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An ideal gas in a closed container is heated so that the final rms speed of the gas particles increases by 2 times the initial rms speed. If the initial gas temperature is $27^{\circ} \mathrm{C}$, then the final temperature of the ideal gas is :
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Verified Answer
The correct answer is:
$927^{\circ} \mathrm{C}$
As, rms velocity of ideal gas particles is given,
$$
\begin{array}{rlrl}
\Rightarrow \quad v_{\mathrm{rms}} & =\sqrt{\frac{3 k T}{m}} \\
& v_{\mathrm{rms}} & \propto \sqrt{T}
\end{array}
$$
where, temperature $T$ is measured in kelvin scale. So, the ratio
$$
\frac{v_{1 \mathrm{rms}}^2}{v_{2 \mathrm{rms}}^2}=\frac{T_1}{T_2}
$$
As given, $v_{2 \mathrm{rms}}=2 v_{1 \mathrm{rms}}$ and $T_1=27^{\circ} \mathrm{C}=300 \mathrm{~K}$
$$
\begin{array}{rlrl}
\Rightarrow & & \frac{v_{1 \mathrm{~ms}}^2}{4 v_{1 \mathrm{rms}}^2} & =\frac{300}{x} \\
\Rightarrow & x & =300 \times 4=1200 \mathrm{~K}
\end{array}
$$
Hence, $1200 \mathrm{~K}=(1200-273)^{\circ} \mathrm{C}=927^{\circ} \mathrm{C}$
Hence, the correct option is (2).
$$
\begin{array}{rlrl}
\Rightarrow \quad v_{\mathrm{rms}} & =\sqrt{\frac{3 k T}{m}} \\
& v_{\mathrm{rms}} & \propto \sqrt{T}
\end{array}
$$
where, temperature $T$ is measured in kelvin scale. So, the ratio
$$
\frac{v_{1 \mathrm{rms}}^2}{v_{2 \mathrm{rms}}^2}=\frac{T_1}{T_2}
$$
As given, $v_{2 \mathrm{rms}}=2 v_{1 \mathrm{rms}}$ and $T_1=27^{\circ} \mathrm{C}=300 \mathrm{~K}$
$$
\begin{array}{rlrl}
\Rightarrow & & \frac{v_{1 \mathrm{~ms}}^2}{4 v_{1 \mathrm{rms}}^2} & =\frac{300}{x} \\
\Rightarrow & x & =300 \times 4=1200 \mathrm{~K}
\end{array}
$$
Hence, $1200 \mathrm{~K}=(1200-273)^{\circ} \mathrm{C}=927^{\circ} \mathrm{C}$
Hence, the correct option is (2).
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