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IV. Numerical Problems · Q1

Q.A fresh air is composed of nitrogen N2N_2 (78%) and oxygen O2O_2 (21%). Find the rms speed of N2N_2 and O2O_2 at 20∘20^\circC.

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✓ Free question

Step 1. T=20∘C=293T=20^\circ\text{C}=293 K, R=8.314 J mol−1K−1R=8.314\ \text{J mol}^{-1}\text{K}^{-1}. The rms speed depends only on T and molar mass, not on the percentage composition given.

Step 2. For N2N_2, M=28×10−3 kg mol−1M=28\times10^{-3}\ \text{kg mol}^{-1}: vrms=3×8.314×29328×10−3=7308.00.028=261000≈511 m s−1v_{rms}=\sqrt{\dfrac{3\times8.314\times293}{28\times10^{-3}}}=\sqrt{\dfrac{7308.0}{0.028}}=\sqrt{261000}\approx511\ \text{m s}^{-1}.

Step 3. For O2O_2, M=32×10−3 kg mol−1M=32\times10^{-3}\ \text{kg mol}^{-1}: vrms=3×8.314×29332×10−3=7308.00.032=228375≈478 m s−1v_{rms}=\sqrt{\dfrac{3\times8.314\times293}{32\times10^{-3}}}=\sqrt{\dfrac{7308.0}{0.032}}=\sqrt{228375}\approx478\ \text{m s}^{-1}.

Step 4. As expected, the lighter N2N_2 molecule has a higher rms speed than the heavier O2O_2 molecule at the same temperature.

✓Final answer

vrms,N2≈511 m s−1v_{rms,N_2}\approx511\ \text{m s}^{-1}, vrms,O2≈478 m s−1v_{rms,O_2}\approx478\ \text{m s}^{-1}.

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