1.

The K.E. of N molecule of O_(2) is x Joules at -123^(@)C. Another sample of O_(2)" at "27^(@)C has a KE of 2x Joules. The latter sample contains.

Answer»

N MOLECULE of `O_(2)`
2N molecule of `O_(2)`
N/2 molecule of `O_(2)`
N/4 molecule of `O_(2)`

Solution :`KE=(3)/(2) RT, T=-123+273=+150 K`
`(3)/(2)xxRxx150=(3)/(2)xx8.314xx75=xJ =225xx8.314=xJ`
At `27^(@)C=27+223=300K`
KE for =2X Joule `=(3)/(2)xx8.314xx300`
N molecule
`THEREFORE x" Joule "=3xx8.314xx75`
In both the cases x Joules correspond to N MOLECULES.
Hence, (A) is the correct answer.


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