Q15 / 180 · KCET 2010 · Chemistry
For one mole of an ideal gas, increasing the temperature from $10^{\circ} \mathrm{C}$ to $20^{\circ} \mathrm{C}$
- Aincreases the average kinetic energy by two times
- Bincreases the rms velocity by $\sqrt{2}$ times
- Cincreases the rms velocity by two times
- Correct answerDincreases both the average kinetic energy and rms velocity, but not significantly
Solution
Given, $\mathrm{T}_{1}=273+10=283 \mathrm{~K}$
$$
\mathrm{T}_{2}=273+20=293 \mathrm{~K}
$$
Average $\mathrm{KE}=\frac{3}{2} \mathrm{KT}$
$$
\frac{\left(\mathrm{KE}_{1}\right)}{\left(\mathrm{KE}_{2}\right)}=\frac{283}{293}=0.96
$$
Root mean square (rms) velocity,
$$
\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}
$$
$$
\begin{aligned}
\frac{\mathrm{v}_{(\mathrm{rms})_{1}}}{\mathrm{v}_{(\mathrm{rms})_{2}}} &=\sqrt{\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}} \\
&=\sqrt{\frac{283}{293}}=0.98
\end{aligned}
$$
Thus, both average kinetic energy and root mean square velocity increase but not significantly when temperature is increased from $10^{\circ} \mathrm{C}$ to $20^{\circ} \mathrm{C}$.
Hence, the correct answer is option: 'D'.