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The r.m.s. velocity of $\mathrm{CO}_2$ at a temperature $T$ (in kelvin) is $x \mathrm{~cm} \mathrm{~s}^{-1}$. At what temperature (in kelvin), the r.m.s. velocity of nitrous oxide would be $4 x \mathrm{~cm} \mathrm{~s}^{-1}$ ?
(Atomic weights of $\mathrm{C}, \mathrm{N}$ and $\mathrm{O}$ are respectively 12,14 and 16 )
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(Atomic weights of $\mathrm{C}, \mathrm{N}$ and $\mathrm{O}$ are respectively 12,14 and 16 )
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Verified Answer
The correct answer is:
$16 T$
$r \mathrm{CO}_2=\sqrt{\frac{3 R T_{\mathrm{CO}_2}}{M_{\mathrm{CO}_2}}}$
$\begin{aligned} & r \mathrm{~N}_2 \mathrm{O}= \sqrt{\frac{3 R T_{\mathrm{N}_2 \mathrm{O}}}{M_{\mathrm{N}_2 \mathrm{O}}}} \\ & M_{\mathrm{CO}_2}= 44, M_{\mathrm{N}_2 \mathrm{O}}=44 \\ & \frac{r \mathrm{CO}_2}{r \mathrm{~N}_2 \mathrm{O}}=\sqrt{\frac{T_{\mathrm{CO}_2}}{T_{\mathrm{N}_2 \mathrm{O}}}} \\ & \frac{x}{4 x}=\sqrt{\frac{T_{\mathrm{C}_2}}{T_{\mathrm{N}_2 \mathrm{O}}}}\end{aligned}$
Squaring on both the sides,
$T_{\mathrm{N}_2 \mathrm{O}}=16 T_{\mathrm{OO}_2}$
$\begin{aligned} & r \mathrm{~N}_2 \mathrm{O}= \sqrt{\frac{3 R T_{\mathrm{N}_2 \mathrm{O}}}{M_{\mathrm{N}_2 \mathrm{O}}}} \\ & M_{\mathrm{CO}_2}= 44, M_{\mathrm{N}_2 \mathrm{O}}=44 \\ & \frac{r \mathrm{CO}_2}{r \mathrm{~N}_2 \mathrm{O}}=\sqrt{\frac{T_{\mathrm{CO}_2}}{T_{\mathrm{N}_2 \mathrm{O}}}} \\ & \frac{x}{4 x}=\sqrt{\frac{T_{\mathrm{C}_2}}{T_{\mathrm{N}_2 \mathrm{O}}}}\end{aligned}$
Squaring on both the sides,
$T_{\mathrm{N}_2 \mathrm{O}}=16 T_{\mathrm{OO}_2}$
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