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Gas laws from KTG

Let A and B b...

Question

Let A and B be two gases and given $\frac{T_{A}}{M_{A}} = 4. \frac{T_{B}}{M_{B}};$ where T is the temperature and M is the molecular mass. If $C_{A}$ and $C_{B}$ are the rms speed, then the ration $\frac{C_{A}}{C_{B}}$ will be equal to

A

2

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B

4

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C

1

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D

0.5

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Solution

The correct option is A 2
As $ν_{r m s} = \sqrt{\frac{3 R T}{M}} ∴ \frac{C_{A}}{C_{B}} = \sqrt{\frac{T_{A} / T_{B}}{M_{A} / M_{B}}} = \sqrt{4} = 2 [A s \frac{T_{A}}{T_{B}} = 4 \frac{M_{A}}{M_{B}} g i v e n]$

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