A sounding balloon is a rubber bag filled with \(\mathrm{H}_{2}(\mathrm{g})\) and carrying a set of instruments (the payload). Because this combination of bag, gas, and payload has a smaller mass than a corresponding volume of air, the balloon rises. As the balloon rises, it expands. From the table below, estimate the maximum height to which a spherical balloon can rise given the mass of balloon, \(1200 \mathrm{g} ;\) payload, \(1700 \mathrm{g}\) : quantity of \(\mathrm{H}_{2}(\mathrm{g})\) in balloon, \(120 \mathrm{ft}^{3}\) at \(0.00^{\circ} \mathrm{C}\) and \(1.00 \mathrm{atm}\); diameter of balloon at maximum height, 25 ft. Air pressure and temperature as functions of altitude are: $$\begin{array}{ccl} \hline \text { Altitude, km } & \text { Pressure, mb } & \text { Temperature, } \mathrm{K} \\ \hline 0 & 1.0 \times 10^{3} & 288 \\ 5 & 5.4 \times 10^{2} & 256 \\ 10 & 2.7 \times 10^{2} & 223 \\ 20 & 5.5 \times 10^{1} & 217 \\ 30 & 1.2 \times 10^{1} & 230 \\ 40 & 2.9 \times 10^{0} & 250 \\ 50 & 8.1 \times 10^{-1} & 250 \\ 60 & 2.3 \times 10^{-1} & 256 \\ \hline \end{array}$$

Step by step solution

01

Calculate the mass of hydrogen

The quantity of hydrogen in the balloon is given as \(120 \mathrm{ft}^{3}\) at \(0^{\circ} \mathrm{C}\) and \(1.00\mathrm{atm}\). First, convert \(120 \mathrm{ft}^{3}\) to liters using the conversion factor \((1 \mathrm{ft}^{3} = 28.317 \mathrm{L})\), giving us \(3398 \mathrm{L}\). We can then use the ideal gas law (\(PV=nRT\)) to solve for the number of moles of hydrogen \(n = \frac{PV}{RT}\). Assuming that \(R=0.0821 \mathrm{L \cdot atm/K \cdot mol}\) and that the volume of hydrogen \(V\) is \(3398 \mathrm{L}\), the pressure \(P\) is \(1 \mathrm{atm}\), and the temperature \(T\) is \(273 \mathrm{K}\), the number of moles of hydrogen \(n\) is \(14.2 \mathrm{mol}\). The molar mass of hydrogen \(H_2\) is approximately \(2 \mathrm{g/mol}\), so the mass of hydrogen in the balloon is \(28.4 \mathrm{g}\).

02

Calculate the total mass of the balloon

Add the masses of the hydrogen, balloon, and payload to get the total mass \(m_{total}\) of the system. So, \(m_{total} = m_{H_2} + m_{balloon} + m_{payload} = 28.4 \mathrm{~g} + 1200 \mathrm{~g} + 1700 \mathrm{~g} = 2928.4 \mathrm{~g} = 2.9284 \mathrm{~kg}\).

03

Calculate the balloon's volume at its maximum height

Use the volume formula for a sphere \(V = \frac{4\pi r^{3}}{3}\) to find the volume of the balloon when it is at its maximum diameter of \(25 \mathrm{ft}\). First, convert the diameter to radius (so \(r = 12.5 \mathrm{ft}\)) and then to meters (where \(1 \mathrm{ft} = 0.3048 \mathrm{m}\)), giving us \(r = 3.81 \mathrm{m}\). Substituting this into the volume formula gives \(V = 231m^{3}\).

04

Calculate the maximum height

Find the altitude where the weight of the displaced air equals the total weight of the balloon system. The weight of the air \(w_{air} = m_{air}g\) is given by \(w_{air} = ρVg\), where \(ρ\) is the density of air, \(V\) is the volume of the balloon, and \(g = 9.8 m/s^{2}\) is the acceleration due to gravity. Substituting and rearranging gives us the density of air at the maximum altitude: \(ρ = \frac{m_{total}}{V} = \frac{2.9284 \mathrm{~kg}}{231 \mathrm{~m}^{3}} = 0.0127 \mathrm{~kg/m}^{3}\). Looking at the table, this density corresponds (approximately) to an altitude of \(30 \mathrm{km}\). Thus, the balloon can rise to around \(30 \mathrm{km}\).

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