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 $_{A}=R M S_{B}$

$\sqrt{\frac{8 RT }{\pi M _{ A }}}=\sqrt{\frac{3 RT }{ M _{ B }}}$

$\sqrt{\frac{8}{\pi M_{A}}}=\sqrt{\frac{3}{M_{B}}}$

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 $_{B}$

$\sqrt{\frac{8 RT _{ A }}{ M _{ A }}} \cong \sqrt{\frac{8 RT _{ B }}{ M _{ B }}}$

Since, $M_{B}>M_{A}$,

$ \therefore $ either $T_{B}$ should be raised or $T_{A}$ be lowered to make them equal