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AP EAMCET · Chemistry · States of Matter

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. A 2000
  2. B 3000
  3. C 6000
  4. D 1500
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Answer & Solution

Correct Answer

(B) 3000

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Detailed explanation

\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} \\ &…

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