Chapter 6, Problem 6.180AP

Interpretation Introduction

Interpretation: The amount of sodium azide needed to inflate a $40\times 40\times 20\text{cm}$ bag to a pressure of 1.25 atm at a temperature of $20°\text{C}$ and $10°\text{C}$ is to be calculated.

Concept introduction: The ideal gas law states that “the volume of given amount of gas is directly proportional to the number of moles of gas, temperature and inversely proportional to the pressure”.

To determine: The amount of sodium azide needed to inflate a $40\times 40\times 20\text{cm}$ bag to a pressure of 1.25 atm at a temperature of $20°\text{C}$ and $10°\text{C}$.

Answer to Problem 6.180AP

Solution

The amount of sodium azide needed to inflate a $40\times 40\times 20\text{cm}$ bag to a pressure of 1.25 atm at a temperature of $20°\text{C}$ and $10°\text{C}$ is $\underset{\_}{10.8\times 10^{5}g}$ and $\underset{\_}{11.1\times 10^{5}g}$ respectively.

Explanation of Solution

Explanation

The ideal gas equation is,

$PV=n\text{R}T$ (1)

Where,

• $P$ is the pressure $1.25\text{atm}$.
• $V$ is the volume $320,000\text{L}$.
• $T$ is the absolute temperature.
• $n$ is the numbers of moles.
• $\text{R}$ is the universal gas constant $\left(\text{0}\text{.08206}\text{L}\cdot \text{atm/(mol}\cdot \text{K)}\right)$.

Rearrange the above expression to calculate the number of moles.

$n=\frac{PV}{\text{R}T}$ (2)

The conversion of volume of gas from centimeters to liters is done as,

$\begin{array}{c}V=\left(40\times 40\times 20\right)\text{cm}\\ =32000{\text{cm}}^3\\ =320{\text{m}}^3\end{array}$

$\begin{array}{c}1\text{cubic}\text{meter}=1000\text{L}\\ 320\text{cubic}\text{meter}\left(V\right)=320\text{cubic}\text{meter}\times \frac{1000\text{L}}{1\text{cubic}\text{meter}}\\ =320,000\text{L}\end{array}$

The conversion of temperature from $°\text{C}\text{to}\text{K}$ is done as,

$K=°C+273.15$

Substitute the given value of temperature in $°C$ in the above expression.

$\begin{array}{c}K=°C+273.15\\ =20°\text{C}+273.15\\ =293.15\text{K}\\ \approx 293\text{K}\end{array}$

Substitute these values of $P$, $T$ $\text{R}$ and $V$ in equation (1).

$\begin{array}{c}PV=n\text{R}T\\ n=\frac{PV}{\text{R}T}\\ =\frac{1.25\text{atm}\times \text{320000}\text{L}}{\text{0}\text{.08206}\text{L}\cdot \text{atm/(mol}\cdot \text{K)}\times 293\text{K}}\\ =16,636.45\text{mol}\end{array}$

The molar mass of ${\text{NaN}}_{\text{3}}$ is calculated by the formula,

$\begin{array}{c}\text{Molar}\text{mass}\text{of}{\text{NaN}}_{\text{3}}=\text{Na}+3\text{N}\\ =\left(1\times 22.98\right)\text{g}+\left(3\times 14.01\right)\text{g}\\ =65.00\text{mol}\end{array}$

The conversion of number of moles into grams is done as,

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{into}\text{grams}=\text{Molar}\text{mass}\times \text{Number}\text{of}\text{moles}\\ =65.00\text{mol}\times 16,636.45\text{mol}\\ =\underset{\_}{10.8\times 10^{5}g}\end{array}$

Similarly for $T=10°\text{C}$ the amount of azide is calculated by the formula,

$PV=n\text{R}T$ (1)

Where,

• $P$ is the pressure $1.25\text{atm}$.
• $V$ is the volume $320,000\text{L}$.
• $T$ is absolute temperature $283\text{K}$.
• $n$ is the numbers of moles.
• $\text{R}$ is universal gas constant $0.08206$.

Rearrange the above expression,

$n=\frac{PV}{\text{R}T}$ (2)

The conversion of volume of gas in Centimeters to liters

$\begin{array}{c}V=\left(40\times 40\times 20\right)\text{cm}\\ =32000{\text{cm}}^3\\ =320{\text{m}}^3\end{array}$

For $T=10°\text{C}$

The conversion of temperature from $°\text{C}\text{to}\text{K}$ is done as,

$K=°C+273.15$

Substitute the given value of temperature in $°C$ in the above expression.

$\begin{array}{c}K=°C+273.15\\ =10°\text{C}+273.15\\ =283.15\text{K}\\ \approx 283\text{K}\end{array}$

Substitute these values of $P$, $T$ $\text{R}$ and $V$ in the equation (2).

$\begin{array}{c}PV=n\text{R}T\\ n=\frac{PV}{\text{R}T}\\ =\frac{1.25\text{atm}\times \text{320000}\text{L}}{\text{0}\text{.08206}\text{L}\cdot \text{atm/(mol}\cdot \text{K)}\times 283\text{K}}\\ =17,224.30\text{mol}\end{array}$

Number of moles into grams $\begin{array}{c}=\text{Molar}\text{mass}\times \text{Number}\text{of}\text{moles}\\ =65.00\text{mol}\times 17,224.30\text{mol}\\ =\underset{\_}{11.1\times 10^{5}g}\end{array}$

Therefore $\underset{\_}{10.8\times 10^{5}g}$ and $\underset{\_}{11.1\times 10^{5}g}$ of sodium azide are needed to inflate $40\times 40\times 20\text{cm}$ bag to pressure of $1.25\text{atm}$ at temperature of $20°\text{C}\text{and}\text{10}°\text{C}$ respectively.

Conclusion

The amount of sodium azide needed to inflate a $40\times 40\times 20\text{cm}$ bag to a pressure of 1.25 atm at a temperature of $20°\text{C}$ and $10°\text{C}$ is $\underset{\_}{10.8\times 10^{5}g}$ and $\underset{\_}{11.1\times 10^{5}g}$ respectively