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?

Given that the pressure (P) is 745 torr and the temperature (T) is 65°C, we can use the ideal gas law to calculate the moles of air inside the balloon. The ideal gas law is given as \[PV = nRT\] where P = pressure, V = volume, n = number of moles, R = gas constant, and T = temperature. Note: We need to convert the pressure from torr to atm and the temperature from Celsius to Kelvin. Converted values: - Pressure: \(P = \frac {745}{760}\) atm. - Temperature: \(T = 65 + 273.15 = 338.15K\) Now we rearrange the ideal gas law to find n: \[n = \frac{PV}{RT}\] Plug in the values and calculate n: \[n = \frac{(\frac{745}{760})(65.4)}{0.0821(338.15)} = 2.425 \, moles\] #Step 3: Calculate the mass of the hot_air inside the balloon#:

The surrounding air is at 21°C and 745 torr. Calculate the number of moles of the cool air displaced using the same ideal gas law formula and the new temperature and pressure values: - Temperature: \(T = 21 + 273.15 = 294.15K\) \[n = \frac{PV}{RT}\] Plug in the values and calculate n: \[n = \frac{(\frac{745}{760})(65.4)}{0.0821(294.15)} = 2.634 \, moles\] #Step 5: Calculate the mass of the cool_air displaced by the balloon#:

The total mass that can be lifted by the hot_air balloon is the difference between the mass of the cool air displaced and the mass of the hot air inside the balloon: \[mass_{lift\_hot} = mass_{cool\_air} - mass_{hot\_air}\] \[mass_{lift\_hot} = 76.4 - 70.3 = 6.1 g\] So, the total mass that can be lifted by the hot_air balloon is 6.1 g. #Step 7: Calculate the number of moles of helium inside the balloon in part b#:

Using the molar mass of helium (4.0 g/mol), calculate the mass of the helium inside the balloon: \[mass_{He} = n \times molar \,mass_{He}\] \[mass_{He} = (2.634)(4.0) = 10.5 g\] #Step 9: Calculate the total mass that can be lifted by the helium-filled balloon in part b#:

In Denver, Colorado, the atmospheric pressure is 630 torr. Calculate the new number of moles of hot_air inside the balloon and the cool_air displaced using the same temperature values as before and the new pressure value: - Pressure: \(P = \frac {630}{760}\) atm. Calculate the number of moles of hot_air inside the balloon: \[n_{hot\_air} = \frac{(\frac{630}{760})(65.4)}{0.0821(338.15)} = 2.057 \, moles\] Now, calculate the mass of hot_air: \[mass_{hot\_air} = (2.057)(29.0) = 59.6 g\] Next, calculate the number of moles of the cool_air displaced: \[n_{cool\_air} = \frac{(\frac{630}{760})(65.4)}{0.0821(294.15)} = 2.238 \, moles\] Now, calculate the mass of cool_air: \[mass_{cool\_air} = (2.238)(29.0) = 64.9 g\] Finally, calculate the total mass that can be lifted by the hot_air balloon in Denver: \[mass_{lift\_hot\_Denver} = mass_{cool\_air} - mass_{hot\_air}\] \[mass_{lift\_hot\_Denver} = 64.9 - 59.6 = 5.3 g\] So, the total mass that can be lifted by the hot_air balloon in Denver, Colorado, is 5.3 g.