Question

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:

• A. A is at temperature $T_1$ and $B$ at $T_2$ and $T_1>T_2$
• B. $A$ is lowered to a temperature $T_2<T$ while $B$ is at $T$
• C. Both $A$ and $B$ are lowered in temperature
• D. Both $A$ and $B$ are raised in temperature

Answer 1

Answer: (B)

Avg vel ${}_A=RM{S}_B$

$\sqrt{\frac{8RT}{\pi M_A}}=\sqrt{\frac{3RT}{M_B}}$

$\sqrt{\frac{8}{\pi M_A}}=\sqrt{\frac{3}{M_B}}$

Squaring both sides -

$\frac{8}{\pi M_A}=\frac{3}{M_B}$

$\frac{8}{3\pi }=\frac{M_A}{M_B}$

$\therefore M_B>M_A$

We wish,

Av. vel ${}_A=$ Av. vel ${}_B$

$\sqrt{\frac{8R{T}_A}{M_A}}\cong \sqrt{\frac{8R{T}_B}{M_B}}$

Since, $M_B>M_A$,

$\therefore$ either $T_B$ should be raised or $T_A$ be lowered to make them equal