(a) Place the following gases in order of increasing average molecular speed at \(300 \mathrm{~K}: \mathrm{CO}, \mathrm{SF}_{6}, \mathrm{H}_{2} \mathrm{~S}, \mathrm{Cl}_{2}, \mathrm{HBr}\). (b) Calcu- late and compare the rms speeds of \(\mathrm{CO}\) and \(\mathrm{Cl}_{2}\) molecules at \(300 \mathrm{~K} .(\mathbf{c})\) Calculate and compare the most probable speeds of \(\mathrm{CO}\) and \(\mathrm{Cl}_{2}\) molecules at \(300 \mathrm{~K}\).

First, we need to calculate the molar mass of each gas molecule: - CO: 12 g/mol (C) + 16 g/mol (O) = 28 g/mol - SF₆: 32 g/mol (S) + 6 * 19 g/mol (F) = 146 g/mol - H₂S: 2 g/mol (H) + 32 g/mol (S) = 34 g/mol - Cl₂: 2 * 35.5 g/mol = 71 g/mol - HBr: 1 g/mol (H) + 80 g/mol (Br) = 81 g/mol

Since the average molecular speed is inversely proportional to the square root of the molar mass, we can order the gases: SF₆ > Cl₂ > HBr > H₂S > CO (in decreasing molecular speeds) Therefore, the gases ordered in increasing molecular speeds are: CO, H₂S, HBr, Cl₂, SF₆. For part (b), we have to calculate and compare the root-mean-square (rms) speeds of CO and Cl₂ at a temperature of 300 K. The formula for the root-mean-square speed is as follows: \[ v_{rms} = \sqrt{\frac{3RT}{M}} \]

First, convert the molar masses of CO and Cl₂ from g/mol to kg/mol: - CO: 28 g/mol = 0.028 kg/mol - Cl₂: 71 g/mol = 0.071 kg/mol Next, use the given temperature, T = 300 K, and the universal gas constant, R = 8.314 J/(mol∙K). Calculate the root-mean-square speeds for both gases: \( v_{rms,CO} = \sqrt{\frac{3(8.314)(300)}{0.028}} \) \( v_{rms,CO} = 13.83\,\mathrm{m/s} \) \( v_{rms,Cl_2} = \sqrt{\frac{3(8.314)(300)}{0.071}} \) \( v_{rms,Cl_2} = 8.051\,\mathrm{m/s} \) So, the root-mean-square speed of CO is approximately 13.83 m/s, and the root-mean-square speed of Cl₂ is approximately 8.051 m/s. For part (c), we need to calculate and compare the most probable speeds of CO and Cl₂ molecules at 300 K. The formula for the most probable speed is as follows: \[ v_p = \sqrt{\frac{2RT}{M}} \]

Using the given temperature, T = 300 K, and the universal gas constant, R = 8.314 J/(mol∙K). Calculate the most probable speeds for both gases: \( v_{p,CO} = \sqrt{\frac{2(8.314)(300)}{0.028}} \) \( v_{p,CO} = 12.12\,\mathrm{m/s} \) \( v_{p,Cl_2} = \sqrt{\frac{2(8.314)(300)}{0.071}} \) \( v_{p,Cl_2} = 7.096\,\mathrm{m/s} \) So, the most probable speed of CO is approximately 12.12 m/s, and the most probable speed of Cl₂ is approximately 7.096 m/s.