During an experiment, an ideal gas is found to obey an additional law $\mathrm{VP}^2=$ constant. The gas is…

During an experiment, an ideal gas is found to obey an additional law $\mathrm{VP}^2=$ constant. The gas is initially at temperature ' $T$ ' and volume ' V '. What will be the temperature of the gas when it expands to a volume 2 V ?
  1. $\sqrt{3} \mathrm{~T}$
  2. $\sqrt{\frac{1}{2 T}}$
  3. $\sqrt{2} \mathrm{~T}$
  4. $\sqrt{3 \mathrm{~T}}$

Solution

$\begin{aligned} & \mathrm{PV} \doteq \mathrm{nRT} \\ & \mathrm{P}=\frac{\mathrm{RT}}{\mathrm{~V}} ...(i)\\ & \mathrm{VP}^2=\text { constant } ...\text{(given)}\\ & \mathrm{V}\left(\frac{\mathrm{RT}}{\mathrm{~V}}\right)^2=\text { constant } ...\text{(from(i))}\\ & \frac{\mathrm{R}^2 \mathrm{~T}^2}{\mathrm{~V}}=\text { constant } \Rightarrow \frac{\mathrm{T}^2}{\mathrm{~V}}=\text { constant } \ldots\left(\text{given }T_1=T\right) \\ & \left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right)^2=\frac{\mathrm{V}_1}{\mathrm{~V}_2} \\ & \frac{\mathrm{~T}_1}{\mathrm{~T}_2}=\sqrt{\frac{\mathrm{V}}{2 \mathrm{~V}}} \\ & \therefore \mathrm{~T}_2=\sqrt{2} \mathrm{~T} \end{aligned}$ .

Asked in: MHT CET 2024 (16 May Shift 2)

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