Problem 66 - University Physics with Modern Physics

Question

Calculate the root-mean-square speed of air molecules at room temperature \(\left(22.0^{\circ} \mathrm{C}\right)\) from the kinetic theory of an ideal gas.

Step-by-Step Solution

Answer

Answer: The root-mean-square speed of air molecules at room temperature (22.0°C) is approximately 492 m/s.

Step 1: Write down the given information

Temperature (T) is given as \(22.0^{\circ} \mathrm{C}\). In order to use this in the rms speed formula, we need to convert it to Kelvin, by adding 273.15 to the Celsius value: \(T = 22.0 + 273.15 = 295.15 \, K\). Also, the molar mass (\(M\)) of air needs to be determined. Dry air is composed of approximately 78% nitrogen (\(N_2\)), 21% oxygen (\(O_2\)), and 1% argon (\(Ar\)). We will assume an average molar mass for air based on these percentages, with \(M_{N_2} = 28.02 \, g/mol\), \(M_{O_2} = 32.00 \, g/mol\), and \(M_{Ar} = 39.95 \, g/mol\).

Step 2: Calculate the average molar mass of air

Based on the percentages of nitrogen, oxygen, and argon in air, we can calculate the average molar mass of air using a weighted average: \(M_{air} = 0.78 \times M_{N_2} + 0.21 \times M_{O_2} + 0.01 \times M_{Ar}\) \(M_{air} = 0.78 \times 28.02 + 0.21 \times 32.00 + 0.01 \times 39.95\) \(M_{air} \approx 28.97 \, g/mol\)

Step 3: Use the molar mass to calculate the mass of one air molecule

The formula for root-mean-square speed requires the mass of one molecule (\(m\)). We can obtain this value by dividing the molar mass by the Avogadro constant (\(N_A\)): \(m = \frac{M_{air}}{N_A}\) \(m = \frac{28.97 \, g/mol}{6.022 \times 10^{23} \, molecules/mol}\) \(m \approx 4.81 \times 10^{-26} \, kg\)

Step 4: Write the rms speed formula and identify known values

The root-mean-square (rms) speed, denoted as \(v_{rms}\), is calculated by the following formula: \(v_{rms} = \sqrt{\frac{3kT}{m}}\) Here, \(k\) is the Boltzmann constant (1.38 × 10^{-23} J/K) and \(T\) is the temperature in Kelvin (which we have already converted, and \(m\) is the mass of one molecule, already calculated in previous steps.

Step 5: Calculate the rms speed

By substituting the known values into the equation, we can calculate the rms speed of air molecules at room temperature: \(v_{rms} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \, J/K \times 295.15 \, K}{4.81 \times 10^{-26} \, kg}}\) \(v_{rms} \approx 492 \, m/s\) The root-mean-square speed of air molecules at room temperature (22.0°C) is approximately 492 m/s.