The rms speed of an ideal diatomic gas at tem | vedclass

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(D) The $rms$ speed of a gas is given by the formula $v = \sqrt{\frac{3RT}{M}}$,where $M$ is the molar mass of the gas.For the diatomic gas at tempera

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$2T$

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(D) The $rms$ speed of a gas is given by the formula $v = \sqrt{\frac{3RT}{M}}$,where $M$ is the molar mass of the gas.For the diatomic gas at temperature $T$,the speed is $v = \sqrt{\frac{3RT}{M}}$.When the gas dissociates into atoms,the molar mass becomes $M' = \frac{M}{2}$.Let the new temperature be $T'$. The new $rms$ speed is $2v = \sqrt{\frac{3RT'}{M/2}}$.Squaring both sides of the equation for the new speed: $(2v)^2 = \frac{3RT'}{M/2} \Rightarrow 4v^2 = \frac{6RT'}{M}$.Since $v^2 = \frac{3RT}{M}$,we substitute this into the equation: $4 \left( \frac{3RT}{M} \right) = \frac{6RT'}{M}$.Canceling common terms $\frac{3R}{M}$ from both sides,we get $4T = 2T'$,which simplifies to $T' = 2T$.

The $rms$ speed of an ideal diatomic gas at temperature $T$ is $v$. When the gas dissociates into atoms,its new $rms$ speed becomes double. At what temperature does the dissociation of the gas occur?

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