Question

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

• A. 2
• B. 4
• C. 1
• D. 0.5

Answer 1

Answer: (A)

2

Answer 2

$\frac{T_{A}}{M_{A}} = 4\frac{T_{B}}{M_{B}}$Ž$\sqrt{\frac{T_{A}}{M_{A}}} = 2\sqrt{\frac{T_{B}}{M_{B}}}$

$\Rightarrow \sqrt{\frac{3RT_{A}}{M_{A}}} = 2\sqrt{\frac{3RT}{M_{B}}} \Rightarrow C_{A} = 2C_{B} \Rightarrow \frac{C_{A}}{C_{B}} = 2$