Heat and Thermodynamics 4 Question 7
8. For a given gas at $1 atm$ pressure, rms speed of the molecules is $200 m / s$ at $127^{\circ} C$. At $2 atm$ pressure and at $227^{\circ} C$, the rms speed of the molecules will be
(2019 Main, 9 April I)
(a) $100 \sqrt{5} m / s$
(b) $80 m / s$
(c) $100 m / s$
(d) $80 \sqrt{5} m / s$
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Solution:
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
$$ \frac{v _{rms, 2}}{v _{rms, 1}}=\sqrt{\frac{T _2}{T _1}} $$
Substituting these given values in Eq. (i), we get
$$ \begin{aligned} \therefore v _{rms, 2} & =\sqrt{\frac{500}{400}} \times 200 m / s \\ & =\frac{1}{2} \sqrt{5} \times 200 m / s=100 \sqrt{5} m / s \end{aligned} $$
