Ratio of `C_(p) and C_(v)` of a gas `'X' is 1.4`. The number of atoms of the gas `'X'` presents in `11.2 "litres"` of it a `NTP` is

Options

A. `6.02 xx 10^(23)`
B. `1.2 xx 10^(24)`
C. `3.01 xx 10^(23)`
D. `2.01 xx 10^(23)`

Answer

A

Explanation

Since `(C_(P))/(C_(V)) = 1.4`, the gas should be diatomic.
If volume is `11.2` litre then, no. of moles `= (1)/(2)`
`:.` No. of molecules `= (1)/(2) xx` Avogadro's no.
No. of atoms `= 2 xx` no. molecules
`2 xx (1)/(2) xx "Avogadro's no." = 6.0223 xx 10^(23)`