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]
- 60 mm s-1
- 12 mm s-1
- 15 mm s-1
- 36 mm s-1
The Correct Option is C
Approach Solution - 1
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:
- Substitute the values into the formula:
\(v_{\text{rms}} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 273}{5 \times 10^{-17}}}\)
- Calculate the numerator:
\(3 \times 1.38 \times 10^{-23} \times 273 = 1.13034 \times 10^{-20}\)
- Calculate the entire expression under the square root:
\(\frac{1.13034 \times 10^{-20}}{5 \times 10^{-17}} = 2.26068 \times 10^{-4}\)
- Now, find the square root of the result:
\(v_{\text{rms}} = \sqrt{2.26068 \times 10^{-4}} \approx 15 \text{ m s}^{-1}\)
- 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.
Approach Solution -2
At NTP, T=298 K
⇒vrms=\(\sqrt\frac{3RT}{M}\)
\(=\sqrt{\frac{3kN_A×298}{5×10^{−17}×N_A}}\)
≃15 mm/s
So, the correct option is (C): 15 mm s-1
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Concepts Used:
Speed and Velocity
The rate at which an object covers a certain distance is commonly known as speed.
The rate at which an object changes position in a certain direction is called velocity.
Difference Between Speed and Velocity:
Read More: Difference Between Speed and Velocity