MHT CET · Physics · Kinetic Theory of Gases
The temperature of a gas is \(-80^{\circ} \mathrm{C}\). To what temperature the gas should be heated so that the r.m.s. speed is increased by 2 times?
- A \(499^{\circ} \mathrm{C}\)
- B \(772^{\circ} \mathrm{C}\)
- C \(1464^{\circ} \mathrm{C}\)
- D \(1737^{\circ} \mathrm{C}\)
Answer & Solution
Correct Answer
(A) \(499^{\circ} \mathrm{C}\)
Step-by-step Solution
Detailed explanation
\(T_1=-80^{\circ} \mathrm{C}=-80+273 \mathrm{~K}=193 \mathrm{~K}\)
R.M.S. velocity, \(\mathrm{v}_{\mathrm{rms}} \propto \sqrt{\mathrm{T}}\)
As the speed is increased by two times, final speed becomes \(\left(2 \mathrm{~V}_{\mathrm{rms}}\right)_1\)
\(\begin{array}{ll}
& \frac{\left(v_{\mathrm{mss}}\right)_2}{\left(v_{\mathrm{ms}}\right)_1}=\sqrt{\frac{T_2}{T_1}}=2 \\
\therefore \quad & \frac{T_2}{T_1}=4 \\
\therefore \quad & T_2=4 \times 193=772 \mathrm{~K}=499^{\circ} \mathrm{C}
\end{array}\)
R.M.S. velocity, \(\mathrm{v}_{\mathrm{rms}} \propto \sqrt{\mathrm{T}}\)
As the speed is increased by two times, final speed becomes \(\left(2 \mathrm{~V}_{\mathrm{rms}}\right)_1\)
\(\begin{array}{ll}
& \frac{\left(v_{\mathrm{mss}}\right)_2}{\left(v_{\mathrm{ms}}\right)_1}=\sqrt{\frac{T_2}{T_1}}=2 \\
\therefore \quad & \frac{T_2}{T_1}=4 \\
\therefore \quad & T_2=4 \times 193=772 \mathrm{~K}=499^{\circ} \mathrm{C}
\end{array}\)
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