Chapter 23, Problem 81AP

Interpretation Introduction

Interpretation:

The pressure is to be calculated.

Concept Introduction:

The pressure is an expression of force exerted on a surface per unit area.

The relation between pressure and density of a gas is as

$P=\frac{dRT}{M}$.

Here, $P$ is the pressure, $d$ is the density, $R$ is the universal gas constant, $M$ is molar mass, and $T$ is the temperature. The conversion of temperature from degree Celsius to Kelvin can be done by using the formula: $T\left(\text{K}\right)=T\left(°\text{C}\right)+273$.

The density of ${\text{O}}_{\text{2}}$ in ${\text{KO}}_{\text{2}}$ using the mass percentage of ${\text{O}}_{\text{2}}$ in the compound as:

$\frac{\text{mass}\text{of}{\text{O}}_{\text{2}}}{\text{mass}\text{of}{\text{KO}}_{\text{2}}}\text{×}\text{density}$.

The relationship between liters and cubic meters can be expressed as:

$\text{1}\text{L}\text{=}\text{}\text{1000}{\text{cm}}^3$.

To convert cubic meters to liters, conversion factor is $\frac{\text{1}\text{L}}{\text{1000}{\text{cm}}^3}$.

Answer to Problem 81AP

Solution: $727\text{ atm}$

Explanation of Solution

Given information:

Density, $d=2.15\text{ g}/{\text{cm}}^{\text{3}}$

Temperature, $T={20}^{\text{o}}\text{C}$

The value of the gas constant is $\text{0}\text{.0821 L}\cdot \text{atm}/\text{mol}\cdot \text{K}$.

The molar mass of oxygen is $32.00\text{ g}/\text{mol}$.

The molar mass of ${\text{KO}}_{\text{2}}$ is $\text{71}\text{.10 g}/\text{mol}$.

The temperature is $80.1°\text{C}$.

The conversion of temperature from degree Celsius to Kelvin can be done by using the formula given below:

$\begin{array}{c}T\left(\text{K}\right)=T\left(°\text{C}\right)+273\\ =\left(20+273\right)\\ =293\text{ K}\text{.}\end{array}$

Calculate the density of ${\text{O}}_{\text{2}}$ in ${\text{KO}}_{\text{2}}$ using the mass percentage of ${\text{O}}_{\text{2}}$ in the compound as:

$\frac{\text{mass}\text{of}{\text{O}}_{\text{2}}}{\text{mass}\text{of}{\text{KO}}_{\text{2}}}\text{×}\text{density}\text{.}$

$\frac{\text{32}\text{.00}\text{g}{\text{O}}_{\text{2}}}{\text{71}\text{.10}\cancel{\text{g}{\text{KO}}_{\text{2}}}}\text{×}\frac{\text{2}\text{.15}\cancel{{\text{g KO}}_{\text{2}}}}{\text{1}{\text{cm}}^{\text{3}}}\text{=}\text{0}\text{.968}{\text{g O}}_{\text{2}}/{\text{cm}}^{\text{3}}.$

In one liter, there are thousand cubic centimeters.

Convert cubic centimeter to liter as:

$\text{0}\text{.968}\text{g }/{\text{cm}}^{\text{3}}=\left(\frac{\text{0}\text{.968}\text{g }}{\text{1}\cancel{{\text{cm}}^{\text{3}}}}\right)\left(\frac{\text{1000}\cancel{{\text{cm}}^{\text{3}}}}{\text{1}\text{L}}\right)\text{=}\text{968}\text{g }/\text{L}\text{.}$

Calculate the pressure from the relationship between pressure and molar mass as

$P=\frac{dRT}{M}.$

Substitute $\text{0}\text{.0821 L}\cdot \text{atm}/\text{mol}\cdot \text{K}$ for $R$, $\text{968}\text{g }/\text{L}$ for $d$, $293\text{ K}$ for $T$, and $32.00\text{ g}/\text{mol}$ for $M$ in the above equation.

$\begin{array}{c}P=\frac{\text{968}\cancel{\text{g }}/\cancel{\text{L}}\left(\text{0}\text{.0821 }\cancel{\text{L}}\cdot \text{atm}/\cancel{\text{mol}}\cdot \cancel{\text{K}}\right)\text{(293}\cancel{\text{K}}\text{)}}{32.00\cancel{\text{g}}/\cancel{\text{mol}}}\\ =\text{727}\text{atm}\text{.}\end{array}$

Conclusion

The calculated pressureis $\text{727}\text{atm}$.