Q138 / 160 · TS EAMCET 2023 (14 May Shift 2) · Physics

If the temperature of a gas is increased from $27^{\circ} \mathrm{C}$ to $159^{\circ} \mathrm{C}$, then the percentage increase in the rms speed of the gas molecules is

  • A
    5
  • B
    10
  • C
    15
  • Correct answer
    D
    20

Solution

$V_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}$ At $27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$ $\begin{aligned} & \therefore \mathrm{V}_{\mathrm{rms}_1}=\sqrt{\frac{3 \mathrm{R}(300)}{\mathrm{M}}} \text { and } \mathrm{V}_{\mathrm{rms}_2}=\sqrt{\frac{3 \mathrm{R}(432)}{\mathrm{M}}} \\ & \text { at } \mathrm{T}^{\prime}=159+273=432\end{aligned}$ $\therefore$ Percentage increase in the $\mathrm{r} \mathrm{ms}$ speed $=\left(\frac{\mathrm{V}_{\mathrm{mms}_2}-\mathrm{V}_{\mathrm{rms}_1}}{\mathrm{~V}_{\mathrm{rms}_1}}\right) \times 100$ $=\frac{\sqrt{\frac{3 R}{M}}(\sqrt{432}-\sqrt{300})}{\sqrt{\frac{3 R}{M}}(\sqrt{300})} \times 100=\left(\frac{20.78-17.32}{17.32}\right) \times 100 \approx 20 \%$

Hence, the correct answer is option: 'D'.

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