At Party City, you purchase a helium-filled balloon with a diameter of \(40.0 \mathrm{~cm}\) at \(20.0^{\circ} \mathrm{C}\) and at \(1.00 \mathrm{~atm} .\) a) How many helium atoms are inside the balloon? b) What is the average kinetic energy of the atoms? c) What is the root-mean-square speed of the atoms?

To calculate the volume of the balloon, we can use the formula for the volume of a sphere: \(V = \frac{4}{3} \pi r^3\) where \(V\) is the volume of the sphere, and \(r\) is the radius. Since the diameter of the balloon is given, we can calculate the radius as half of the diameter: \(r = \frac{40}{2}=20 \mathrm{~cm}\) Now plug in the value of \(r\) into the volume formula: \(V = \frac{4}{3} \pi (20)^3\) Calculating the volume, we get: \(V \approx 33510.3 \mathrm{~cm^3}\)

To work with the ideal gas law, we need to have the volume in SI units (cubic meters) and the temperature in Kelvin. To do this, we can convert from centimeters to meters, and from celsius to Kelvin: \(V = 33510.3 \mathrm{~cm^3}\times\frac{1\,\mathrm{m^{3}}}{1000000\,\mathrm{cm^{3}}} \approx 3.35103\times10^{-2}\,\mathrm{m^3}\) \(T = 20.0^{\circ} \mathrm{C} + 273.15 = 293.15 \,\mathrm{K}\)

Now we can use the ideal gas law to find the number of moles (n) of helium in the balloon: \(PV = nRT\) where \(P\) is the pressure (in atm), \(V\) is the volume (in meters), \(n\) is the number of moles, \(R\) is the ideal gas constant (0.08206 L·atm/mol·K), and \(T\) is the temperature (in Kelvin). We are given \(P = 1 \,\mathrm{atm}\), and we have found \(V\) and \(T\). Now we can rearrange the equation to solve for \(n\) and convert volume to liters: \(n = \frac{PV}{RT}\) \(n = \frac{1 \,\mathrm{atm} \times (3.35103 \times 10^{-2} \,\mathrm{m^3} * \frac{1000\,\mathrm{L}}{1\,\mathrm{m^{3}}})}{0.08206 \mathrm{L\cdot atm/mol\cdot K} \times 293.15 \,\mathrm{K}}\) Calculating the number of moles, we get: \(n \approx 0.0141 \,\mathrm{mol}\)

Now that we have the number of moles, we can use Avogadro's number to find the number of helium atoms inside the balloon: \(N = n \times N_A\) where \(N\) is the number of helium atoms, \(n\) is the number of moles, and \(N_A\) is Avogadro's number (\(6.022\times 10^{23} \,\mathrm{atoms/mol}\)): \(N = 0.0141 \,\mathrm{mol} \times 6.022 \times 10^{23} \,\mathrm{atoms/mol}\) Calculating the number of helium atoms, we get: \(N \approx 8.47\times 10^{21} \,\mathrm{atoms}\)

The average kinetic energy per atom can be found using the formula: \(KE_{avg} = \frac{3}{2} kT\) where \(KE_{avg}\) is the average kinetic energy per atom, \(k\) is Boltzmann's constant (\(1.381\times10^{-23}\,\mathrm{J/K}\)), and \(T\) is the temperature (in Kelvin): \(KE_{avg} = \frac{3}{2} \times 1.381 \times 10^{-23} \,\mathrm{J/K} \times 293.15 \,\mathrm{K}\) Calculating the average kinetic energy per atom, we get: \(KE_{avg} \approx 6.07 \times 10^{-21} \,\mathrm{J}\)

Finally, we can find the root-mean-square speed for helium atoms using the formula: \(v_{rms} = \sqrt{\frac{3kT}{m}}\) where \(v_{rms}\) is the root-mean-square speed, \(k\) is Boltzmann's constant, \(T\) is the temperature (in Kelvin), and \(m\) is the mass of a single helium atom. To find the mass of a helium atom, we can use the molar mass of helium (\(4.00\,\mathrm{g/mol}\)) and Avogadro's number: \(m = \frac{4.00\,\mathrm{g/mol}}{6.022\times 10^{23}\,\mathrm{atoms/mol}} = 6.64\times 10^{-24}\,\mathrm{g}\) Now, convert the mass of a helium atom to kilograms: \(m = 6.64\times 10^{-24}\,\mathrm{g} \times \frac{1\,\mathrm{kg}}{1000\,\mathrm{g}} = 6.64\times 10^{-27}\,\mathrm{kg}\) Plug the values into the equation for root-mean-square speed: \(v_{rms} = \sqrt{\frac{3\times1.381\times 10^{-23}\,\mathrm{J/K} \times 293.15\,\mathrm{K}}{6.64\times 10^{-27}\,\mathrm{kg}}}\) Calculating the root-mean-square speed, we get: \(v_{rms} \approx 1371.68\,\mathrm{m/s}\) So, the answers are: a) There are approximately \(8.47\times10^{21}\) helium atoms inside the balloon. b) The average kinetic energy of the atoms is approximately \(6.07\times10^{-21}\,\mathrm{J}\). c) The root-mean-square speed of the atoms is approximately \(1371.68\,\mathrm{m/s}\).