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A thermally insulated vessel containing monatomic gas is moving with a speed of $30 \mathrm{~m} / \mathrm{s}$. If the vessel suddenly stops, the increase in gas temperature is (Molar mass of gas $=83 \mathrm{~g} / \mathrm{mol}$ and $R=8.3 \mathrm{~J} / \mathrm{K} \mathrm{mol}$ )
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Verified Answer
The correct answer is:
$3 \mathrm{~K}$
Given, velocity of monatomic gas,
$$
v=30 \mathrm{~m} / \mathrm{s}
$$
molar mass of gas, $m=83 \mathrm{~g} / \mathrm{mole}$
$$
=0.083 \mathrm{~kg} / \mathrm{mole}
$$
and gas constant, $R=8.3 \mathrm{~J} / \mathrm{K}-\mathrm{mol}$
$\therefore$ As we know that $v=\sqrt{\frac{3 R T}{m}}$
$$
\begin{aligned}
30 & =\sqrt{\frac{3 \times 8.3 \times T}{0.083}} \\
30 & =\sqrt{300 T} \\
\Rightarrow \quad 900 & =300 T \Rightarrow T=\frac{900}{300} \\
T & =3 \mathrm{~K}
\end{aligned}
$$
$$
v=30 \mathrm{~m} / \mathrm{s}
$$
molar mass of gas, $m=83 \mathrm{~g} / \mathrm{mole}$
$$
=0.083 \mathrm{~kg} / \mathrm{mole}
$$
and gas constant, $R=8.3 \mathrm{~J} / \mathrm{K}-\mathrm{mol}$
$\therefore$ As we know that $v=\sqrt{\frac{3 R T}{m}}$
$$
\begin{aligned}
30 & =\sqrt{\frac{3 \times 8.3 \times T}{0.083}} \\
30 & =\sqrt{300 T} \\
\Rightarrow \quad 900 & =300 T \Rightarrow T=\frac{900}{300} \\
T & =3 \mathrm{~K}
\end{aligned}
$$
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