Question

The approximate temperature at which the rms speed of Nitrogen gas molecule is \( 500 \text{ ms}^{-1} \). Gas constant \( R=8.314 \text{ J mol}^{-1} \text{K}^{-1} \), Mass number of Nitrogen = 28:

• A. \( 280 \text{ K} \)
• B. \( 300 \text{ K} \)
• C. \( 350 \text{ K} \)
• D. \( 250 \text{ K} \)

Hint:

Always convert molar mass from grams per mole to kilograms per mole when using the standard gas constant \( R \).

Answer 1

The Correct Option is A

Step 1: Recall the rms speed formula.
For gas molecules, $v_{rms} = \sqrt{\dfrac{3RT}{M}}$, where $M$ is the molar mass in kilograms per mole. We want the temperature $T$.
Step 2: Square to remove the root.
$v_{rms}^2 = \dfrac{3RT}{M}$.
Step 3: Make $T$ the subject.
\[ T = \frac{v_{rms}^2\,M}{3R}. \]
Step 4: List the values.
$v_{rms} = 500\,\text{m s}^{-1}$, $M = 28\times10^{-3}\,\text{kg/mol}$, $R = 8.314\,\text{J mol}^{-1}\text{K}^{-1}$.
Step 5: Substitute.
\[ T = \frac{(500)^2 \times (28\times10^{-3})}{3 \times 8.314} = \frac{250000 \times 0.028}{24.942} = \frac{7000}{24.942}. \]
Step 6: Evaluate.
$T \approx 280.6\,\text{K}$, which is about $280\,\text{K}$.
\[ \boxed{280\ \text{K}} \]