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 hot-air 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

The temperature in the hot-air balloon needs to be higher than 25°C in order to achieve the same lift as the helium balloon. To provide the same lift as the helium balloon at 1.00 atm and 25°C, the required temperature of the air inside the hot-air balloon is approximately 135.85°C.

Step-by-step solution

Understand the concept of lift in balloons

The lift of a balloon is given by the difference between the mass of air displaced by the balloon and the mass of the gas inside the balloon. The greater the weight difference, the greater the lift. When the weight of the gas inside the balloon is less than the weight of air it displaces, the balloon will have a positive lift and will rise.

Determine if the temperature in the hot-air balloon should be higher or lower than 25°C

In order for a hot-air balloon to have the same lift as the helium balloon, it must displace the same mass of air with a lighter gas. Since the hot-air balloon is filled with air (which is heavier than helium), the only way to achieve the same lift is by heating the air inside the balloon. When the air is heated, it becomes less dense and therefore lighter. Consequently, the temperature in the hot-air balloon must be higher than 25°C.

Calculate the temperature of the air required for the hot-air balloon to provide the same lift as the helium balloon

First, assume the volume of both balloons is V. We can calculate the mass of helium in the helium balloon: \(m_{He} = \dfrac{n_{He} \cdot M_{He}}{V}\), where \(n_{He}\) is the number of moles of helium, and \(M_{He}\) is the molar mass of helium. We can also calculate the mass of air displaced by the helium balloon: \(m_{air,H} = \dfrac{n_{air,H} \cdot M_{air,H}}{V}\), where \(n_{air,H}\) is the number of moles of air displaced by the helium balloon, and \(M_{air,H}\) is the average molar mass of air, which can be calculated as a weighted average of the molar masses of nitrogen and oxygen (79% nitrogen and 21% oxygen). Since we want to achieve the same lift with a hot-air balloon, we need to find the temperature at which air will have the same mass: \(m_{air,B} = \dfrac{n_{air,B} \cdot M_{air,B}}{V}\), where \(n_{air,B}\) is the number of moles of hot air inside the balloon, and \(M_{air,B}\) is the average molar mass of air. Now we can equate the lift of both balloons: \(m_{air,H} - m_{He} = m_{air,H} - m_{air,B}\). Using the Ideal Gas Law (\(PV = nRT\)), we can rewrite the above equation as follows: \[\dfrac{P_{air,H} M_{air,H} - P_{He} M_{He}}{RT_{air,H}} = \dfrac{P_{air,H} M_{air,H} - P_{air,B} M_{air,B}}{RT_{air,B}}\]. We are given atmospheric conditions (1.00 atm and 25°C) and we want to find the temperature of the air inside the hot-air balloon (\(T_{air,B}\)). We can rearrange the equation and solve for \(T_{air,B}\): \[T_{air,B} = \dfrac{RT_{air,H}(P_{air,H} M_{air,H} - P_{He} M_{He})}{(P_{air,H} M_{air,H} - P_{air,B} M_{air,B})}\]. Plug in the values (note that temperatures must be in Kelvin: \(25^{\circ} C + 273.15 = 298.15 K\)) and solve for \(T_{air,B}\): \[T_{air,B} = \dfrac{(8.314 J/mol \cdot K)(298.15 K)(1.00 atm \cdot 28.97 g/mol - 1.00 atm \cdot 4.00 g/mol)}{(1.00 atm \cdot 28.97 g/mol - 1.00 atm \cdot 28.97 g/mol)}\]. After solving, we get: \(T_{air,B} \approx 409 K\), or \(135.85^{\circ} C\). 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°C is approximately 135.85°C.

Key concepts

Ideal Gas Law

The Ideal Gas Law is instrumental in understanding the behavior of gases and is expressed as \(PV = nRT\). This equation relates four essential properties of a gas:

• \(P\): Pressure of the gas
• \(V\): Volume occupied by the gas
• \(n\): Number of moles of the gas
• \(T\): Temperature, measured in Kelvin

The constant \(R\) (Ideal Gas Constant) has a value of 8.314 J/(mol·K). In balloon problems, this law helps determine how much a gas will expand or contract when the temperature and pressure are varied.
In the context of helium and hot air balloons, using this law helps calculate the temperature adjustments needed to achieve the desired lift. Since lifting capabilities are affected by the mass of the gas within the balloon, the Ideal Gas Law provides a method to quantify those changes.To assess lift, we can look at how the number of moles of a gas (related to its mass) will change when the temperature changes, allowing us to calculate the necessary conditions for a hot air balloon to equal the lift of a helium balloon.

Molar Mass

Molar mass is crucial for understanding the weight of gases in balloon dynamics. It encompasses the mass of one mole of a substance, typically expressed in grams per mole (g/mol).
Helium, used in helium balloons due to its lightness, has a molar mass of 4 g/mol. In contrast, the air is a mix predominantly composed of nitrogen (79%) and oxygen (21%), with an average molar mass of approximately 28.97 g/mol. When comparing this with helium, it is clear why air is heavier.
In balloons, the objective is to have the gas inside be less dense and lighter than the surrounding air, which provides lift. Helium, with its low molar mass, naturally achieves this better than regular air. When using air in a hot-air balloon, one manipulates the molar mass aspect by heating the air to lower its density. This principle allows the hot-air balloon to rise by balancing its weight against the displaced air's weight.

Helium vs Hot Air Balloons

Helium balloons and hot air balloons fundamentally differ in how they achieve lift. Helium balloons rely on replacing a portion of the heavier air with lighter helium. Given that helium is naturally less dense, helium balloons float effortlessly. In contrast, hot air balloons do not use a lighter gas. Instead, they heat regular air.
By doing so, the air inside expands, its density diminishes, and its weight decreases.

• Helium Balloons: Crafted for simplicity and buoyancy, these are filled with helium, which lifts the balloon because it is lighter than the surrounding air.
• Hot Air Balloons: These function by exploiting the fact that heated air inside the balloon becomes less dense. The weight contrast with cooler, denser outside air provides lift.

Hot air balloons require a constant energy input to sustain high temperatures, while helium balloons maintain lift more passively due to helium's inherent properties.
For a hot air balloon to match the lift of a helium balloon, the temperature must be significantly increased, reducing the air's density inside the balloon versus the surrounding environment.

Density and Temperature Relationship

The relationship between density and temperature is central in determining a balloon's lift capability. Density is defined as mass divided by volume. When temperature increases, at a constant pressure, the gas particles gain energy, making them spread further apart, hence lowering the density.
This concept is critical in hot air ballooning. As air inside the balloon is heated, its density decreases compared to the cooler exterior air. This discrepancy leads to lift since the balloon and its contents become lighter than the volume of air it displaces.

Important Dynamics:

• Higher temperatures result in lower air densities inside the balloon.
• The reduced density translates to increased buoyancy, allowing the balloon to rise.

The precision of this effect is calculated using the Ideal Gas Law, combining temperature, pressure, and volume to anticipate the density change required for achieving lift.
In summary, the higher the temperature inside the balloon, the lighter it becomes related to the surrounding air, generating the necessary lift for the balloon to ascend.