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Two gases $A$ and $B$ are contained in two separate, but otherwise identical containers. Gas $A$ consists of monatomic molecules, each with atomic mass of $4 u$ whereas Gas $B$ consists of rigid diatomic molecules, each with atomic mass of $20 u$. If gas $A$ is kept at $27^{\circ} \mathrm{C}$, at what temperature should gas $B$ be kept so that both have the same rms speed?
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Verified Answer
The correct answer is:
$270^{\circ} \mathrm{C}$
For monatomic gas $A$,
atomic mass of molecules of gas $A, m_A=4 u$
and temperature of gas, $T_A=27^{\circ} \mathrm{C}$
For diatomic gas $B$,
atomic mass of molecules of gas $B$,
$m_B=2 \times 20 \mathrm{u}=40 \mathrm{u}$
Temperature of gas, $B=\mathrm{T}_B$
Since, $r m s$ speed of the molecules of gas, $A=\mathrm{rms}$ speed of molecules of gas, $B$
$$
\begin{aligned}
\left(v_{\mathrm{rms}}\right)_A & =\left(v_{\mathrm{rms}}\right)_B \\
\Rightarrow \quad \sqrt{\frac{3 R T_A}{m_A}} & =\sqrt{\frac{3 R T_B}{m_B}} \\
\Rightarrow \quad \frac{\mathrm{T}_A}{m_A} & =\frac{\mathrm{T}_B}{m_B} \Rightarrow \frac{27}{4 u}=\frac{\mathrm{T}_B}{40 \mathrm{u}} \\
\mathrm{T}_B & =270^{\circ} \mathrm{C}
\end{aligned}
$$
atomic mass of molecules of gas $A, m_A=4 u$
and temperature of gas, $T_A=27^{\circ} \mathrm{C}$
For diatomic gas $B$,
atomic mass of molecules of gas $B$,
$m_B=2 \times 20 \mathrm{u}=40 \mathrm{u}$
Temperature of gas, $B=\mathrm{T}_B$
Since, $r m s$ speed of the molecules of gas, $A=\mathrm{rms}$ speed of molecules of gas, $B$
$$
\begin{aligned}
\left(v_{\mathrm{rms}}\right)_A & =\left(v_{\mathrm{rms}}\right)_B \\
\Rightarrow \quad \sqrt{\frac{3 R T_A}{m_A}} & =\sqrt{\frac{3 R T_B}{m_B}} \\
\Rightarrow \quad \frac{\mathrm{T}_A}{m_A} & =\frac{\mathrm{T}_B}{m_B} \Rightarrow \frac{27}{4 u}=\frac{\mathrm{T}_B}{40 \mathrm{u}} \\
\mathrm{T}_B & =270^{\circ} \mathrm{C}
\end{aligned}
$$
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