Chapter U3.64, Problem 7E

Interpretation Introduction

Interpretation: The balloon that has greater volume amongst oxygen, ${\text{O}}_{\text{2}}$ and helium, $\text{He}$ under given conditions should be determined.

Concept Introduction: The expression relating the pressure, P, volume, V with number of moles, n and temperature, T of an ideal gas is:

  $\text{PV = nRT}$ - (1)

Where R is Universal gas constant.

Answer to Problem 7E

The two balloons have an equal volume.

Explanation of Solution

Given:

Temperature = ${\text{25 }}^{\text{o}}\text{C}$

Pressure = $\text{1}\text{.2 atm}$

Mass of oxygen, ${\text{O}}_{\text{2}}$ is $\text{16 g}$ .

Mass of helium, $\text{He}$ is $\text{2}\text{.0 g}$ .

Converting the temperature from degree Celsius to Kelvin:

  $\text{25 + 273}\text{.15 = 298}\text{.15 K}$

The molar mass of gas oxygen, ${\text{O}}_{\text{2}}$ and helium, $\text{He}$ is $\text{32 g/mol}$ and $\text{4 g/mol}$ respectively. The number of moles of each gas is calculated using formula:

  $\text{number of moles = }\frac{\text{given mass}}{\text{Molar mass}}$

Number of moles of oxygen, ${\text{O}}_{\text{2}}$

• :

  $\begin{array}{l}{\text{number of moles of O}}_{\text{2}}\text{ = }\frac{\text{16}\text{.0 g}}{\text{32 g/mol}}\\ {\text{number of moles of O}}_{\text{2}}\text{ = 0}\text{.5 mol}\end{array}$

Number of moles of helium, $\text{He}$

• :

  $\begin{array}{l}\text{number of moles of He = }\frac{\text{2}\text{.0 g}}{\text{4 g/mol}}\\ \text{number of moles of He = 0}\text{.5 mol}\end{array}$

Now, in order to determine the volume of each gas, the equation (1) is rearranged as:

  $\text{V = }\frac{\text{nRT}}{\text{P}}$

Using above equation, the volume of each gas is calculated as:

Volume of oxygen, ${\text{O}}_{\text{2}}$

• :

  $\begin{array}{l}{\text{Volume of O}}_{\text{2}}\text{ = }\frac{\text{0}\text{.5 mol}\times \text{0}\text{.0821 Latm/molK}\times \text{298}\text{.15 K}}{1.2\text{ atm}}\\ {\text{Volume of O}}_{\text{2}}\text{ = }\frac{12.239\text{ Latm}}{1.2\text{ atm}}\\ {\text{Volume of O}}_{\text{2}}\text{ = }10.199\text{ L }\simeq \text{ 10}\text{.2 L}\end{array}$

Volume of helium, $\text{He}$

• :

  $\begin{array}{l}\text{Volume of He = }\frac{\text{0}\text{.5 mol}\times \text{0}\text{.0821 Latm/molK}\times \text{298}\text{.15 K}}{1.2\text{ atm}}\\ \text{Volume of He = }\frac{12.239\text{ Latm}}{1.2\text{ atm}}\\ \text{Volume of He = }10.199\text{ L }\simeq \text{ 10}\text{.2 L}\end{array}$

Hence, the two balloons have an equal volume.

Conclusion

The volume occupied by two gases is same.