Question

Let, the average molar mass of air is $24.9g$. An open vessel at $27{}^{o}C$ is heated upto $t{}^{o}C$ until $1/3^{rd}$ of the air measured at final temperature excapes out. The rms velocity of air molecules at $t{}^{o}C$ is: (Take $R=8.3J{K}^{-1}mo{l}^{-1}$)

• A. $340.12mse{c}^{-1}$
• B. $420.35mse{c}^{-1}$
• C. $515.25mse{c}^{-1}$
• D. $632.45mse{c}^{-1}$

Answer 1

The correct option is D $632.45mse{c}^{-1}$

Initial temperature =$300K$
Let new temperature corresponding to $t^{o}C$ is $TK$.
Let number of moles present at $TK=n$
Moles of air excaped at $TK=n/3$
So, number of moles at $300K$ $\Rightarrow n+\frac{n}{3}=\frac{4n}{3}$
Volume and pressure are kept constant
$\therefore \frac{n_1}{n_2}=\frac{T_2}{T_1}$
$\frac{4n}{3}\times 300=nT$
$\therefore T=400K$
Now, r.m.s velocity $=\sqrt{\frac{3RT}{M}}=\sqrt{\frac{3\times 8.3\times 400}{24.9\times {10}^{-3}}}=632.45mse{c}^{-1}$.