}\\ =5.493\text{L}\end{array}$

Therefore, the volume occupied by nitrogen gas at $\text{546}\text{K}$ is $\text{5}\text{.493}\text{L}$.

(e)

Interpretation Introduction

Interpretation:

Volume that will be occupied by nitrogen gas at $\text{0}\text{°C}$ has to be calculated if the sample of nitrogen at $\text{25}\text{°C}$ has a volume of $\text{3}\text{.00}\text{L}$.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given initial volume ($V_i$) of nitrogen gas is $\text{3}\text{.00}\text{L}$ and initial temperature ($T_i$) is $\text{25}\text{°C}$.  Final temperature ($T_f$) is given as $\text{100}\text{°C}$.

Initial and final temperature conversion into Kelvin scale:

    $\begin{array}{c}{T}_i=\left(25+273.15\right)\text{K}\\ \text{=298}\text{.15}\text{K}\\ T_f=\left(0+273.15\right)\text{K}\\ \text{=273}\text{.15}\text{K}\end{array}$

Charles’s law relating the volume and temperature can be given as,

    $\frac{V_i}{T_i}=\frac{V_f}{T_f}$

Rearranging the above equation to calculate the final volume,

    $V_f=\frac{V_i{T}_f}{T_i}$        (1)

Substituting the values in equation (1),

    $\begin{array}{c}{V}_f=\frac{\left(\text{3}\text{.00}\text{L}\right)\left(\text{273}\text{.15}\text{K}\right)}{\left(\text{298}\text{.15}\text{K}\right)}\\ =2.748\text{L}\end{array}$

Therefore, the volume occupied by nitrogen gas at $\text{0}\text{°C}$ is $\text{2}\text{.748}\text{L}$.

(f)

Interpretation Introduction

Interpretation:

Volume that will be occupied by nitrogen gas at $\text{373}\text{K}$ has to be calculated if the sample of nitrogen at $\text{25}\text{°C}$ has a volume of $\text{3}\text{.00}\text{L}$.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given initial volume ($V_i$) of nitrogen gas is $\text{3}\text{.00}\text{L}$ and initial temperature ($T_i$) is $\text{25}\text{°C}$.  Final temperature ($T_f$) is given as $\text{373}\text{K}$.

Initial temperature conversion into Kelvin scale:

    $\begin{array}{c}{T}_i=\left(25+273.15\right)\text{K}\\ \text{=298}\text{.15}\text{K}\end{array}$

Charles’s law relating the volume and temperature can be given as,

    $\frac{V_i}{T_i}=\frac{V_f}{T_f}$

Rearranging the above equation to calculate the final volume,

    $V_f=\frac{V_i{T}_f}{T_i}$        (1)

Substituting the values in equation (1),

    $\begin{array}{c}{V}_f=\frac{\left(\text{3}\text{.00}\text{L}\right)\left(\text{373}\text{K}\right)}{\left(\text{298}\text{.15}\text{K}\right)}\\ =3.753\text{L}\end{array}$

Therefore, the volume occupied by nitrogen gas at $\text{373}\text{K}$ is $\text{3}\text{.753}\text{L}$.