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?

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 \(\displaystyle \boxed{6699\mathrm{g}}\).

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 \(\displaystyle \boxed{62025\mathrm{g}}\).

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 \(\displaystyle \boxed{377\mathrm{g}}\).