The ratio between RMS velocities of $\mathrm{H}_2$ at $50 \mathrm{~K}$ and $\mathrm{O}_2$ at $800…
The ratio between RMS velocities of $\mathrm{H}_2$ at $50 \mathrm{~K}$ and $\mathrm{O}_2$ at $800 \mathrm{~K}$ is
- 4 : 1
- 2 : 1
- 1 : 1
- 1 : 4
Solution
Root mean square speed $=\sqrt{\frac{3 R T}{M}}$
Given,
$
\begin{aligned}
M_1 & =\text { molar mass of } \mathrm{H}_2=2 \\
M_2 & =\text { molar mass of } \mathrm{O}_2=32 \\
T_1 & =50 \mathrm{~K}, T_2=800 \mathrm{~K}
\end{aligned}
$
On taking ratio of $r_1$ and $r_2$, we get
$
\begin{aligned}
& \frac{r_1}{r_2}=\sqrt{\frac{M_2 \times T_1}{M_1 \times T_2}}=\sqrt{\frac{32}{2}} \times \sqrt{\frac{T_1}{T_2}} \\
& \frac{r_1}{r_2}=\sqrt{16} \times \sqrt{\frac{50}{800}} \text { or } \frac{r_1}{r_2}=4 \times \frac{1}{4}
\end{aligned}
$
or
$
\frac{r_1}{r_2}=\frac{1}{1}=1: 1
$
Hence, option (c) is correct
Asked in: AP EAMCET 2019 (20 Apr Shift 2)
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