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(A) The root mean square velocity $(v_{rms})$ of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.Since $R$ and $M$ are constant

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$-123$

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(A) The root mean square velocity $(v_{rms})$ of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.Since $R$ and $M$ are constants for a given gas,we have $v_{rms} \propto \sqrt{T}$.This implies $T \propto v_{rms}^2$.Given that the final velocity $v_2$ is half of the initial velocity $v_1$,i.e.,$v_2 = \frac{v_1}{2}$.Therefore,$\frac{T_2}{T_1} = \left(\frac{v_2}{v_1}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.The initial temperature $T_1 = 327^{\circ} C = 327 + 273 = 600 \ K$.Thus,$T_2 = \frac{T_1}{4} = \frac{600 \ K}{4} = 150 \ K$.Converting back to Celsius: $T_2 = 150 - 273 = -123^{\circ} C$.

To what temperature should the hydrogen at $327^{\circ} C$ be cooled at constant pressure,so that the root mean square velocity of its molecules becomes half of its previous value (in $^{\circ} C$)?

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