1.

Ratio of C_(p) and C_(v) of a gas 'X' is 1.4. The number ofatoms of the gas 'X' present in 11.2litres of it at N.T.P. is

Answer»

`6.02 xx 10^(23)`
`1.2 xx 10^(24)`
`3.01 xx 10^(23)`
`2.01 xx 10^(23)`

Solution :Since `(C_(p))/(C_(v))=1.4` , the GAS should be diatomic.
If VOLUME is 11.2It then , no. of moles `=(1)/(2)`
` :. ` no. of MOLECULES `=(1)/(2) xx` Avogadro's No.
no. of atoms `= 2 xx `no. of molecules
`2 xx (1)/(2) xx `AVAGADRO's No.
`= 6.023 x 10^(23)`


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