One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen? (Take the size of hydrogen molecule to be about 1Å). Why is this ratio so large?
Solution 1
Volume of one mole of ideal gas, Vg = 22.4 litre = 22.4 x 10-3 m3
Radius of hydrogen molecule = 1A/2 = 0.5 A = 0.5 x 10-10 m
Volume of hydrogen molecule = 4/3 πr3
=4/3 x 22/7 (0.5 x 10-10)3 m3
= 0.5238 x 10-30 m3
One mole contains 6.023 x 1023 molecules.
Volume of one mole of hydrogen, VH
= 0.5238 x 10-30 x 6.023 x 1023 m3 = 3.1548 x 10-7 m3
Now Vg/VH=22.4 x 10-3/3.1548 x 10-7 =7.1 x 104
The ratio is very large. This is because the interatomic separation in the gas is very large compared to the size of a hydrogen molecule.
Solution 2
Radius of hydrogen atom, r = 0.5 Å = 0.5 × 10–10 m
Volume of hydrogen atom = `4/3pir^3`
= `4/3xx22/7xx(0.5xx10^(-10))^3`
= `0.524 xx 10^(-30) m^3`
Now, 1 mole of hydrogen contains 6.023 × 1023 hydrogen atoms.
∴ Volume of 1 mole of hydrogen atoms, Va = 6.023 × 1023 × 0.524 × 10–30
= 3.16 × 10–7 m3
Molar volume of 1 mole of hydrogen atoms at STP
Vm = 22.4 L = 22.4 × 10–3 m3
`:.V_m/V_a = (22.4xx10^(-3))/(3.16 xx10^(-7)) = 7.08 xx 10^4`
Hence, the molar volume is 7.08 × 104 times higher than the atomic volume. For this reason, the inter-atomic separation in hydrogen gas is much larger than the size of a hydrogen atom
