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, V_{g }= 22.4 litre = 22.4 x 10^{-3} m^{3}

Radius of hydrogen molecule = 1A/2 = 0.5 A = 0.5 x 10^{-10} m

Volume of hydrogen molecule = 4/3 πr^{3}

=4/3 x 22/7 (0.5 x 10^{-10})^{3} m^{3}

= 0.5238 x 10^{-30} m^{3}

One mole contains 6.023 x 10^{23 }molecules.

Volume of one mole of hydrogen, VH

= 0.5238 x 10^{-30} x 6.023 x 10^{23} m^{3} = 3.1548 x 10^{-7} m^{3}

Now V_{g}/V_{H}=22.4 x 10^{-3}/3.1548 x 10^{-7} =7.1 x 10^{4}

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 × 10^{23} hydrogen atoms.

∴ Volume of 1 mole of hydrogen atoms, *V*_{a} = 6.023 × 10^{23} × 0.524 × 10^{–30}

= 3.16 × 10^{–7} m^{3}

Molar volume of 1 mole of hydrogen atoms at STP

*V*_{m} = 22.4 L = 22.4 × 10^{–3} m^{3}

`:.V_m/V_a = (22.4xx10^(-3))/(3.16 xx10^(-7)) = 7.08 xx 10^4`

Hence, the molar volume is 7.08 × 10^{4} 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