Question

Ratio of $C_p$ and $C_v$ of a gas X is 1.4, the number of atom of the gas 'X' present in 11.2 litres of it at NTP will be
(1) $6.02\times {10}^{23}$
(2) $1.2\times {10}^{23}$
(3) $3.01\times {10}^{23}$
(4) $2.01\times {10}^{23}$

Answer 1

The correct option is 1.

The ratio of $C_p$ and $C_v$ of gas gives the atomicity of gas if

$\frac{C_p}{C_v}=1.66$ -- monoatomic

$\frac{C_p}{C_v}=1.40$ ----- diatomic

Since, here $\frac{C_p}{C_v}=1.4$ so, the gas "X" is diatomic like $H_2$, $N_2$ etc.

Now, we know that, at NTP 1 mole of gas $=22.4L$

So, 11.2 L of gas X $=$ 0.5 mole of X gas

Now, 1 mole will have $=6.02\times {10}^{23}$ molecules of gas $=6.02\times {10}^{23}\times 2$ atoms of gas

So, 0.5 mole of gas X will have $=6.02\times {10}^{23}\times 2\times 0.5$ of atoms gas $=6.02\times {10}^{23}$ atoms.