Question

The velocity of the nitrogen molecule in room temperature air is:

• A. zero
• B. 10 m/s
• C. 100 m/s
• D. 500 m/s
• E. 5000 m/s

Hint:

The velocity of molecules in gases can be estimated using the ideal gas law and kinetic theory. Typical molecular speeds are often in the range of hundreds of meters per second.

Answer 1

The Correct Option is D

The root mean square (rms) speed of a gas molecule is given by: 

$$ v_{\text{rms}} = \sqrt{\frac{3kT}{m}} $$ or alternatively, $$ v_{\text{rms}} = \sqrt{\frac{3RT}{M}} $$ where:
$R$ = gas constant,
$T$ = temperature in Kelvin,
$M$ = molar mass of the gas.

For nitrogen ($\text{N}_2$), molar mass $M \approx 28 \, \text{g/mol} = 0.028 \, \text{kg/mol}$,
At room temperature, $T \approx 300\, \text{K}$ 

Now, $$ v_{\text{rms}} = \sqrt{\frac{3 \times 8.314 \times 300}{0.028}} \approx \sqrt{26614.29} \approx 516\, \text{m/s} $$ 
Correct answer: (d) 500 m/s

Answer 2

1. Concept: The velocity of gas molecules is related to their kinetic energy, which in turn is related to the temperature. A common measure of this velocity is the root-mean-square (RMS) speed, given by the formula from the kinetic theory of gases.
2. RMS Speed Formula: The RMS speed (\(v_{rms}\)) is given by: \[ v_{rms} = \sqrt{\frac{3 k_B T}{m}} \] where:
   • \(k_B\) is the Boltzmann constant (\(1.38 \times 10^{-23}\) J/K)
   • \(T\) is the absolute temperature in Kelvin
   • \(m\) is the mass of a single molecule in kilograms
3. Parameters:
   • Temperature (T): "Room temperature" is typically around 20-25 °C. Let's use \(T = 20^\circ C = 293\) K (or approximately 300 K for estimation).
   • Molecule: Nitrogen (N₂).
   • Mass of N₂ molecule (m):
     • Molar mass of N₂ is approx. 28 g/mol = 0.028 kg/mol.
     • Avogadro's number (\(N_A\)) is approx. \(6.022 \times 10^{23}\) molecules/mol.
     • Mass \(m = \frac{\text{Molar mass}}{N_A} = \frac{0.028 \text{ kg/mol}}{6.022 \times 10^{23} \text{ mol}^{-1}} \approx 4.65 \times 10^{-26}\) kg.
4. Calculation (using T ≈ 293 K):

\[ v_{rms} = \sqrt{\frac{3 \times (1.38 \times 10^{-23} \text{ J/K}) \times (293 \text{ K})}{4.65 \times 10^{-26} \text{ kg}}} \]

\[ v_{rms} = \sqrt{\frac{1.213 \times 10^{-20}}{4.65 \times 10^{-26}}} \approx \sqrt{0.261 \times 10^6} \approx \sqrt{2.61 \times 10^5} \text{ m/s} \]

\[ v_{rms} \approx 510 \text{ m/s} \]

1. Calculation (using T ≈ 300 K):

\[ v_{rms} = \sqrt{\frac{3 \times (1.38 \times 10^{-23} \text{ J/K}) \times (300 \text{ K})}{4.65 \times 10^{-26} \text{ kg}}} \]

\[ v_{rms} = \sqrt{\frac{1.242 \times 10^{-20}}{4.65 \times 10^{-26}}} \approx \sqrt{0.267 \times 10^6} \approx \sqrt{2.67 \times 10^5} \text{ m/s} \]

\[ v_{rms} \approx 517 \text{ m/s} \]

1. Compare with Options: The calculated RMS speed is approximately 510-517 m/s. This value represents a typical speed for the molecules.
   • (a) zero: Incorrect. Molecules are in constant random motion unless at absolute zero.
   • (b) 10 m/s: Too slow.
   • (c) 100 m/s: Too slow.
   • (d) 500 m/s: This is the closest value to our calculation.
   • (e) 5000 m/s: Too fast (comparable to orbital speeds).

The velocity of nitrogen molecules in room temperature air is closest to 500 m/s.

The correct option is (d) 500 m/s.