Problem Statement
Find the rms speed of nitrogen molecules at $T = 300\ \text{K}$.
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.
- See the step-by-step solution for the specific equations applied.
- All quantities are in SI units unless otherwise stated.
Step-by-Step Solution
Step 1 — Identify given quantities and set up the problem: $$v_{rms} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3\times8.314\times300}{0.028}} = \sqrt{267,321} \approx 517\ \text{m/s}$$
Step 2 — Apply the relevant physical law or equation: Result: $v_{rms} \approx 517\ \text{m/s}$.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
$$v_{rms} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3\times8.314\times300}{0.028}} = \sqrt{267,321} \approx 517\ \text{m/s}$$
$$\text{Numerical result} = \text{given expression substituted with values}$$
$$\boxed{v_{rms} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3\times8.314\times300}{0.028}} = \sqrt{267,321} \approx 517\ \text{m/s}}$$
$$v_{rms} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3\times8.314\times300}{0.028}} = \sqrt{267,321} \approx 517\ \text{m/s}$$
Result: $v_{rms} \approx 517\ \text{m/s}$.
Answer
$$\boxed{v_{rms} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3\times8.314\times300}{0.028}} = \sqrt{267,321} \approx 517\ \text{m/s}}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
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