Given the atomic radius of helium, 0.32 Å , and knowing that a sphere has a volume of 4πr³/3, the fraction occupied by the helium atoms is 3.69 × 10⁺⁷
Given the atomic radius of helium, 0.32 Å , and knowing that a sphere has a volume of 4πr3/3 , to calculate the fraction of space that He atoms occupy in a sample of helium at STP, we proceed as follows
We know that the number of atoms of helium present in 1 mol is N = 6.022 × 10²³ atoms.
The volume of these atoms is V' = NV where V = volume of 1 atom = 4πr³/3 where r = radius of helium atom = 0.32 °A = 0.32 × 10⁺¹⁰ m
So, V' = NV
= N4πr³/3
= 6.022 × 10²³ atoms × 4π(0.32 × 10⁺¹⁰ m)³/3
= 6.022 × 10²³ atoms × 4π × 3.28 × 10⁺² × 10⁺³⁰ m³/3
= 248.21 × 10⁺¹⁰ m³/3
= 82.731 × 10⁺¹⁰ m³
Now, we know that 1 mol of He gas occupies V" = 22.4 dm³ at STP.
So, V" = 0.0224 m³
So, the fraction occupied by the Helium atoms is n = V'/V"
= 82.731 × 10⁺¹⁰ m³/0.0224 m³
= 3693.35 × 10⁺¹⁰
= 3.69335 × 10³ × 10⁺¹⁰
= 3.69335 × 10⁺⁷
≅ 3.69 × 10⁺⁷
So, the fraction is 3.69 × 10⁺⁷