Q.

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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a

12 J

b

6 J

c

9 J

d

3 J

answer is B.

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Detailed Solution

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

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