Let A and B be two gases and given: frac{T_A} | vedclass

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(A) The $r.m.s.$ speed of a gas is given by the formula $C = \sqrt{\frac{3RT}{M}}$.For gas $A$,$C_A = \sqrt{\frac{3RT_A}{M_A}}$.For gas $B$,$C_B = \sq

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$2$

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(A) The $r.m.s.$ speed of a gas is given by the formula $C = \sqrt{\frac{3RT}{M}}$.For gas $A$,$C_A = \sqrt{\frac{3RT_A}{M_A}}$.For gas $B$,$C_B = \sqrt{\frac{3RT_B}{M_B}}$.The ratio of the $r.m.s.$ speeds is $\frac{C_A}{C_B} = \frac{\sqrt{\frac{3RT_A}{M_A}}}{\sqrt{\frac{3RT_B}{M_B}}} = \sqrt{\frac{T_A}{M_A} \cdot \frac{M_B}{T_B}}$.Given that $\frac{T_A}{M_A} = 4 \cdot \frac{T_B}{M_B}$,we have $\frac{T_A/M_A}{T_B/M_B} = 4$.Substituting this into the ratio formula: $\frac{C_A}{C_B} = \sqrt{4} = 2$.

Let $A$ and $B$ be two gases and given: $\frac{T_A}{M_A} = 4 \cdot \frac{T_B}{M_B}$,where $T$ is the temperature and $M$ is the molecular mass. If $C_A$ and $C_B$ are the $r.m.s.$ speeds,then the ratio $\frac{C_A}{C_B}$ will be equal to:

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