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Numerical · Q17

Q.At what temperature will the RMS speed of nitrogen (N2N_2, molar mass 28 g/mol28\ \text{g/mol}) molecules equal 500 m/s500\ \text{m/s}?

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✓ Free question

Given vrms=500 m/sv_{rms}=500\ \text{m/s}, M=28 g/mol=0.028 kg/molM=28\ \text{g/mol}=0.028\ \text{kg/mol}, R=8.31 J/(mol K)R=8.31\ \text{J/(mol\,K)}. Squaring vrms=3RT/Mv_{rms}=\sqrt{3RT/M} and solving for TT,

T=vrms2M3R=(500)2×0.0283×8.31=250,000×0.02824.93=700024.93≈281 KT = \frac{v_{rms}^2 M}{3R} = \frac{(500)^2 \times 0.028}{3 \times 8.31} = \frac{250{,}000 \times 0.028}{24.93} = \frac{7000}{24.93} \approx 281\ \text{K}

✓Final answer

Nitrogen molecules reach an RMS speed of 500 m/s500\ \text{m/s} at a temperature of about 281 K281\ \text{K} (roughly 8 ∘C8\,^\circ\text{C}).

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