A gas X has Cₚ and Cᵥ ratio as 1.4, at NTP 11.2 L of gas X will contain_______ number of atoms

The correct option is 6.02 × 10²³.

Let us break down the solution step-by-step to understand why this is the correct answer:

Step 1: Identify the atomicity of gas X
The ratio of specific heat capacities at constant pressure ($C_p$) and constant volume ($C_v$) is given as:

$\gamma =\frac{C_p}{C_v}=1.4$

A gas with a ratio of specific heats ($\gamma$) equal to 1.4 is a diatomic gas. Therefore, each molecule of gas X contains 2 atoms (atomicity = 2).

Step 2: Calculate the number of moles of gas X at NTP
At Normal Temperature and Pressure (NTP), 1 mole of any ideal gas occupies a molar volume of 22.4 L.
We are given a volume of 11.2 L of gas X. We can find the number of moles ($n$) using the formula:

$n=\frac{\text{Given Volume}}{\text{Molar Volume}}$

$n=\frac{11.2}{22.4}=0.5\text{moles}$

Step 3: Calculate the number of molecules of gas X
One mole of a substance contains Avogadro's number of molecules ($N_A\approx 6.02\times {10}^{23}$).
Thus, the number of molecules in 0.5 moles of gas X is:

$\text{Number of molecules}=n\times N_A$

$\text{Number of molecules}=0.5\times 6.02\times {10}^{23}=3.01\times {10}^{23}\text{molecules}$

Step 4: Calculate the total number of atoms
Since gas X is diatomic, each molecule consists of 2 atoms. Therefore, the total number of atoms is:

$\text{Total number of atoms}=\text{Atomicity}\times \text{Number of molecules}$

$\text{Total number of atoms}=2\times 3.01\times {10}^{23}=6.02\times {10}^{23}\text{atoms}$