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The root mean square (rms) speed of \(X_2\) gas is \(\mathrm{x} \mathrm{m/s}\) at a given temperature. When the temperature is doubled, the \(\mathrm{X}_2\) molecules dissociated completely into atoms. The root mean square speed of the sample of gas then becomes (in \(\mathrm{m} / \mathrm{s}\) )
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Verified Answer
The correct answer is:
\(2 x\)
Hint : \(C_{m m s}=\sqrt{\frac{3 R T}{M}}\)
\(\begin{array}{ll}
\mathrm{T}_1=\mathrm{T} & \mathrm{T}_2=2 \mathrm{~T} \\
\mathrm{M}_1=\mathrm{M} & \mathrm{M}_2=\mathrm{M} / 2 \\
\mathrm{C}_1=\mathrm{X} & \mathrm{C}_2=?
\end{array}\)
\(\frac{C_1}{C_2}=\sqrt{\frac{T_1}{M_1} \times \frac{\mathrm{M}_2}{\mathrm{~T}_2}}=\sqrt{\frac{\mathrm{T}}{\mathrm{M}} \times \frac{\mathrm{M} / 2}{2 T}}=\frac{1}{2}\)
\(\therefore \frac{\mathrm{x}}{\mathrm{C}_2}=\frac{1}{2}\), Hence \(\mathrm{C}_2=2 \mathrm{x~} \mathrm{m} / \mathrm{s}\)
\(\begin{array}{ll}
\mathrm{T}_1=\mathrm{T} & \mathrm{T}_2=2 \mathrm{~T} \\
\mathrm{M}_1=\mathrm{M} & \mathrm{M}_2=\mathrm{M} / 2 \\
\mathrm{C}_1=\mathrm{X} & \mathrm{C}_2=?
\end{array}\)
\(\frac{C_1}{C_2}=\sqrt{\frac{T_1}{M_1} \times \frac{\mathrm{M}_2}{\mathrm{~T}_2}}=\sqrt{\frac{\mathrm{T}}{\mathrm{M}} \times \frac{\mathrm{M} / 2}{2 T}}=\frac{1}{2}\)
\(\therefore \frac{\mathrm{x}}{\mathrm{C}_2}=\frac{1}{2}\), Hence \(\mathrm{C}_2=2 \mathrm{x~} \mathrm{m} / \mathrm{s}\)
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