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(D) The $RMS$ velocity of a gas is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.Since $v_{rms} \propto \sqrt{T}$, reducing the $RMS$ velocity to half mea

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$32$

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(D) The $RMS$ velocity of a gas is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.Since $v_{rms} \propto \sqrt{T}$, reducing the $RMS$ velocity to half means the final temperature $T_f$ must be $\frac{1}{4}$ of the initial temperature $T_i$ $(T_f = \frac{T_i}{4})$.For a reversible adiabatic process, $T_i V_i^{\gamma-1} = T_f V_f^{\gamma-1}$.For a diatomic gas, $\gamma = \frac{7}{5}$, so $\gamma - 1 = \frac{2}{5}$.Substituting the values: $T_i V_i^{2/5} = (\frac{T_i}{4}) V_f^{2/5}$.This simplifies to $4 = (\frac{V_f}{V_i})^{2/5}$.Raising both sides to the power of $\frac{5}{2}$: $4^{5/2} = \frac{V_f}{V_i}$.$V_f / V_i = (2^2)^{5/2} = 2^5 = 32$.

How many times the volume of a diatomic gas should be increased reversibly and adiabatically in order to reduce its $RMS$ velocity to half of its initial value?

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