Chemistry - Ideal Gas Question with Solution | TestHub

Given the relationship:

$$\sqrt{\frac{2RT}{M_A}}=\sqrt{\frac{3RT}{M_B}}$$

This implies that $M_B=\frac{3}{2}{M}_A$, which means $M_B>M_A$.

For equal mean velocities, we need:

$$\sqrt{\frac{T_A}{M_A}}=\sqrt{\frac{T_B}{M_B}}$$

From this, we can derive:

$$\frac{T_B}{T_A}=\frac{M_B}{M_A}=\frac{3}{2}>1$$

Therefore, substance B must be hotter than substance A.

If we start with both substances at temperature $T$, we can lower the temperature of A to $T^{\prime }=\frac{2}{3}T<T$ and keep B at temperature $T$. This matches option (b).