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A gas mixture consists of molecules of $A, B$ and $C$ with masses $m_A>m_B>m_C$. Rank the three types of molecules in decreasing order of
(a) average KE
(b) rms speeds
(a) average KE
(b) rms speeds
Solution:
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Verified Answer
(a) The average KE will be same as condition of temperature and pressure are the same.
(b) For the rms speed, the temperature and pressure are the same.
$$
\begin{aligned}
v_{\mathrm{rms}} &=\sqrt{\frac{3 p V}{M}}=\sqrt{\frac{3 R T}{M}} \\
&=\sqrt{\frac{3 R T}{m N}}=\sqrt{\frac{3 k_B T}{m}}
\end{aligned}
$$
where, $M=$ molar mass of the gas
$m=$ mass of each molecular of the gas
$R=$ gas constant
$T=$ absolute temperature (same for all)
So, $v_{\mathrm{rms}} \propto \sqrt{\frac{1}{m}}$
$m_A>m_B>m_C \quad$ (given)
Hence, $\left(v_{\text {ms }}\right)_C>\left(v_{\text {rms }}\right)_B>\left(v_{\text {ms }}\right)_A$
(b) For the rms speed, the temperature and pressure are the same.
$$
\begin{aligned}
v_{\mathrm{rms}} &=\sqrt{\frac{3 p V}{M}}=\sqrt{\frac{3 R T}{M}} \\
&=\sqrt{\frac{3 R T}{m N}}=\sqrt{\frac{3 k_B T}{m}}
\end{aligned}
$$
where, $M=$ molar mass of the gas
$m=$ mass of each molecular of the gas
$R=$ gas constant
$T=$ absolute temperature (same for all)
So, $v_{\mathrm{rms}} \propto \sqrt{\frac{1}{m}}$
$m_A>m_B>m_C \quad$ (given)
Hence, $\left(v_{\text {ms }}\right)_C>\left(v_{\text {rms }}\right)_B>\left(v_{\text {ms }}\right)_A$
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