Using v_rms = sqrt(3 k_B T / m), compute v_rms for T = 300 K and molecular mass m = 4.65 × 10^-26 kg.

Using v_rms = sqrt(3 k_B T / m), compute v_rms for T = 300 K and molecular mass m = 4.65 × 10^-26 kg.

• A. 517 m/s
• B. 530 m/s
• C. 480 m/s
• D. 600 m/s

Answer: (A)

The speed of particles in a gas at a given temperature is tied to their thermal energy through v_rms = sqrt(3 k_B T / m). Here, T is 300 K and m is the mass per molecule, 4.65 × 10^-26 kg, with k_B = 1.38 × 10^-23 J/K. Compute the inside: 3 k_B T = 3 × (1.38 × 10^-23) × 300 ≈ 1.242 × 10^-20 J. Divide by m: (1.242 × 10^-20) / (4.65 × 10^-26) ≈ 2.67 × 10^5 (m^2/s^2). Take the square root: v_rms ≈ sqrt(2.67 × 10^5) ≈ 5.17 × 10^2 m/s, i.e., about 517 m/s. So the speed corresponding to these conditions is roughly 517 m/s.

The speed of particles in a gas at a given temperature is tied to their thermal energy through v_rms = sqrt(3 k_B T / m). Here, T is 300 K and m is the mass per molecule, 4.65 × 10^-26 kg, with k_B = 1.38 × 10^-23 J/K.

Compute the inside: 3 k_B T = 3 × (1.38 × 10^-23) × 300 ≈ 1.242 × 10^-20 J. Divide by m: (1.242 × 10^-20) / (4.65 × 10^-26) ≈ 2.67 × 10^5 (m^2/s^2). Take the square root: v_rms ≈ sqrt(2.67 × 10^5) ≈ 5.17 × 10^2 m/s, i.e., about 517 m/s.

So the speed corresponding to these conditions is roughly 517 m/s.