Question

The root mean square speed of smoke particles of mass 5 × 10^-17 in their Brownian motion in air at NTP is approximately. [Given k = 1.38 × 10^-23 JK^-1]

• A. 60 mm s^-1
• B. 12 mm s-1
• C. 15 mm s-1
• D. 36 mm s-1

Answer 1

The Correct Option is C

To determine the root mean square (RMS) speed of smoke particles in Brownian motion at normal temperature and pressure (NTP), we utilize the formula for the RMS speed of particles in a gas:

\(v_{\text{rms}} = \sqrt{\frac{3kT}{m}}\) 

where:

• \(v_{\text{rms}}\) is the root mean square speed,
• \(k = 1.38 \times 10^{-23} \text{ JK}^{-1}\) is the Boltzmann constant,
• \(T\) is the temperature in Kelvin, specifically 273 K for NTP,
• \(m = 5 \times 10^{-17} \text{ kg}\) is the mass of the smoke particle.

Let's calculate \(v_{\text{rms}}\) step-by-step:

1. Substitute the values into the formula:

\(v_{\text{rms}} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 273}{5 \times 10^{-17}}}\)

1. Calculate the numerator:

\(3 \times 1.38 \times 10^{-23} \times 273 = 1.13034 \times 10^{-20}\)

1. Calculate the entire expression under the square root:

\(\frac{1.13034 \times 10^{-20}}{5 \times 10^{-17}} = 2.26068 \times 10^{-4}\)

1. Now, find the square root of the result:

\(v_{\text{rms}} = \sqrt{2.26068 \times 10^{-4}} \approx 15 \text{ m s}^{-1}\)

1. Convert the answer to mm/s:

\(15 \text{ m s}^{-1} = 15 \times 10^{3} \text{ mm s}^{-1} = 15 \text{ mm s}^{-1}\)

Thus, the correct answer is 15 mm s^-1.

Answer 2

At NTP, T=298 K
⇒v_rms=\(\sqrt\frac{3RT}{M}\)
\(=\sqrt{\frac{3kN_A×298}{5×10^{−17}×N_A}}\)
≃15 mm/s
So, the correct option is (C): 15 mm s^-1