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Let A and \(\mathrm{B}\) the two gases and given :
\(\frac{T_A}{M_A}=4 \cdot \frac{T_B}{M_B}\) Where \(T\) is the temperature and \(M\) is molecular mass. If \(C_A\) and \(C_B\) are the r.m.s. speed, then the ratio \(\frac{C_A}{C_B}\) will be equal to:
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\(\frac{T_A}{M_A}=4 \cdot \frac{T_B}{M_B}\) Where \(T\) is the temperature and \(M\) is molecular mass. If \(C_A\) and \(C_B\) are the r.m.s. speed, then the ratio \(\frac{C_A}{C_B}\) will be equal to:
Solution:
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Verified Answer
The correct answer is:
$2$
Let \(A\) and \(B\) two gases, then,
\(\frac{T_A}{M_A}=\frac{4 T_B}{M_B}\)
Now,
\(\begin{aligned}
& \frac{C_A}{C_B}=\frac{\sqrt{\frac{3 R T_A}{M_A}}}{\sqrt{\frac{3 R T_B}{M_B}}} \\
& \frac{C_A}{C_B}=2
\end{aligned}\)
\(\frac{T_A}{M_A}=\frac{4 T_B}{M_B}\)
Now,
\(\begin{aligned}
& \frac{C_A}{C_B}=\frac{\sqrt{\frac{3 R T_A}{M_A}}}{\sqrt{\frac{3 R T_B}{M_B}}} \\
& \frac{C_A}{C_B}=2
\end{aligned}\)
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