Question

Helium-filled balloons rise in the air because the density of helium is less than the density of air. (a) If air has an average molar mass of \(29.0 \mathrm{~g} / \mathrm{mol}\), what is the density of air at \(25^{\circ} \mathrm{C}\) and \(770 \mathrm{~mm} \mathrm{Hg}\) ? (b) What is the density of helium at the same temperature and pressure? (c) Would a balloon filled with carbon dioxide at the same temperature and pressure rise?

Short answer

Answer: No, a carbon dioxide-filled balloon will not rise as it has a higher density (1.967 g/L) compared to air (1.225 g/L).

Step-by-step solution

Use the ideal gas law equation

To find the density of the gases, we will use the ideal gas law equation, which is \(PV=nRT\). Here, we need to first convert the pressure from \(\mathrm{mm~Hg}\) to \(\mathrm{kPa}\).

Calculate the density formula

Density is defined as mass divided by volume. The density of a gas can be found by using the formula: $$ \rho = \frac{n \times M}{V} $$ where \(\rho\) is the density, \(n\) is the amount of substance in moles, \(M\) is the molar mass, and \(V\) is the volume.

Find the density of air

We are given that the air has an average molar mass of \(29.0~\mathrm{g/mol}\). The temperature is \(25^{\circ}\mathrm{C}\) and pressure is \(770~\mathrm{mm~Hg}\). We now convert the pressure to kPa: $$ P = 770~\mathrm{mm~Hg} \times \frac{101.325~\mathrm{kPa}}{760~\mathrm{mm~Hg}} = 103.507~\mathrm{kPa} $$ We also convert the temperature to Kelvin: $$ T = 25 + 273.15 = 298.15~\mathrm{K} $$ Now we can use the ideal gas law, rearranged to solve for the density: $$ \rho = \frac{PM}{RT} $$ Then, $$ \rho_\mathrm{air} = \frac{103.507~\mathrm{kPa} \times 29.0~\mathrm{g/mol}}{8.314~\mathrm{J/(mol\cdot K)} \times 298.15~\mathrm{K}} = 1.225~\mathrm{g/L} $$ So, the density of air is approximately \(1.225~\mathrm{g/L}\).

Find the density of helium

For helium, we are given that the molar mass is \(4.00~\mathrm{g/mol}\). Using the same temperature and pressure values as in Step 3, we calculate the density of helium: $$ \rho_\mathrm{He} = \frac{103.507~\mathrm{kPa} \times 4.00~\mathrm{g/mol}}{8.314~\mathrm{J/(mol\cdot K)} \times 298.15~\mathrm{K}} = 0.169~\mathrm{g/L} $$ The density of helium is approximately \(0.169~\mathrm{g/L}\).

Find the density of carbon dioxide

We are given that the molar mass of carbon dioxide is \(44.0~\mathrm{g/mol}\). Using the same temperature and pressure conditions as in Step 3, we calculate the density of carbon dioxide: $$ \rho_\mathrm{CO_2} = \frac{103.507~\mathrm{kPa} \times 44.0~\mathrm{g/mol}}{8.314~\mathrm{J/(mol\cdot K)} \times 298.15~\mathrm{K}} = 1.967~\mathrm{g/L} $$ The density of carbon dioxide is approximately \(1.967~\mathrm{g/L}\).

Compare densities to determine if a carbon dioxide-filled balloon would rise

We have now calculated the densities of air, helium, and carbon dioxide. To determine if a balloon filled with carbon dioxide would rise at the same temperature and pressure, we need to compare its density with that of air. As we can see, the density of carbon dioxide is higher than that of air: $$ \rho_\mathrm{CO_2} > \rho_\mathrm{air} $$ 1.967 g/L > 1.225 g/L Since carbon dioxide is denser than air, a balloon filled with carbon dioxide would not rise under the same temperature and pressure conditions as helium balloons do.

Key concepts

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry and physics that relates the pressure \textbf{(P)}, volume \textbf{(V)}, temperature \textbf{(T)} in Kelvin, and amount of substance in moles \textbf{(n)} for an ideal gas. The equation is expressed as \(PV = nRT\), where \(R\) is the universal gas constant, with a value of \(8.314 \frac{J}{(mol \times K)}\). This equation allows us to determine the properties of a gas under various conditions.

The beauty of the ideal gas law lies in its ability to model how gases will behave under different conditions. When one variable, such as temperature or pressure, changes, it affects the others. This is how we can calculate the density of a gas — by rearranging the ideal gas law equation to isolate the desired variable.

Molar Mass

Molar mass is a property of substances that links the macroscopic world we experience to the microscopic world of atoms and molecules. It is the mass of one mole of a substance, typically expressed in \(g/mol\). In our example with air, helium, and carbon dioxide, knowing the molar mass is essential for calculating the densities of these gases.

One mole contains Avogadro's number of entities, about \(6.022 \times 10^{23}\). This means that the molar mass tells us how many grams one mole of a substance weighs, which is equivalent to the mass of \(6.022 \times 10^{23}\) molecules of that substance. For air, which is a mixture of gases, the average molar mass is calculated based on the proportion of each gas present.

Converting Pressure Units

Why Convert Pressure Units?

When dealing with gases, pressures can be measured in various units like atmospheres (atm), millimeters of mercury (mmHg), or kilopascals (kPa), to name a few. It's essential to use the correct units when applying the ideal gas law. Sometimes, this requires converting units to ensure they match those of the gas constant (R).

How to Convert Pressure Units

Conversion ratios, also known as conversion factors, are used to change one unit of pressure to another. In our exercise, we convert pressure from mmHg to kPa since the gas constant is given in \(J/(mol \times K)\), and 1 joule per second is equivalent to 1 Pascal. The conversion factor between mmHg and kPa is \(1 atm = 760 mmHg = 101.325 kPa\), and by multiplying the pressure value by this ratio, we obtain the pressure in the desired units.

Temperature Conversion to Kelvin

Temperature plays a critical role in gas behavior and calculations. However, to use the ideal gas law, we must express the temperature in Kelvin (K), which is the SI unit for thermodynamic temperature. To convert a temperature from Celsius to Kelvin, we add 273.15 to the Celsius temperature.

Keeping temperature in Kelvin ensures consistency when plugged into the ideal gas law, as the gas constant is based on the Kelvin scale. This is because Kelvin is an absolute temperature scale, which means its zero point, 0K, is set at absolute zero, the theoretical lowest possible temperature where particles theoretically stop moving. Hence, temperature in Kelvin reflects the absolute motion of particles, directly related to their kinetic energy. Converting to Kelvin is straightforward: \(T(K) = T(°C) + 273.15\). In the exercises, this conversion is essential to finding the correct density values for the gases in question.