The kinetic energy of ‘N’ molecules of H2 is 3J at –73°C. The kinetic energy of the same sample of H2 at 127°C is

KE=32RNaT ⇒KE α T KE1KE2=n1T1n2T2 3KE2=N×200N×400 KE2=6J

The kinetic energy of ‘N’ molecules of H2 is 3J at –73°C. The kinetic energy of the same sample of H2 at 127°C is

KE=32RNaT ⇒KE α T KE1KE2=n1T1n2T2 3KE2=N×200N×400 KE2=6J

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The kinetic energy of ‘N’ molecules of H₂ is 3J at –73°C. The kinetic energy of the same sample of H₂ at 127°C is

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12 J

6 J

9 J

3 J

answer is B.

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$KE = \frac{3}{2} \frac{R}{N_{a}} T \Rightarrow KE α T \frac{{KE}_{1}}{{KE}_{2}} = \frac{n_{1} T_{1}}{n_{2} T_{2}} \frac{3}{{KE}_{2}} = \frac{N \times 200}{N \times 400} {KE}_{2} = 6 J$

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The kinetic energy of ‘N’ molecules of H2 is 3J at –73°C. The kinetic energy of the same sample of H2 at 127°C is

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The kinetic energy of ‘N’ molecules of H₂ is 3J at –73°C. The kinetic energy of the same sample of H₂ at 127°C is