When the temperature of a gas is raised from $27^{\circ} \mathrm{C}$ to $90^{\circ} \mathrm{C}$. The…
When the temperature of a gas is raised from $27^{\circ} \mathrm{C}$ to $90^{\circ} \mathrm{C}$. The increase in the rms velocity of the gas molecules is
$10 \%$
$15 \%$
$20 \%$
$17.5 \%$
Solution
The rms speed of the gas molecule is
$v=\sqrt{\frac{3 R T}{M}} \Rightarrow v \propto \sqrt{T}$
$\therefore \frac{\mathrm{v}_2}{\mathrm{v}_1}=\sqrt{\frac{\mathrm{T}_2}{\mathrm{~T}_1}}=\sqrt{\frac{(273+90)}{(273+27)}}=\frac{11}{10}$
$\therefore \%$ change in rms speed is
$\% \Delta \mathrm{v}=\frac{\mathrm{v}_2-\mathrm{v}_1}{\mathrm{v}_1} \times 100=\frac{\frac{11}{10} \mathrm{v}_1-\mathrm{v}_1}{\mathrm{v}_1} \times 100=10 \%$