3) [24] Atmospheric pressure is approximately 1E5 Pa (one bar). Use this fact to estimate the mass of gas in the entire atmosphere. You will need to look up the diameter of the earth, which should be treated as a sphere. a} First nd one mistake in the ChatG PT response to this question listed at the bottom of the problem set. Explain why you think it is wrong, and how you would correct it. Please do not nd ALL the errors. b) Solve the problem yourself. Hint: use a force balance. The following is the ChatGPT response to problem 3. Atmospheric pressure is approximately 185 Pa (one bar). Use this fact to estimate the mass of gas in the entire atmosphere. You will need to look up the diameter of the earth, which should be treated as a sphere. ChatGPT To estimate the mass of gas in the entire atmosphere, we can use the ideal gas law along with the given atmospheric pressure and the volume of the Earth's atmosphere. The ideal gas law states: PV = nRT, where: P is pressure, V is volume, n is the number of moles of gas, R is the ideal gas constant, T is temperature. For the entire atmosphere, we'll consider the volume of the Earth's atmosphere, which can be approximated as the volume of a sphere with the radius equal to the Earth's radius. The volume of a sphere is given by: V = 4/3 * n * r, where r is the radius of the sphere. The mass (m) of a gas can be related to the number of moles (n) using the molar mass (M) of the gas: m= n * M. Given that the atmospheric pressure is approximately 185 Pa and the ideal gas constant R is approximately 8.31 J/(mol-K), and knowing that the Earth's radius (diameter / 2) is approximately 6371 km = 6.37156 m, we can calculate the mass of the gas in the entire atmosphere: 1. Calculate the volume of the Earth's atmosphere: V = 4/3 * n * (6.37186 m)?. 2. Use the ideal gas law to find the number of moles of gas (n): n = PV / RT. 3. Estimate the mass of the gas using the molar mass of air (which is approximately 0.029 kg/mol): m = n * M. Let's perform the calculations: 1. V=4/3 * n* (6.37186 m) = 1.083E21 m'. 2. n = (165 Pa) * (1.083E21 m') / (8.31 J/(mol-K) * 288 K) = 4.138E18 mol. 3. m= (4.138E18 mol) * (0.029 kg/mol) = 1.199E17 kg. So, the estimated mass of gas in the entire atmosphere is approximately 1.199E17 kilograms