Problem 159 - Chemistry An Atoms First Approach

Question

You have a helium balloon at \(1.00\) atm and \(25^{\circ} \mathrm{C}\). You want to make a hot-air balloon with the same volume and same lift as the helium balloon. Assume air is \(79.0 \%\) nitrogen and \(21.0 \%\) oxygen by volume. The "lift" of a balloon is given by the difference between the mass of air displaced by the balloon and the mass of gas inside the balloon. a. Will the temperature in the hot-air balloon have to be higher or lower than \(25^{\circ} \mathrm{C} ?\) Explain. b. Calculate the temperature of the air required for the hotair balloon to provide the same lift as the helium balloon at \(1.00\) atm and \(25^{\circ} \mathrm{C}\). Assume atmospheric conditions are \(1.00\) atm and \(25^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

The temperature in the hot-air balloon has to be higher than \(25^{\circ} \mathrm{C}\) to provide the same lift as the helium balloon. The required temperature of the hot-air balloon to achieve the same lift is approximately 340.01 K or \(66.86^{\circ} \mathrm{C}\).

Step 1: Consider the lift equation

The lift of a balloon is given by the difference between the mass of air displaced by the balloon and the mass of gas inside the balloon. Let's denote the mass of helium inside the helium balloon as m_He, the mass of air displaced by the helium balloon as m_air_he, the mass of the gas mixture of the hot-air balloon as m_ha, and the mass of the air displaced by the hot-air balloon as m_air_ha. To have the same lift, the mass difference between the helium balloon and the air it displaces should be equal to the mass difference between the hot-air balloon and the air it displaces. m_air_he - m_He = m_air_ha - m_ha

Step 2: Helium balloon is lighter than air

We know that helium is lighter than air, which means that if we want to make a hot-air balloon with the same lift, we need to make the mass of the gas inside the hot air balloon lighter as well. This means that the temperature of the hot-air balloon needs to be higher than the helium balloon's temperature. Answer: The temperature in the hot-air balloon has to be higher than \(25^{\circ} \mathrm{C}\). #b. Calculate the required temperature for the hot-air balloon to provide the same lift as the helium balloon#

Step 3: Use the Ideal Gas Law

We will use the Ideal Gas Law: \(PV = nRT\) Here, P is the pressure, V is the volume, n is the amount of gas in moles, R is the ideal gas constant, and T is the temperature. Since we are looking to have the same lift and volume, the pressure and volume of both the helium balloon and the hot-air balloon will be the same.

Step 4: Determine the molar mass of the gases

We'll start by finding the molar mass of both helium and the gas mixture in the hot-air balloon. Helium has a molar mass M_He = 4 g/mol, while Nitrogen is 79.0% and has a molar mass of 28 g/mol, and Oxygen is 21.0% and has a molar mass of 32 g/mol. Mixture_molar_mass = 0.79 * 28 + 0.21 * 32

Step 5: Equate the mass difference

We can rewrite the lift equation in terms of the amount of gas (n) and their molar masses (M): (PV / RT_ha) * Mixture_molar_mass - (PV / RT_he) * M_He = 0 Where T_ha and T_he are the temperatures of the hot-air balloon and the helium balloon, respectively.

Step 6: Solve for the required temperature

We can solve the equation for T_ha to find the required temperature: T_ha = (M_He / Mixture_molar_mass) * T_he Now, we can plug in the known values and solve for T_ha. T_ha = (4 / (0.79 * 28 + 0.21 * 32)) * (25+273.15) T_ha ≈ 340.01 K The required temperature of the hot-air balloon to provide the same lift as the helium balloon at 1.00 atm and \(25^{\circ} \mathrm{C}\) is approximately 340.01 K or \(66.86^{\circ} \mathrm{C}\).

Key Concepts

Balloon Lift

Lift in a balloon is all about buoyancy, which means how a lighter gas or air inside the balloon can cause it to float in the surrounding heavier atmosphere. Imagine placing a lightweight ball in a tub of water. If the ball is lighter than the water it displaces, it will float. Similarly, a balloon lifts because the gas inside it is lighter than the air it pushes away. For a helium balloon, it is the difference between the mass of helium inside the balloon and the air it displaces. If you want to achieve the same lift with a hot-air balloon, the hot air has to be lighter than the surrounding cool air. The magic here is in the temperature. By making the air inside the hot-air balloon warmer, it becomes less dense. When it becomes less dense, the balloon can lift, just like our lightweight ball floating in water.

Molar Mass Calculation

Molar mass is like the weight of one mole of a gas. To compare different gases, and their lifting abilities, we need to know how heavy they are per mole. Helium has a very simple and light structure with a molar mass of 4 g/mol. When comparing it to air, which is mostly nitrogen and oxygen, we need to calculate a combined molar mass for these two gases. For example, nitrogen makes up 79% of air and has a molar mass of 28 g/mol. Oxygen is 21% and its molar mass is 32 g/mol. Therefore, the molar mass of air as a mixture can be calculated using the formula:

• Molar mass of air = 0.79 * 28 + 0.21 * 32

This helps us understand the overall weight of air, and compare it with helium, especially when aiming to adjust the balloon to have equal lifting power.

Temperature Calculation

Calculating the temperature for the hot-air balloon involves the Ideal Gas Law: \(PV = nRT\). To maintain the same lift as a helium balloon at a given temperature and pressure, we must adjust the temperature of the hot-air balloon. Given that pressure (P) and volume (V) are constants, we focus on the temperatures and the gas amounts (n). The challenge is making the gas inside the hot-air balloon as light as helium but by heating it up instead. By adjusting the Ideal Gas Law to our needs, we find:

• \(T_{ha} = \left( \frac{M_{He}}{\text{Mixture molar mass}} \right) \times T_{he}\)

Plug in the values to rearrange for the required hot air balloon temperature, and you’ll find it needs to be much hotter than 25°C, roughly around 340.01 K or 66.86°C.

Helium Properties

Helium is a fascinating and useful gas due to its unique properties. It is lighter than air, colorless, odorless, tasteless, non-toxic, and inert, meaning it doesn't react easily with other substances. These properties make helium an ideal choice for lifting objects like balloons. Its low density compared to the surrounding air allows it to float naturally. To understand why it's so effective, consider its low molar mass of 4 g/mol. This very low density means helium can displace a lot of air compared to its weight, making it highly efficient for balloons despite having less lifting capacity than hydrogen. However, helium is much safer in terms of inflammability, making it the preferred choice for filling balloons in public and industrial settings.