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 N.T.P. 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

Answer: (A)

$6.02\times {10}^{23}$

Since $\frac{C_p}{C_V}=1.4$ the gas should be diatomic.

If volume is $11.2lt$ then, no. of moles $=\frac{1}{2}$

$\because$ 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.

$=6.0223\times {10}^{23}$