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 the molecular mass. If $C_{A}$ and $C_{B}$ are the rms 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

As $v_{rms} = \sqrt{\frac{3RT}{M}}$ ∴ $\frac{C_{A}}{C_{B}} = \sqrt{\frac{T_{A}/T_{B}}{M_{A}/M_{B}}} = \sqrt{4} = 2$ $\left\lbrack \text{As }\frac{T_{A}}{T_{B}} = 4\frac{M_{A}}{M_{B}}\text{ given} \right\rbrack$