MHT CET · Physics · Kinetic Theory of Gases
The temperature of a gas is \(-68^{\circ} \mathrm{C}\). To what temperature should it be heated, so that the r.m.s. velocity of the molecules be doubled?
- A \(357^{\circ} \mathrm{C}\)
- B \(457^{\circ} \mathrm{C}\)
- C \(547^{\circ} \mathrm{C}\)
- D \(820^{\circ} \mathrm{C}\)
Answer & Solution
Correct Answer
(C) \(547^{\circ} \mathrm{C}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{T}_1=-68^{\circ} \mathrm{C}=-68+273 \mathrm{~K}=205 \mathrm{~K}\)
R.M.S. velocity, \(\mathrm{v}_{\mathrm{rms}} \propto \sqrt{\mathrm{T}}\)
\(
\begin{array}{ll}
& \frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_2}{\left(\mathrm{v}_{\mathrm{rms}}\right)_1}=\sqrt{\frac{\mathrm{T}_2}{\mathrm{~T}_1}}=2 \\
\therefore & \frac{\mathrm{T}_2}{\mathrm{~T}_1}=4 \\
\therefore & \mathrm{T}_2=4 \times 205=820 \mathrm{~K} \\
\therefore & \mathrm{T}_2=547^{\circ} \mathrm{C}
\end{array}
\)
R.M.S. velocity, \(\mathrm{v}_{\mathrm{rms}} \propto \sqrt{\mathrm{T}}\)
\(
\begin{array}{ll}
& \frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_2}{\left(\mathrm{v}_{\mathrm{rms}}\right)_1}=\sqrt{\frac{\mathrm{T}_2}{\mathrm{~T}_1}}=2 \\
\therefore & \frac{\mathrm{T}_2}{\mathrm{~T}_1}=4 \\
\therefore & \mathrm{T}_2=4 \times 205=820 \mathrm{~K} \\
\therefore & \mathrm{T}_2=547^{\circ} \mathrm{C}
\end{array}
\)
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