Kinetic energy in Kj of 280 g of `N_(2) at 27^(@)C ` is approximately `(R = 8. 314 J mol ^(-1) K ^(-1))`

Options

A. `18.7`
B. `37.4`
C. `56.1`
D. `74.8`

Answer

B

Explanation

`280 g of N_(2) = (280)/(28) =10` mles of `N_(2)`
`27^(@) C = (27+ 273) K =300K`
K.E `3xx` number of mol `xx 3 xx` moles of gas
`=(3nRT)/(2) = (xx "gas constant " xx "temperature")/(2)`
`= (3xx10 xx8.314xx300)/(2) = 37.4xx10^(3)J=37.4kJ`