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For two gases, A and B with molecular weights M_A and M_B, it is observed that at a certain temperature, T, the mean velocity of A is equal to the root mean square velocity of B. Thus the mean velocity of A can be made equal to the mean velocity of B, if:

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A is at temperature, T₁ and B and $T_{2} T_{1} > T_{2}$

A is lowered to a temperature $T_{2} < T$ while B is at T

Both A and B are raised to a higher temperature

Both A and B are lowered in temperature

answer is B.

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Detailed Solution

It is known that,

$\frac{8 RT}{{πM}_{A}} = \frac{3 RT}{M_{B}}$

$\Rightarrow M_{A} = \frac{8}{3 π} \times M_{B}$

$\Rightarrow M_{B}$

Hence, temperature of A must be lowered.

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