Q. 173.8( 18 Votes )

# 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 1 Å). Why is this ratio so large?

Answer :

Given,

Volume of one mole of ideal gas at STP, V_{m}=22.4L=22.4 × 10^{-3} m^{3}

Radius of hydrogen atom, r = 1 Å / 2 = 0.5 Å = 0.5 × 10^{-10} m

Volume of the hydrogen atom, V = (4/3)πr^{3}

⇒ V = (4/3)×3.14×(0.5×10^{-10} m)^{3}

⇒ V = = 5.24 × 10^{-31} m^{3}

One mole of hydrogen atom contains N_{A} = 6.023 × 10^{23} atoms.

Where, N_{A} is Avogadro’s number.

So, Volume of one mole of hydrogen atom,

V_{t} = 6.023 × 10^{23} × 5.24 × 10^{-31} m^{3}

⇒ V_{t} = 3.16 × 10^{-7} m^{3}

The ratio of molar volume to atomic volume is

⇒ V_{m �/}V_{t} = 7.089 × 10^{4}

This ratio is very high due to the fact that the inter-atomic distance is very high as compared to the size of atoms in hydrogen gas.

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