Question

Ratio of $C_p$ and $C_v$ of a gas '$X$' is $1.4$. The number of atoms of the gas '$X$' present in $11.2\, L$ of it at NTP will be

• A. $6.02 \times 10^{23}$
• B. $1.2 \times 10^{23}$
• C. $3.01 \times 10^{23}$
• D. $2.01 \times 10^{23}$

Answer 1

Answer: (A)

For the gas $X$ ratio of $C_{p} / C_{V}=1: 4$
So, the gas $X$ is diatomic.
At NTP, volume of $1$ mole of a gas $=22.4 \,L$
$1$ mole of a gas $=6.023 \times 10^{23}$ molecules
Thus, at NTP $22.4 L$ contains $=6.023 \times 10^{23}$ molecules
So, at NTP $11.2 L$ contains $=\frac{6.023 \times 10^{23} \times 11.2}{22.4}$ molecules
$=3.01 \times 10^{23}$ molecules
Hence, number of atoms of gas ' $X$ ' (diatomic)
$=3.01 \times 10^{23} \times 2$ atoms
$6.02 \times 10^{23}$ atoms