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The kinetic energy in J of 1 mole of N2 ...

The kinetic energy in J of 1 mole of `N_2` at `27^@ C` is
`(R= 8.314 mol^(-1) k^(-1))`

A

2494

B

18706

C

7482

D

3741

Text Solution

Verified by Experts

The correct Answer is:
D
Kinetic energy `= (3)/(2) n RT`
`= (3)/(2) (1) (8.314) J mol^(-1) k^(-1) ( 27 + 273) k`
`= 450 (8.314)`
`therefore KE =3,741.3 J`
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