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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 of gas `=(3)/(2)nRT`
Given,
n = 1 mol, `R=8.314" J mol"^(-1)K^(-1)`,
`T=27^(@)C=300" K"`
`:." Kinetic energy of "N_(2)` gas
`=(3)/(2)xx1xx8.314xx300`
`=3741.3" J"=3.741" kJ"`
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