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

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.

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.

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#

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

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.

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.

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.

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

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

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: \[T \cong 297.5 \, K\]

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.