Chapter 3, Problem 3B.5E

(a)

Interpretation Introduction

Interpretation:

Pressure of $0.15\text{mol}$ of argon in flask of volume $350\text{mL}$ has to be determined.

Concept Introduction:

An ideal gas contains a large number of randomly moving particles that are supposed to have perfectly elastic collisions among themselves. It is a theoretical concept. Gases that show perfect elastic collision are practically not possible. At higher $T$ and lower $P$ gases behave almost ideally.

Ideal gas law is an equation of state for any hypothetical gas. Expression for ideal gas equation is as follows:

  $PV=nRT$

Here,

$P$ is pressure of gas.

$V$ is volume of gas.

$n$ denotes moles of gas.

$R$ is gas constant.

$T$ is temperature of gas.

Answer to Problem 3B.5E

Pressure of $0.15\text{mol}$ of argon in flask of volume $350\text{mL}$ is $1058.8462\text{kPa}$.

Explanation of Solution

Ideal gas equation is as follows:

  $PV=nRT$        (1)

Rearrange equation (1) to calculate $P$.

  $P=\frac{nRT}{V}$        (2)

The conversion factor to convert $24\text{°C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{temperature}\text{in}\text{Kelvin}=\text{temperature}\text{in}\text{celsius}+273.15\\ =24\text{°C}+273.15\\ =297.15\text{K}\end{array}$

Substitute $0.15\text{mol}$ for $n$, $8.31446\text{L}\cdot \text{kPa}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$, $350\text{mL}$ for $V$ and $297.15\text{K}$ for $T$ in equation (2) to calculate $P$.

  $\begin{array}{c}P=\frac{\left(0.15\text{mol}\right)\left(8.31446\text{L}\cdot \text{kPa}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(297.15\text{K}\right)}{350\text{mL}\left(\frac{0.001\text{L}}{1\text{mL}}\right)}\\ =1058.8462\text{kPa}\end{array}$

Hence, pressure of $0.15\text{mol}$ of argon in flask of volume $350\text{mL}$ is $1058.8462\text{kPa}$.

(b)

Interpretation Introduction

Interpretation:

Volume of container in $\text{mL}$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 3B.5E

Volume of container in $\text{mL}$ is $404.891\text{mL}$.

Explanation of Solution

Expression to calculate moles of ${\text{BrF}}_3$ is as follows:

  ${\text{moles of BrF}}_3=\left(\frac{{\text{mass of BrF}}_3}{\text{molecular weight }\text{}{\text{of BrF}}_3}\right)$        (3)

Substitute $23.9\text{mg}$ for mass of ${\text{BrF}}_3$ and $136.9 \text{g/mol}$ for molar mass of ${\text{BrF}}_3$ in equation (3).

  $\begin{array}{c}{\text{moles of BrF}}_3=\left(\frac{23.9\text{mg}\left(\frac{0.001\text{g}}{\text{1}\text{mg}}\right)}{136.9 \text{g/mol}}\right)\\ =0.000174\text{mol}\end{array}$

Rearrange equation (1) to calculate $V$.

  $V=\frac{nRT}{P}$        (4)

The conversion factor to convert $100\text{°C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{temperature}\text{in}\text{Kelvin}=\text{temperature}\text{in}\text{celsius}+273.15\\ =100\text{°C}+273.15\\ =373.15\text{K}\end{array}$

Substitute $0.000174\text{mol}$ for $n$, $62.36\text{L}\cdot \text{Torr}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$, $10\text{torr}$ for $P$ and $373.15\text{K}$ for $T$ in equation (3) to calculate $P$.

  $\begin{array}{c}V=\frac{\left(0.000174\text{mol}\right)\left(62.36\text{L}\cdot \text{Torr}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(373.15\text{K}\right)}{10\text{torr}}\\ =0.404891\text{L}\left(\frac{1000\text{mL}}{1\text{L}}\right)\\ =404.891\text{mL}\end{array}$

Volume of container in $\text{mL}$ is $404.891\text{mL}$.

(c)

Interpretation Introduction

Interpretation:

Mass of ${\text{SO}}_2$ present has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 3B.5E

Mass of ${\text{SO}}_2$ present in flask is $\text{198}\text{.2842}\text{g}$.

Explanation of Solution

The conversion factor to convert $30\text{°C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{temperature}\text{in}\text{Kelvin}=\text{temperature}\text{in}\text{celsius}+273.15\\ =30\text{°C}+273.15\\ =303.15\text{K}\end{array}$

Rearrange equation (1) to calculate $n$.

  $n=\frac{PV}{RT}$        (5)

Substitute $100\text{L}$ for $V$, $8.205\times {10}^{-2}\text{L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$, $0.77\text{atm}$ for $P$ and $303.15\text{K}$ for $T$ in equation (5) to calculate $n$.

  $\begin{array}{c}n=\frac{\left(0.77\text{atm}\right)\left(100\text{L}\right)}{\left(8.205\times {10}^{-2}\text{L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(303.15\text{K}\right)}\\ =3.095\text{mol}\end{array}$

Moles of ${\text{SO}}_2$ is $3.095\text{mol}$.

Expression to calculate moles of ${\text{SO}}_2$ is as follows:

  ${\text{moles of SO}}_2=\left(\frac{{\text{mass of SO}}_2}{\text{molecular weight }\text{}{\text{of SO}}_2}\right)$        (6)

Rearrange equation (6) to calculate mass of ${\text{SO}}_2$.

  ${\text{mass of SO}}_2=\left({\text{moles of SO}}_2\right)\left(\text{molecular weight }\text{}{\text{of SO}}_2\right)$        (7)

Substitute $3.095\text{mol}$ for moles of ${\text{SO}}_2$ and $64.066 \text{g/mol}$ for molecular weight of ${\text{SO}}_2$ in equation (7).

  $\begin{array}{c}{\text{mass of SO}}_2=\left(3.095\text{mol}\right)\left(64.066 \text{g/mol}\right)\\ =\text{198}\text{.2842}\text{g}\end{array}$

Hence, mass of ${\text{SO}}_2$ is $\text{198}\text{.2842}\text{g}$.

(d)

Interpretation Introduction

Interpretation:

Moles of ${\text{CH}}_4$ at $129\text{kPa}$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 3B.5E

Moles of ${\text{CH}}_4$ is $324206.8\text{mol}$.

Explanation of Solution

The conversion factor to convert $14\text{°C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{temperature}\text{in}\text{Kelvin}=\text{temperature}\text{in}\text{celsius}+273.15\\ =14\text{°C}+273.15\\ =287.15\text{K}\end{array}$

Substitute $6\times {10}^3{\text{m}}^3$ for $V$, $8.314{\text{m}}^3\cdot \text{Pa}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$, $129\text{kPa}$ for $P$ and $287.15\text{K}$ for $T$ in equation (7) to calculate $n$.

  $\begin{array}{c}n=\frac{\left(129\text{kPa}\right)\left(\frac{{10}^3\text{Pa}}{1\text{kPa}}\right)\left(6\times {10}^3{\text{m}}^3\right)}{\left(8.314{\text{m}}^3\cdot \text{Pa}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(287.15\text{K}\right)}\\ =324206.8\text{mol}\end{array}$

Hence, moles of ${\text{CH}}_4$ is $324206.8\text{mol}$.