At 240.55 K , for one mole of an ideal gas, a graph of P (on $y$-axis) and $\mathrm{V}^{-1}$ (on $x$-axis)…

At 240.55 K , for one mole of an ideal gas, a graph of P (on $y$-axis) and $\mathrm{V}^{-1}$ (on $x$-axis) gave a straight line passing through origin. Its slope $(m)$ is $2000 \mathrm{~J} \mathrm{~mol}^{-1}$. What is the kinetic energy (in $\mathrm{J} \mathrm{mol}^{-1}$ ) of ideal gas?
  1. 2000
  2. 3000
  3. 6000
  4. 1500

Solution

$\begin{aligned} & \text { Kinetic energy }(\mathrm{K} . \mathrm{E})=\frac{3}{2} \mathrm{PV} \\ & \therefore \quad \mathrm{P}=\frac{2 \mathrm{KE}}{3} \times \frac{1}{\mathrm{~V}} \\ & \therefore \quad \mathrm{KE}=3000 \mathrm{~J} \mathrm{~mol}^{-1} \\ & \mathrm{Y}=\mathrm{mX}+\mathrm{c} \\ & \mathrm{m}(\mathrm{Slope})=2000 \mathrm{~J} \mathrm{~mol}^{-1} \\ & \frac{2 \mathrm{KE}}{3}=2000 \mathrm{~J} \mathrm{~mol}^{-1}\end{aligned}$

Asked in: AP EAMCET 2024 (22 May Shift 1)

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