Chapter 10, Problem 10.82QA

Interpretation Introduction

To find:

The number of balloons that will float in air.

Answer to Problem 10.82QA

Solution:

Only one balloon filled with $He$ gas will float in the air.

Explanation of Solution

1) Concept:

We are given the density of the air, the four gases, the temperature, and the pressure. The ideal gas equation is

$PV=nRT$

$n=\frac{m}{\mathcal{M}}$

where $n=$ number of moles of a gas, $m=$ mass in grams of a gas, $\mathcal{M}=\mathcal{ }$ molar mass of the gas. If we plug the value of $n$ in the ideal gas equation, the equation will be modified into

$PV=\frac{m}{\mathcal{M}} RT$

Rearranging the equation for $m$, we get

$m=\frac{PV\mathcal{M}}{RT}$

So, using this equation, we need to find the mass of each of the gas in the balloons. Adding the mass of empty balloons to the mass of the gas, we will get the mass of the balloon filled with a gas. Using the density formula, we calculate the density of each balloon filled with gas. Comparing the density of each balloon filled with a gas with the density of air will give number of balloons that can float in air. If the density of the balloon is equal or less than the given density of air, the balloon will float. If the density of the balloon filled with a gas is greater than the density of air, the balloon will not float. We note that among the given gases, Helium $\left(He\right)$ has the lowest molar mass. So, the mass of the balloon filled with the gas $He$ is expected to have a density less than the density of air.

2) Formula:

i. $d=\frac{\mathcal{M}\times P}{RT}$

ii.

$m=\frac{P\mathcal{M}V}{RT}$

iii.

$d=\frac{m}{V}$

3) Given:

i) Mass of empty balloon = $10.0 g$

ii) Volume of a balloon = $20.0 L$

iii) Four gases are: $He, Ne, C{O}_{2}and CO$

iv) Density of air = $1.17 g/L$

v) Temperature = $25 ℃=\left(25+273\right)=295 K$

4) Calculations:

Using the above modified formula for mass in grams, we find mass of each gas in the balloon, and then find the mass of balloon as

i) Calculations for Helium balloon:

$m=\frac{PV\mathcal{M}}{RT}$

$m=\frac{1.00 atm \times 20.0 L\times 4.003 g/mol}{0.08206 L·\frac{atm}{mol}·K\times 298K}=3.273 g$

Mass of the balloon filled with Helium gas can be calculated as

$Total mass of balloon=Mass of empty balloon+Mass of balloon filled with the gas$

$Total mass of balloon=10g+3.273g=13.273g$

Calculating density of each of the given gases as

$d=\frac{m}{V}$

$d=\frac{13.273g}{20.0 L}=0.66365 g/L$

Thus, density of $He$ balloon is less than the given density of air. So, helium balloon will float in the air.

ii) Calculations for Neon balloon:

$m=\frac{PV\mathcal{M}}{RT}$

$m=\frac{1.00 atm \times 20.0 L\times 20.18 g/mol}{0.08206 L·\frac{atm}{mol}·K\times 298K}=16.504 g$

Mass of the balloon filled with Neon gas can be calculated as

$Total mass of balloon=Mass of empty balloon+Mass of balloon filled with the gas$

$Total mass of balloon=10g+16.504g=26.504g$

Calculating density of each of the given gases as

$d=\frac{m}{V}$

$d=\frac{26.504g}{20.0 L}=1.3252 g/L$

Thus, density of $Ne$ balloon is greater than the given density of air. So, $Ne$ balloon will not float in air.

iii) Calculations for $C{O}_2$ balloon:

$m=\frac{PV\mathcal{M}}{RT}$

$m=\frac{1.00 atm \times 20.0 L\times 44.01 g/mol}{0.08206 L·\frac{atm}{mol}·K\times 298K}=35.9942 g$

Mass of the balloon filled with carbon dioxide gas can be calculated as

$Total mass of balloon=Mass of empty balloon+Mass of balloon filled with the gas$

$Total mass of balloon=10g+35.9942g=45.9942g$

Calculating density of each of the given gases as

$d=\frac{m}{V}$

$d=\frac{45.9942g}{20.0 L}=2.299 g/L$

Thus, density of $C{O}_2$ balloon is greater than the given density of air. So, $C{O}_2$ balloon will not float in air.

iv) Calculations for $CO$ balloon:

$m=\frac{PV\mathcal{M}}{RT}$

$m=\frac{1.00 atm \times 20.0 L\times 28.01 g/mol}{0.08206 L·\frac{atm}{mol}·K\times 298K}=22.9084 g$

Mass of the balloon filled with carbon monoxide gas can be calculated as

$Total mass of balloon=Mass of empty balloon+Mass of balloon filled with the gas$

$Total mass of balloon=10g+22.9084=32.9084g$

Calculating density of each of the given gases as

$d=\frac{m}{V}$

$d=\frac{32.9084g}{20.0 L}=1.6045 g/L$

Thus, density of $C{O}_{ }$ balloon is greater than the given density of air. So, $C{O}_{ }$ balloon will not float in air.

From above calculated densities of gases filled with different gases, the balloon filled with $He$ gas can only float in air as its density is smaller than the density of the air at the mentioned conditions of the temperature and the pressure.

Conclusion:

From the modified ideal gas equation, we calculated the densities of balloons filled with each gas and compared them to the given density of air. Balloons with the density equal to or less than that of air will float in air.