The kinetic energy of `4 mol` of nitrogen gas at `127^(@) C` is …….. `cal(R = 2 cal mol^(-1) K^(-1))`

Options

A. 4400
B. 3200
C. 4800
D. 1524

Answer

C

Explanation

K.E. of 1 mole of a gas
`=(3)/(2) RT = (3)/(2) xx 2 cal K^(-1) xx (127 + 273) K`
`= 3 xx 400 cal mol^(-1)`
`= 1200 cal mol^(-1)`
K.E. of 4 moles of gas `= 4 mol xx (1200 cal mol^(-1))`
`=4800 cal`