The root mean square speed of molecules of nitrogen gas at 27°C is approximately : (Given mass of a nitrogen molecule = \(4.6 × 10^{–26}\) kg and take Boltzmann constant KB = \(1.4 × 10^{–23}\) JK–1 )
- 1260 m/s
- 91 m/s
- 27.4 m/s
- 523 m/s
The Correct Option is D
Solution and Explanation
Given:
- Temperature (\( T \)) = \( 27^\circ \text{C} = 300 \, \text{K} \)
- Mass of a nitrogen molecule (\( m \)) = \( 4.6 \times 10^{-26} \, \text{kg} \)
- Boltzmann constant (\( k_B \)) = \( 1.4 \times 10^{-23} \, \text{J/K} \)
Step 1: Formula for Root Mean Square (rms) Speed
The root mean square (rms) speed (\( V_{\text{rms}} \)) of gas molecules is given by:
\[ V_{\text{rms}} = \sqrt{\frac{3k_B T}{m}}, \]
where:
- \( k_B \): Boltzmann constant
- \( T \): Temperature in Kelvin
- \( m \): Mass of one molecule
Step 2: Substitute the Given Values
Substitute \( k_B = 1.4 \times 10^{-23} \, \text{J/K} \), \( T = 300 \, \text{K} \), and \( m = 4.6 \times 10^{-26} \, \text{kg} \):
\[ V_{\text{rms}} = \sqrt{\frac{3 \times 1.4 \times 10^{-23} \times 300}{4.6 \times 10^{-26}}}. \]
Simplify the numerator:
\[ 3 \times 1.4 \times 300 = 1260 \times 10^{-23} = 1.26 \times 10^{-20}. \]
Divide by the denominator:
\[ \frac{1.26 \times 10^{-20}}{4.6 \times 10^{-26}} = 2.73 \times 10^5. \]
Take the square root:
\[ V_{\text{rms}} = \sqrt{2.73 \times 10^5} \approx 523 \, \text{m/s}. \]
Final Answer:
The root mean square speed of nitrogen molecules is approximately \( 523 \, \text{m/s} \).
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