Find the rms speed in air at \(0.0^{\circ} \mathrm{C}\) and 1.00 atm of (a) the \(\mathrm{N}_{2}\) molecules, (b) the \(\mathrm{O}_{2}\) molecules, and (c) the \(\mathrm{CO}_{2}\) molecules.

Short Answer

Expert verified
Answer: The root-mean-square speeds at 0.0°C and 1.00 atm are: (a) for N₂ molecules: 518.0 m/s (b) for O₂ molecules: 481.3 m/s (c) for CO₂ molecules: 393.9 m/s

Step by step solution

01

1. Converting temperature into Kelvin

T(°C) = T(K) - 273.15 T(K) = T(°C) + 273.15 T(K) = 0.0 + 273.15 = 273.15 K
02

2. Calculating the mass of each molecule in kilograms

Mass of one nitrogen (N₂) molecule: Nucleon mass approximated as 1 amu (atomic mass unit) = 1.661 × 10^{-27} kg N₂ mass = 2 × (mass of one nitrogen atom) = 2 × 14 × 1.661 × 10^{-27} kg = 4.6508 × 10^{-26} kg Mass of one oxygen (O₂) molecule: O₂ mass = 2 × (mass of one oxygen atom) = 2 × 16 × 1.661 × 10^{-27} kg = 5.3128 × 10^{-26} kg Mass of one carbon dioxide (CO₂) molecule: CO₂ mass = (mass of one carbon atom) + 2 × (mass of one oxygen atom) = 12 × 1.661 × 10^{-27} kg + 2 × 16 × 1.661 × 10^{-27} kg = 7.3084 × 10^{-26} kg
03

3. Applying the rms speed formula

(a) rms speed of N₂ molecules: \(v_{rms}(\mathrm{N}_{2}) = \sqrt{\frac{3kT}{m_{N_2}}} = \sqrt{\frac{3 \times 1.381 \times 10^{-23} \, \mathrm{J \cdot K^{-1}} \times 273.15 \, \mathrm{K}}{4.6508 \times 10^{-26} \, \mathrm{kg}}} = 518.0 \, \mathrm{m/s}\) (b) rms speed of O₂ molecules: \(v_{rms}(\mathrm{O}_{2}) = \sqrt{\frac{3kT}{m_{O_2}}} = \sqrt{\frac{3 × 1.381 \times 10^{-23} \, \mathrm{J \cdot K^{-1}} \times 273.15 \, \mathrm{K}}{5.3128 \times 10^{-26} \, \mathrm{kg}}} = 481.3 \, \mathrm{m/s}\) (c) rms speed of CO₂ molecules: \(v_{rms}(\mathrm{CO}_{2}) = \sqrt{\frac{3kT}{m_{CO_2}}} = \sqrt{\frac{3 × 1.381 \times 10^{-23} \, \mathrm{J \cdot K^{-1}} \times 273.15 \, \mathrm{K}}{7.3084 \times 10^{-26} \, \mathrm{kg}}} = 393.9 \, \mathrm{m/s}\) Hence, the rms speed at 0.0°C and 1.00 atm are: (a) for N₂ molecules: 518.0 m/s (b) for O₂ molecules: 481.3 m/s (c) for CO₂ molecules: 393.9 m/s

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