Chapter 3, Problem 3B.22E

(a)

Interpretation Introduction

Interpretation:

Pressure of gas in a full cylinder at $20°\text{C}$ when it empty cylinder weighs $0.74\text{kg}$ 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 the gas.

$V$ is volume of gas.

$n$ denotes moles of gas.

$R$ is gas constant.

$T$ is temperature of gas.

Answer to Problem 3B.22E

Pressure of gas in a full cylinder at $20°\text{C}$ is $664.5868\text{bar}$.

Explanation of Solution

Mass of ${\text{CO}}_2$ in cylinder is calculated as follows:

  $\begin{array}{c}\text{weight}\text{of}{\text{CO}}_2=\left(\text{weight}\text{of full}\text{cylinder}\right)-\left(\text{weight}\text{of}\text{empty cylinder}\right)\\ =\left(1.04\text{kg}\right)-\left(0.74\text{kg}\right)\\ =0.3\text{kg}\left(\frac{{10}^3\text{g}}{\text{1}\text{kg}}\right)\\ =300\text{g}\end{array}$

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

  $\left({\text{moles of CO}}_2\right)=\left(\frac{{\text{mass of CO}}_2}{\text{molar mass }\text{}{\text{of CO}}_2}\right)$        (1)

Substitute $300\text{g}$ for mass of ${\text{CO}}_2$ and $44.01 \text{g/mol}$ for molar mass of ${\text{CO}}_2$ in equation (1) to calculate moles of ${\text{CO}}_2$.

  $\begin{array}{c}\left({\text{moles of CO}}_2\right)=\left(\frac{300\text{g}}{44.01 \text{g/mol}}\right)\\ =6.8166\text{mol}\end{array}$

The conversion factor to convert $20\text{°C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{temperature}\text{in}\text{Kelvin}=\text{temperature}\text{in}\text{celsius}+273.15\\ =20\text{°C}+273.15\\ =293.15\text{K}\end{array}$

Ideal gas equation is as follows:

  $PV=nRT$        (2)

Rearrange equation (2) to calculate $P$.

  $P=\frac{nRT}{V}$        (3)

Substitute $250\text{mL}$ for $V$, $8.31446\times {10}^{-2}\text{ L}\cdot \text{bar}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$, $6.8166\text{mol}$ for $n$ and $293.15\text{K}$ for $T$ in equation (3) to calculate $P$.

  $\begin{array}{c}P=\frac{\left(6.8166\text{mol}\right)\left(8.31446\times {10}^{-2}\text{ L}\cdot \text{bar}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(293.15\text{K}\right)}{\left(250\text{mL}\right)\left(\frac{0.001\text{L}}{1\text{mL}}\right)}\\ =664.5868\text{bar}\end{array}$

Hence, pressure of gas in a full cylinder at $20°\text{C}$ is $664.5868\text{bar}$.

(b)

Interpretation Introduction

Interpretation:

Pressure of gas in a full cylinder at $20°\text{C}$ when it weighs $0.87\text{g}$ has to be determined.

Concept Introduction:

Refer to part (a)

Answer to Problem 3B.22E

Pressure of gas in a full cylinder at $20°\text{C}$ when it weighs $0.87\text{g}$ is $376.6010\text{bar}$.

Explanation of Solution

Mass of ${\text{CO}}_2$ in cylinder is calculated as follows:

  $\begin{array}{c}\text{weight}\text{of}{\text{CO}}_2=\left(\text{weight}\text{of full}\text{cylinder}\right)-\left(\text{weight}\text{of}\text{empty cylinder}\right)\\ =\left(1.04\text{kg}\right)-\left(0.87\text{kg}\right)\\ =0.17\text{kg}\left(\frac{{10}^3\text{g}}{\text{1}\text{kg}}\right)\\ =170\text{g}\end{array}$

Substitute $170\text{g}$ for mass of ${\text{CO}}_2$ and $44.01 \text{g/mol}$ for molar mass of ${\text{CO}}_2$ in equation (1) to calculate moles of ${\text{CO}}_2$.

  $\begin{array}{c}\left({\text{moles of CO}}_2\right)=\left(\frac{170\text{g}}{44.01 \text{g/mol}}\right)\\ =3.8627\text{mol}\end{array}$

Substitute $250\text{mL}$ for $V$, $8.31446\times {10}^{-2}\text{ L}\cdot \text{bar}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$, $3.8627\text{mol}$ for $n$ and $293.15\text{K}$ for $T$ in equation (3) to calculate $P$.

  $\begin{array}{c}P=\frac{\left(3.8627\text{mol}\right)\left(8.31446\times {10}^{-2}\text{ L}\cdot \text{bar}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(293.15\text{K}\right)}{\left(250\text{mL}\right)\left(\frac{0.001\text{L}}{1\text{mL}}\right)}\\ =376.6010\text{bar}\end{array}$

Hence, pressure of gas in a full cylinder when it weighs $0.87\text{g}$ is $376.6010\text{bar}$.