Ratio of ${\mathrm{C}}_{\mathrm{P}} \mathrm{and} {\mathrm{C}}_{\mathrm{V}} \mathrm{of} \mathrm{a} \mathrm{gas} \text{'}\mathrm{X}\text{'} \mathrm{is} 1.4.$ The number of atoms of the gas ‘X’ present in 11.2 litres of it at NTP is

Since $\frac{C_P}{C_V}=1.4,$ the gas should be diatomic.

If volume is 11.2 L, then number of moles $\frac{1}{2}$

 $$\begin{gathered} \therefore \mathrm{No}. \mathrm{of} \mathrm{molecules} = \frac{1}{2} \mathrm{x} \mathrm{Avogadro}\text{'}\mathrm{s} \mathrm{No}. \\ \mathrm{No}. \mathrm{of} \mathrm{atoms} = 2 \mathrm{x} \mathrm{no}. \mathrm{of} \mathrm{molecules} \\ =2 \mathrm{x} \frac{1}{2} \mathrm{x} \mathrm{Avogadro}\text{'}\mathrm{s} \mathrm{No}. = 6.0223 \mathrm{x} {10}^{23} \end{gathered}$$