1.

Calculate kinetic energy of `4g N_(2)` at `-13^(@)C`.

Answer» We know,
`KE=(3)/(2)xxnRT`
Number of moles of `4 g N_(2)=(4)/(28)g mol^(-1)`
`T=-13^(@)C=260.15K`
`R=8.314 J K^(-1) mol^(-1)`
`:. K.E.=(3)/(2)xx(4)/(28)xx8.314xx260.15`
`=463.47 J mol^(-1)`


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