Heat And Thermodynamics Ques 124

Answer:

Correct Answer: 124.(a)

Solution:

Formula:

Important Points :

Key Idea For a gas molecule,

$$ \begin{aligned} & v _{rms}=\sqrt{\frac{3 R T}{M}} \\ \therefore \quad & v _{rms} \propto \sqrt{T} \end{aligned} $$

Let unknown rms speed be $v _{\text {rms, } 2}$

at $\quad T _2=227^{\circ} C($ or $500 K)$

and at $T _1=127^{\circ} C$ (or $400 K$ )

$$ v _{rms, 1}=200 m / s $$

$\therefore$ Using the relation $v _{rms} \propto \sqrt{T}$, we can write

$$ \frac{v _{rms, 2}}{v _{rms, 1}}=\sqrt{\frac{T _2}{T _1}} \cdots(i) $$

Substituting these given values in Eq. (i), we get

$$ \begin{aligned} \therefore v _{rms, 2} & =\sqrt{\frac{500}{400}} \times 200 m / s \\ & =\frac{1}{2} \sqrt{5} \times 200 m / s=100 \sqrt{5} m / s \end{aligned} $$