Question

Ratio of Cp and Cv of a gas 'X' is 1.4. Calculate the number of atoms of the gas present in 11.2 liters of it at NTP?

Answer 1

There are 5.61 × 10²³ atoms in the gas.

You know that

`γ = C_P/C_V` = 1.4

You also know that

`γ = 1 + 2/f` or `f = 2/(γ – 1)`, where `f` is the number of degrees of freedom.

So `f = 2/(γ – 1) = 2/(1.4 – 1) = 2/0.4` = 5

A monatomic gas has three translational degrees of freedom (`f` = 3).

A diatomic gas has three translational and two rotational degrees of freedom
(`f` = 5).

So the gas is diatomic and has the formula X₂.

Now, according to the Ideal Gas Law,

`PV = nRT`

At NTP, `P` = 101.325 kPa and `T` = 20 °C = 293.15 K.

`n = (PV)/(RT) = (101.325"kPa" × 11.2"L")/(8.314"kPa·L·K⁻¹mol⁻¹" × 293.15"K")` = 0.4656 mol (3 significant figures + 1 guard digit)

No. of atoms = 0.4563 mol X₂ × `(6.022 × 10²³"molecules X₂")/(1"mol X₂") × (2"atoms")/(1"molecule X₂")` = 5.61 × 10²³ atoms