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Relating Translational KE and T

The temperatu...

Question

The temperature at which most probable speed of $O_{2}$ becomes equal to the root mean square speed of $N_{2}$ is:
[Given: $N_{2}$ at $427^{0} C$ ]

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Solution

$V_{most probable} = \sqrt{\frac{2 R T}{M}}; V_{r m s} = \sqrt{\frac{3 R T}{M}} V_{m p O_{2}} = \sqrt{\frac{2 R T_{1}}{32}}; V_{m p N_{2}} = \sqrt{\frac{3 R T_{2}}{28}} Given : T_{2} = 427 + 273 = 700 K \Rightarrow \sqrt{\frac{2 R T_{1}}{32}} = \sqrt{\frac{3 R T_{2}}{28}} \Rightarrow \frac{2 T_{1}}{32} = \frac{3 T_{2}}{28} T_{1} = \frac{16 \times 3 T_{2}}{28} T_{1} = 1200 K$

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