Question

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 )

• A. 1260 m/s
• B. 91 m/s
• C. 27.4 m/s
• D. 523 m/s

Answer 1

The Correct Option is D

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 \( 523 \, \text{m/s} \).