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} .$

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.

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.

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.