# 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?

Aug 24, 2014

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,

$P V = n R T$

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

$n = \frac{P V}{R T} = \left(101.325 \text{kPa" × 11.2"L")/(8.314"kPa·L·K⁻¹mol⁻¹" × 293.15"K}\right)$ = 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