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 STP is?

• A. $6.02\times {10}^{23}$
• B. $1.2\times {10}^{24}$
• C. $3.01\times {10}^{23}$
• D. $2.01\times {10}^{23}$

Answer 1

The correct option is A $6.02\times {10}^{23}$

For diatomic molecules, $\frac{C_P}{C_V}=1.4$

At NTP, $1$ mole$\equiv 22.4L$

$\therefore 0.5$ mole$\equiv 11.2L$

$\therefore$ Number of atoms= $2\times noofmoles\times 6.022\times {10}^{23}$

= $2\times 0.5\times 6.022\times {10}^{23}$

= $6.022\times {10}^{23}$

Hence, option A is