Q8 / 160 · TS EAMCET 2021 (06 Aug Shift 2) · Chemistry
At $300 \mathrm{~K}$ the average velocity of a gas is $3 \times 10^2 \mathrm{~cm} / \mathrm{s}$. The average velocity (in $\mathrm{cm} / \mathrm{s}$ ) of it at $1200 \mathrm{~K}$ is
- Correct answerA$6 \times 10^2$
- B$4 \times 10^2$
- C$8 \times 10^2$
- D$1 \times 10^3$
Solution
Using this formula, $\mu_{\mathrm{av}}=\sqrt{\frac{8 R T}{\pi M}}$
where, $\mu_{\mathrm{av}}=$ average velocity of a gas
$$
\begin{aligned}
R & =\text { gas constant } \\
T & =\text { temperature } \\
M & =\text { mass } \\
\pi & =3.14
\end{aligned}
$$
At temperature $T_1, \mu_{\mathrm{av}_1}=\sqrt{\frac{8 R T_1}{\pi M}}$
At temperature $T_2, \mu_{\mathrm{av}_2}=\sqrt{\frac{8 R T_2}{\pi M}}$
$\frac{\mu_{\mathrm{av}_1}}{\mu_{\mathrm{av}_2}}=\frac{\sqrt{8 R T_1 / \pi M}}{\sqrt{8 R T_2 / \pi M}}$
$\frac{\mu_{\mathrm{av}_1}}{\mu_{\mathrm{av}_2}}=\sqrt{\frac{T_1}{T_2}}$
Given,
$\mu_{\mathrm{av}_{\mathrm{l}}}=3 \times 10^2 \mathrm{~cm} / \mathrm{s}$
$$
\begin{aligned}
\frac{3 \times 10^2}{\mu_{\mathrm{av}_2}} & =\sqrt{\frac{300}{1200}} \\
\mu_{\mathrm{av}_2} & =2 \times 3 \times 10^2 \\
\mu_{\mathrm{av}_2} & =6 \times 10^2 \mathrm{~cm} / \mathrm{s}
\end{aligned}
$$
Hence, the correct answer is option: 'A'.