Question

if the ratio of CP and CV for a gas is 1.4 the number of atoms present in 11.2 l of it at STP is

Answer 1

There are 5.61 × 10²³ atoms in the gas.

You know that

γ=CPCV = 1.4

You also know that

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

So f=2γ–1=21.4–1=20.4 = 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=PVRT=101.325kPa×11.2L8.314kPa⋅L⋅K⁻¹mol⁻¹×293.15K= 0.4656 mol (3 significant figures + 1 guard digit)

No. of atoms = 0.4563 mol X₂ × 6.022×10²³molecules X₂1mol X₂×2atoms1molecule X₂ = 5.61 × 10²³ atoms

Answer 2

Answer: 6.022×10²³

Explanation: