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

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$6.02 \times 10^{23}$

$1.2 \times 10^{24}$

$3.0 \times 10^{23}$

$2.01 \times 10^{23}$

answer is A.

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Detailed Solution

Since $\frac{C_{P}}{C_{V}} = 1.4,$ the gas should be diatomic.

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

$∴ No . of molecules = \frac{1}{2} x Avogadro' s No . No . of atoms = 2 x no . of molecules = 2 x \frac{1}{2} x Avogadro' s No . = 6.0223 x 10^{23}$

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