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

Would a carbon dioxide-filled balloon rise? Short Answer: Yes, a helium-filled balloon will rise in air because helium has a lower density than air. However, a carbon dioxide-filled balloon will not rise because its density is greater than the density of air.

Step-by-step solution

Information given by the problem

First, let's write down the information given by the problem for air and helium: - Molar mass of air (\(M_{air}\)) = 29.0 g/mol - Temperature (\(T\)) = 25°C = 298 K (converted to Kelvin) - Pressure (\(P\)) = 770 mm Hg = 101.33 kPa (converted to kilopascals) - Molar gas constant (\(R\)) = 8.314 J/mol*K For helium: - Molar mass of helium (\(M_{He}\)) = 4.00 g/mol

Find the volume of one mole of air and helium

We'll use the ideal gas law to find the volume of one mole of air and helium at the given conditions. The ideal gas law is given by: \(PV = nRT\). Since we are dealing with one mole of each gas, n = 1. Rearrange the equation to solve for volume (V). \(V = \dfrac{nRT}{P}\) For air: \(V_{air} = \dfrac{(1)(8.314 \,\text{J/mol*K})(298 \,\text{K})}{(101.33 \,\text{kPa})}\) For helium: \(V_{He} = \dfrac{(1)(8.314 \,\text{J/mol*K})(298 \,\text{K})}{(101.33 \,\text{kPa})}\)

Calculate the mass of air and helium

To find the mass of one mole of air and helium, multiply the number of moles by the molar mass. For air: Mass of air (\(m_{air}\)) = 1 mol × 29.0 g/mol = 29.0 g For helium: Mass of helium (\(m_{He}\)) = 1 mol × 4.00 g/mol = 4.00 g

Calculate the density of air and helium

Now we can find the density by dividing the mass by the volume for each gas. The density formula is: \(\rho = \dfrac{m}{V}\) For air: \(\rho_{air} = \dfrac{29.0 \,\text{g}}{V_{air}}\) For helium: \(\rho_{He} = \dfrac{4.00 \,\text{g}}{V_{He}}\) Since the volume of helium and air are the same in this case, we can directly compare the densities. Helium has a lower mass and the same volume, so its density will be lower.

Determine if a carbon dioxide-filled balloon would rise

For this step, we need to compare the density of air with the density of carbon dioxide at the same temperature and pressure. The molar mass of carbon dioxide (CO2) is 44.01 g/mol. We already calculated the volume for one mole of air at the given conditions; if we use the same reasoning as in Step 3, we can find the mass of one mole of carbon dioxide: Mass of CO2 (\(m_{CO2}\)) = 1 mol × 44.01 g/mol = 44.01 g Now calculate the density of CO2: \(\rho_{CO2} = \dfrac{44.01 \,\text{g}}{V_{air}}\) Since the mass of CO2 is greater than the mass of air and their volumes are the same, the density of CO2 is greater than the density of the air. Therefore, a carbon dioxide-filled balloon would not rise at the same temperature and pressure.

Key concepts

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry and physics that relates the pressure (\( P \)), volume (\( V \)), temperature (\( T \)), and amount of substance (\( n \)) of a gas. Given by the formula \( PV = nRT \), where \( R \) is the ideal gas constant, this law allows us to calculate the state of an ideal gas under various conditions.

To find out the density of a gas like air or helium using the ideal gas law, we need to first determine the volume occupied by one mole of gas under the specified conditions. By setting the number of moles (\( n \)) to 1, the law simplifies to \( V = \frac{RT}{P} \).

Calculating Volume Using the Ideal Gas Law

For example, if the temperature (\( T \)) is given in degrees Celsius, we convert it to Kelvin by adding 273.15. The pressure has to be in a consistent unit too, such as pascals (Pa) or kilopascals (kPa). Then, by plugging the values into the simplified ideal gas law, we can solve for the volume (\( V \)) which will be used to calculate gas density.

Molar Mass

Molar mass is a measurement that reflects the mass of one mole of a substance. It is typically expressed in grams per mole (\( g/mol \)) and is crucial when working with the ideal gas law to calculate densities of gases.

To calculate the density of a gas, you need to know its molar mass, as density (\( \rho \)) is defined as mass (\( m \)) per unit volume (\( V \)). The molar mass of air and helium are given as 29.0 g/mol and 4.00 g/mol respectively in our exercise.

Using Molar Mass to Calculate Mass

To find the mass of one mole, you multiply the molar mass by the number of moles, which is 1 in this scenario. For instance, for helium with a molar mass of 4.00 g/mol, the mass of one mole would simply be 4.00 g. Knowing the mass allows us to then calculate density by comparing it to the volume we obtained from the ideal gas law.

Standard Temperature and Pressure

Standard temperature and pressure (STP) are standard sets of conditions for experimental measurements to be comparable between different sets of data. In chemistry, the most commonly used standard temperature is 0°C (273.15K) and the standard pressure is 1 atmosphere (101.325 kPa). However, it's important to adjust calculations to the actual given conditions, as in the exercise where the temperature is 25°C and the pressure is 770 mm Hg.

The ideal gas law is often used in conjunction with these standard conditions to describe the behavior of an ideal gas. However, when the conditions deviate from STP, as they often do in real-world situations, the gas law can still be applied; the pressure and temperature need to be converted to the right units, and then calculations can continue normally.

Understanding STP is important because it provides a reference point. Questions may ask for comparisons at 'the same temperature and pressure,' indicating you should hold these variables constant while examining changes in other properties, such as density, when you are comparing different gases.