Question

At what temperature the most probable velocity of O2 gas is equal to the RMS velocity of O3 at t°C?​

Answer 1

Answer : At temperature $t^oC$ the most probable velocity of $O_2$ gas is equal to the RMS velocity of $O_3$.

Explanation :

Formula used for root mean square velocity :

$\nu_{rms}=\sqrt{\frac{3RT}{M}}$

where,

$\nu_{rms}$ = rms velocity of the molecule

R = gas constant

T = temperature

M = molar mass of the gas

Formula used for most probable velocity :

$\nu_{mp}=\sqrt{\frac{2RT}{M}}$

where,

$\nu_{mp}$ = most probable velocity of the molecule

R = gas constant

T = temperature

M = molar mass of the gas

As per question,

$\nu_{mp}_{O_2}=\nu_{rms}_{O_3}$

$\sqrt{\frac{2RT_{O_2}}{M_{O_2}}}=\sqrt{\frac{3RT_{O_3}}{M_{O_3}}}$

$\frac{2T_{O_2}}{M_{O_2}}=\frac{3T_{O_3}}{M_{O_3}}$

Molar mass of $O_2$ = 32 g/mole

Molar mass of $O_3$ = 48 g/mole

Temperature of $O_3$ = $t^oC$

Now put all the given values in the above relation, we get:

$\frac{2T_{O_2}}{32}=\frac{3\times t}{48}$

$T_{O_2}=t^oC$

Therefore, at temperature $t^oC$ the most probable velocity of $O_2$ gas is equal to the RMS velocity of $O_3$.