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:
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:
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Always convert molar mass from grams per mole to kilograms per mole when using the standard gas constant \( R \).
Updated On: Jun 9, 2026
- \( 280 \text{ K} \)
- \( 300 \text{ K} \)
- \( 350 \text{ K} \)
- \( 250 \text{ K} \)
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The Correct Option is A
Solution and Explanation
Concept:
The root mean square (rms) speed of gas molecules is given by \( v_{rms} = \sqrt{\frac{3RT}{M}} \), where \( M \) is the molar mass in kg/mol.
Step 1: Prepare the variables.
\( v_{rms} = 500 \text{ ms}^{-1} \). \( M = 28 \times 10^{-3} \text{ kg/mol} \). \( R = 8.314 \text{ J mol}^{-1} \text{ K}^{-1} \).
Step 2: Rearrange for temperature \( T \).
$$ v_{rms}^2 = \frac{3RT}{M} \implies T = \frac{v_{rms}^2 \times M}{3R} $$
Step 3: Calculate the numerical value.
$$ T = \frac{(500)^2 \times (28 \times 10^{-3})}{3 \times 8.314} $$ $$ T = \frac{250,000 \times 0.028}{24.942} = \frac{7000}{24.942} \approx 280.6 \text{ K} $$ The approximate temperature is 280 K. $$\boxed{280 \text{ K}}$$
Step 1: Prepare the variables.
\( v_{rms} = 500 \text{ ms}^{-1} \). \( M = 28 \times 10^{-3} \text{ kg/mol} \). \( R = 8.314 \text{ J mol}^{-1} \text{ K}^{-1} \).
Step 2: Rearrange for temperature \( T \).
$$ v_{rms}^2 = \frac{3RT}{M} \implies T = \frac{v_{rms}^2 \times M}{3R} $$
Step 3: Calculate the numerical value.
$$ T = \frac{(500)^2 \times (28 \times 10^{-3})}{3 \times 8.314} $$ $$ T = \frac{250,000 \times 0.028}{24.942} = \frac{7000}{24.942} \approx 280.6 \text{ K} $$ The approximate temperature is 280 K. $$\boxed{280 \text{ K}}$$
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