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?
- A 2000
- B 3000
- C 6000
- D 1500
Answer & Solution
Correct Answer
(B) 3000
Step-by-step Solution
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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