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 \(25^{\circ} \mathrm{C}\).

Short answer

a. The temperature in the hot-air balloon has to be higher than \(25^{\circ}\)C because the hot-air balloon is filled with a mixture of nitrogen and oxygen, which are heavier than helium. Increasing the temperature of the gas inside the balloon will help achieve a lower density allowing it to provide the same lift as the helium balloon. b. The required temperature for the hot-air balloon to provide the same lift as the helium balloon at 1.00 atm and \(25^{\circ}\)C is approximately 297.5 K or \(24.5^{\circ}\)C.

Step-by-step solution

Given information and observation

The helium balloon has a temperature of 25°C and lifts due to its lower density compared to air. We want to create a hot-air balloon with the same lift, meaning it will be displacing the same mass of air as the helium balloon, but we're going to be using a mixture of nitrogen and oxygen instead of helium.

Analyze the densities of gases

Since the hot-air balloon will be filled with a mixture of nitrogen and oxygen, and these gases are heavier than helium, it follows that for the hot-air balloon to displace the same mass of air and provide the same lift, the gas inside the balloon must be less dense than the surrounding air.

Conclusion for part a

The only way to achieve a lower density inside the hot-air balloon is by increasing the temperature of the gas. So, the temperature in the hot-air balloon has to be higher than \(25^{\circ}\)C. #b: Calculate the required temperature for the hot-air balloon to provide the same lift as the helium balloon#

Eq.1: Density definition

To find the temperature, we can use the Ideal Gas Law and the density definition formula: \[density = \frac{mass}{volume}\]

Eq.2: Ideal Gas Law

The Ideal Gas Law is: \[PV = nRT\] where P is the pressure, V is the volume, n is the number of moles of the gas, R is the gas constant, and T is the temperature.

Relation between Eq.1 and Eq.2

If we combine the density definition with the Ideal Gas Law, we get: \[density = \frac{nM}{V} = \frac{P}{RT}\] where M is the molar mass of the gas.

Calculate the density of air at 25°C

First, we need to calculate the density of air at 25°C: \[density_{air} = \frac{P}{R(T)} = \frac{1.00 \, atm}{(0.0821 \, \frac{L \cdot atm}{K \cdot mol})(298 \, K)}\] \[density_{air} \cong 0.0409 \, \frac{mol}{L}\] Now, we need to find the effective molar mass of air.

Effective molar mass of air

Assuming air is 79.0% nitrogen and 21.0% oxygen by volume, we have: \[M_{air} = 0.79 M_{N_2} + 0.21 M_{O_2} = 0.79 (28 \, g/mol) + 0.21 (32 \, g/mol) = 28.84 \, g/mol\]

Calculate the mass of air displaced by the helium balloon.

Now we can calculate the mass of air displaced by the helium balloon: \[mass_{air} = density_{air} \cdot volume \cdot M_{air} = 0.0409 \frac{mol}{L} \cdot V \cdot 28.84 \frac{g}{mol}\]

Calculate the required temperature for the hot-air balloon

We want mass of the gas in the hot-air balloon, with the same volume as the helium balloon, to be equal to the mass of air displaced. Since the hot-air balloon is filled with pure air, we can use the same density formula, but now we need to find T to have the mass equal to mass_air: \[mass_{air} = density_{air} \cdot V \cdot M_{air} = \frac{P}{R(T)} \cdot V \cdot M_{air}\] Rearranging to solve for T, we get: \[T = \frac{P \cdot V \cdot M_{air}}{mass_{air}R}\] Substituting the known values: \[T = \frac{1.00 \, atm \cdot V \cdot 28.84 \, g/mol}{(0.0409 \, \frac{mol}{L} \cdot V \cdot 28.84 \, g/mol) (0.0821 \, \frac{L \cdot atm}{K \cdot mol})}\] After cancelling terms, we are left with: \[T = \frac{1.00 \, atm}{0.0409 \, \frac{mol}{L} \cdot 0.0821 \, \frac{L \cdot atm}{K \cdot mol}} \] Calculating the value of T: \[T \cong 297.5 \, K\]

Conclusion for part b

So, the temperature of the air required for the hot-air balloon to provide the same lift as the helium balloon at 1.00 atm and \(25^{\circ}\)C is approximately 297.5 K or \(24.5^{\circ}\)C.

Key concepts

Gas Density Calculations

Understanding the concept of gas density is essential when analyzing lift and buoyancy, especially in balloons. The density of a gas can be calculated by the formula:

• Density = \( \frac{mass}{volume} \)

This means the density depends on the mass of gas particles and the amount of space they occupy.

Considering our balloon example, the goal is to achieve a lower density inside the balloon than that of the surrounding air. Gases like helium naturally have a lower density compared to air at the same temperature and pressure due to smaller molar mass. However, in a hot-air balloon, we must manipulate density by increasing temperature as per the Ideal Gas Law.

Molar Mass of Air

Air is primarily composed of nitrogen and oxygen. Knowing the molar mass of air enables calculations related to gas density and buoyancy.

To get the molar mass of air, consider its composition:

• 79% Nitrogen
• 21% Oxygen

Nitrogen has a molar mass of approximately 28 g/mol, and oxygen around 32 g/mol.

The effective molar mass of air can be calculated as:

• \(M_{air} = 0.79 \times 28 \, g/mol + 0.21 \times 32 \, g/mol \)
• This results in approximately 28.84 g/mol.

Understanding this molar mass helps in determining the density of air under various conditions using the Ideal Gas Law:

Lift and Buoyancy in Balloons

The fundamental principle behind balloon lift is the difference in density between the gas inside and the air outside the balloon. The lift is essentially the buoyant force and is calculated by the equation:

• Lift = Mass of Air Displaced - Mass of Gas Inside Balloon

This means for the balloon to lift off or float, it needs to displace a mass of air greater than the mass of the gas inside.

In hot-air balloons, this lift is achieved by heating the air inside the balloon. As the temperature rises, the air becomes less dense due to expansion.

Applying the Ideal Gas Law (\( PV = nRT \)), where T is temperature, it's clear that as T increases, the volume or space that the air occupies inside the balloon increases while mass remains constant, causing density to reduce. Lower density inside compared to outside generates buoyant lift. This principle was used to adjust the hot-air balloon’s temperature to about 297.5 K (or 24.5°C) to provide the same lift as helium, hence achieving buoyancy.