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The kinetic energy of two moles of N2 at...

The kinetic energy of two moles of `N_2` at `27^(@)C` is (R= 8.314 J `K^(-1) "mol"^(-1)`)

A

5491.6 J

B

6491.6 J

C

7482.6J

D

8882.4 J

Text Solution

Verified by Experts

The correct Answer is:
C
`K.E. = (3)/(2) nRT = (3)/(2) xx 2 xx 8.314 xx 300 = 7482.6 J`.
`[T=27^(@)C = (273 + 27)K=300K]`
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