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Q.

At what temperature the most probable velocity of O₂ gas is equal to the RMS velocity of O₃ at ‘t’° C ?

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a

(273 + t) K

b

(273 + t) $\sqrt{\frac{3}{2} K}$

c

(273 + t) $\sqrt{\frac{2}{3} K}$

d

3/2 (273 + t) K

answer is C.

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Detailed Solution

$V_{RMS} = \sqrt{\frac{3 RT}{M}} V_{mp} = \sqrt{\frac{2 RT}{M}} V_{{mp}_{O_{2}}} = V_{{RMS}_{O_{3}}} \sqrt{\frac{2 {RT}_{O_{2}}}{M_{O_{2}}}} = \sqrt{\frac{3 {RT}_{O_{3}}}{M_{O_{3}}}} \frac{2 T_{O_{2}}}{M_{O_{2}}} = \frac{3 T_{O_{3}}}{M_{O_{3}}} \frac{2 T_{O_{2}}}{32} = \frac{3 t}{48} T_{O_{2}} = t ° C = (273 + t) K$

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