Question

The total mass that can be lifted by 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. Consider a hot-air balloon that approximates a sphere \(5.00 \mathrm{m}\) in diameter and contains air heated to \(65^{\circ} \mathrm{C}\). The surrounding air temperature is \(21^{\circ} \mathrm{C}\). The pressure in the balloon is equal to the atmospheric pressure, which is \(745\) torr. a. What total mass can the balloon lift? Assume that the average molar mass of air is \(29.0 \mathrm{g} / \mathrm{mol}\). (Hint: Heated air is less dense than cool air. b. If the balloon is filled with enough helium at \(21^{\circ} \mathrm{C}\) and \(745\) torr to achieve the same volume as in part a, what total mass can the balloon lift? c. What mass could the hot-air balloon in part a lift if it were on the ground in Denver, Colorado, where a typical atmospheric pressure is \(630 .\) torr?

Short answer

The total mass that the hot-air balloon can lift is 6699 g. If filled with helium at the same conditions, it can lift 62025 g. In Denver, Colorado, with an atmospheric pressure of 630 torr, the hot-air balloon can lift 377 g.

Step-by-step solution

a. Mass that the hot-air balloon can lift

First, we need to find the volume of the hot-air balloon using the given diameter. Since the diameter is given as \(\displaystyle 5\mathrm{m}\), the radius will be half of that: \(\displaystyle 2.50\mathrm{m}\). The volume of a sphere can be calculated as: \[V = \frac{4}{3} \pi r^3\] Substitute the radius value: \[\begin{aligned} V& =\frac{4}{3} \pi (2.50)^3\\ &=65.45\mathrm{m^{3}} \end{aligned}\] Now, convert the pressure to atmospheres: \[P = \frac{745\text{torr}}{760\text{atm/torr}} = 0.9803\text{atm}\] At this point, we can use the ideal gas law \(\displaystyle PV=nRT\) to find the number of moles of displaced air as well as heated air in the balloon. To find the mass of each, we need to multiply the number of moles by the molar mass. The difference between the two masses is the mass that the balloon can lift. First, let's find the number of moles of displaced air: \[n_{air} = \frac{PV}{RT}\] Here, \(\displaystyle P=0.9803\mathrm{atm}\), \(\displaystyle V=65.45\mathrm{m^{3}} = 6.545\times 10^{4}\mathrm{L}\), and \(\displaystyle T=21\degree + 273=294\mathrm{K}\). The ideal gas constant, \(\displaystyle R=0.0821\mathrm{atm\cdot L/(mol\cdot K)}\). Plug in these values: \[\begin{aligned} n_{air}& =\frac{(0.9803)(6.545\times 10^{4})}{(0.0821)(294)}\\ &=2481\mathrm{mol} \end{aligned}\] Now, find the mass of the displaced air using the molar mass of air, \(\displaystyle 29.0\mathrm{g/mol}\): \[m_{air} = n_{air} \times M_{air}\] Substitute the values: \[\begin{aligned} m_{air}& =(2481\mathrm{mol})(29.0\mathrm{g/mol})\\ &=71949\mathrm{g} \end{aligned}\] Next, let's find the number of moles of heated air inside the balloon: \[n_{heated} =\frac{PV}{RT}\] Here, \(\displaystyle P=0.9803\mathrm{atm}\), \(\displaystyle V=65.45\mathrm{m^{3}}=6.545\times 10^{4}\mathrm{L}\), and \(\displaystyle T=65\degree +273=338\mathrm{K}\). Plug in these values: \[\begin{aligned} n_{heated}& =\frac{(0.9803)(6.545\times 10^{4})}{(0.0821)(338)}\\ &=2250\mathrm{mol} \end{aligned}\] Now, find the mass of the heated air inside the balloon using the molar mass of air, \(\displaystyle 29.0\mathrm{g/mol}\): \[m_{heated} = n_{heated} \times M_{air}\] Substitute the values: \[\begin{aligned} m_{heated}& =(2250\mathrm{mol})(29.0\mathrm{g/mol})\\ &=65250\mathrm{g} \end{aligned}\] Finally, find the total mass that the hot-air balloon can lift: \[m_{lift} = m_{air} - m_{heated}\] Substitute the values: \[m_{lift} = 71949\mathrm{g} - 65250\mathrm{g} = 6699\mathrm{g}\] The total mass that the hot-air balloon can lift is \(\displaystyle \boxed{6699\mathrm{g}}\).

b. Mass that the helium-filled balloon can lift

First, let's find the number of moles of helium inside the balloon. Since the volume is the same as in part (a), the pressure and temperature are the same as well; we can use the ideal gas law \(\displaystyle PV=nRT\) to find the number of moles of helium. \[n_{He} =\frac{PV}{RT}\] Here, \(\displaystyle P=0.9803\mathrm{atm}\), \(\displaystyle V=65.45\mathrm{m^{3}} = 6.545\times 10^{4}\mathrm{L}\), and \(\displaystyle T=21\degree + 273=294\mathrm{K}\). Plug in these values: \[\begin{aligned} n_{He}& =\frac{(0.9803)(6.545\times 10^{4})}{(0.0821)(294)}\\ &=2481\mathrm{mol} \end{aligned}\] Now, find the mass of the helium using the molar mass of helium, \(\displaystyle 4.00\mathrm{g/mol}\): \[m_{He} = n_{He} \times M_{He}\] Substitute the values: \[\begin{aligned} m_{He}& =(2481\mathrm{mol})(4.00\mathrm{g/mol})\\ &=9924\mathrm{g} \end{aligned}\] Next, find the total mass that the helium-filled balloon can lift: \[m_{lift} = m_{air} - m_{He}\] Substitute the values: \[m_{lift} = 71949\mathrm{g} - 9924\mathrm{g} = 62025\mathrm{g}\] The total mass that the helium-filled balloon can lift is \(\displaystyle \boxed{62025\mathrm{g}}\).

c. Mass that the hot-air balloon can lift at Denver

First, we need to find the number of moles of displaced air in Denver, where the atmospheric pressure is \(\displaystyle 630\text{ torr}\). Convert the pressure to atmospheres: \[P = \frac{630\text{torr}}{760\text{atm/torr}} = 0.8289\text{atm}\] Use the ideal gas law \(\displaystyle PV=nRT\) to find the number of moles of displaced air: \[n_{air}=\frac{PV}{RT}\] Here, \(\displaystyle P=0.8289\mathrm{atm}\), \(\displaystyle V=65.45\mathrm{m^{3}} = 6.545\times 10^{4}\mathrm{L}\), and \(\displaystyle T=21\degree + 273=294\mathrm{K}\). Plug in these values: \[n_{air}=\frac{(0.8289)(6.545\times 10^{4})}{(0.0821)(294)}=2263\mathrm{mol}\] Now, find the mass of the displaced air: \[m_{air} = n_{air} \times M_{air}\] \[m_{air} = (2263\mathrm{mol})(29.0\mathrm{g/mol}) = 65627\mathrm{g}\] Finally, find the total mass that the hot-air balloon can lift at Denver: \[m_{lift} = m_{air} - m_{heated}\] \[m_{lift} = 65627\mathrm{g} - 65250\mathrm{g} = 377\mathrm{g}\] The mass that the hot-air balloon in part (a) could lift if it were on the ground in Denver, Colorado, where a typical atmospheric pressure is \(\displaystyle 630\mathrm{torr}\), would be \(\displaystyle \boxed{377\mathrm{g}}\).

Key concepts

Ideal Gas Law

The Ideal Gas Law is a useful tool for understanding how gases behave under different conditions. The formula is written as \( PV = nRT \), where:

• \( P \) stands for pressure,
• \( V \) stands for volume,
• \( n \) is the number of moles of the gas,
• \( R \) is the ideal gas constant \( 0.0821 \, \text{atm} \cdot \text{L} / (\text{mol} \cdot \text{K}) \), and
• \( T \) is the temperature in Kelvin.

To convert Celsius to Kelvin, simply add 273 to the Celsius temperature. In our hot-air balloon problem, we use this formula to determine the number of moles of air both inside and displaced by the balloon. Knowing \( n \) helps in calculating the mass of gas inside and outside, which is crucial to determine the lifting capacity.

Molar Mass

Molar mass is an important concept when working with gases, especially in calculations involving the Ideal Gas Law. It is the mass of one mole of a particular substance and is usually expressed in grams per mole (\( \text{g/mol} \)). For air, the average molar mass is approximately \( 29.0 \, \text{g/mol} \). This value is influenced by the various gases that make up the air, like nitrogen and oxygen.
To find the mass of a gas from the number of moles, you multiply the number of moles by its molar mass:

• \( m = n \times \text{Molar mass} \)

For instance, after calculating the number of moles of air or helium using the Ideal Gas Law, you use the molar masses of \( 29.0 \, \text{g/mol} \) for air and \( 4.00 \, \text{g/mol} \) for helium to find the total mass. This step is vital in determining how much mass the balloon displaces (and hence can lift).

Density Differences

Density differences are a subtle yet crucial concept for understanding how a hot-air balloon achieves lift. Density is the mass per unit volume of a substance. When you heat air inside a balloon, its density decreases. This means it occupies the same volume but with less mass compared to cooler, surrounding air.
Why does this matter? Because the balloon will weigh less when filled with heated air compared to the cooler air outside. This difference in density allows the balloon to rise, as the heavier, cooler air is displaced by the lighter, heated air. In mathematical terms, we see this through the formula \( V = \frac{m}{\text{density}} \), showing that for a given volume, less mass (i.e., lower density) will be in the heated air, contributing to lift.

Pressure Conversion

Pressure conversion is an essential skill when dealing with gases under varying atmospheric conditions. Atmospheric pressure is often measured in units like atmospheres (atm), torr, or pascals. In many chemistry problems, it's common to convert these to atmospheres because the ideal gas constant \( R \) is most frequently used with atm.
For our hot-air balloon problem, we need to convert the given pressure from torr to atm to use in the Ideal Gas Law. The conversion is simple:

• \( 1 \, \text{atm} = 760 \, \text{torr} \)

So, if the atmospheric pressure is 745 torr, dividing by 760 gives approximately \( 0.9803 \, \text{atm} \). This conversion ensures that our calculations remain consistent and accurate, allowing us to use the ideal gas constant without errors.