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 $11.2$ litres of it at $NTP$ is

• A. $6.02 \times 10^{23}$
• B. $1.2 \times 10^{24}$
• C. $3.0\times 10^{23}$
• D. $2.01 \times 10^{23}$

Answer 1

Answer: (A)

Since $\frac{C_{P}}{C_{V}}=1.4$, the gas should be diatomic
If volume is 11.2 L, then no. of moles $=\frac{1}{2}$
$\therefore $ No. of molecules $=\frac{1}{2}$ $\times$ Avogadro’s No
No. of atoms $=2 \times$ no. of molecules
$2 \times \frac{1}{2} \times$ Avogadro’s No. $=6.0223\times 10^{23}$