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Exercises · 12.1

Q.Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3 A˚3\ \text{\AA}.

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The fraction of molecular volume to actual volume for oxygen gas at STP is found by comparing the volume of one mole of molecules (treated as hard spheres) to the molar volume of an ideal gas at STP. The result is about 4×10−44 \times 10^{-4}, meaning only about 0.04% of the gas volume is actually filled by molecules.

Why this question matters

This problem is a classic illustration of how much "empty space" exists in a gas. At STP, a mole of any ideal gas occupies 22.4 litres — but the molecules themselves take up only a tiny fraction of that. The rest is vacuum. The ratio we calculate here is the molecular volume fraction, which tells us how tightly packed the molecules would be if they were touching, compared to how spread out they actually are.

The key idea: treat each oxygen molecule as a tiny hard sphere of given diameter. The "molecular volume" of one mole is just the number of molecules times the volume of one sphere. The "actual volume" is the molar volume at STP. Divide the two.

Step-by-step solution

1. Find the volume of a single oxygen molecule

The molecule is modelled as a sphere of diameter d=3 A˚=3×10−10 md = 3\ \text{Å} = 3 \times 10^{-10}\ \text{m}.

Volume of one sphere:

Vmolecule=43πr3=43π(d2)3=43πd38=πd36V_{\text{molecule}} = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi \left( \frac{d}{2} \right)^3 = \frac{4}{3} \pi \frac{d^3}{8} = \frac{\pi d^3}{6}

Substitute dd:

Vmolecule=π(3×10−10)36=π×27×10−306=27π6×10−30=9π2×10−30 m3V_{\text{molecule}} = \frac{\pi (3 \times 10^{-10})^3}{6} = \frac{\pi \times 27 \times 10^{-30}}{6} = \frac{27\pi}{6} \times 10^{-30} = \frac{9\pi}{2} \times 10^{-30}\ \text{m}^3

Numerically:

Vmolecule≈9×3.14162×10−30≈14.137×10−30 m3=1.4137×10−29 m3V_{\text{molecule}} \approx \frac{9 \times 3.1416}{2} \times 10^{-30} \approx 14.137 \times 10^{-30} \ \text{m}^3 = 1.4137 \times 10^{-29}\ \text{m}^3

Tip

Keep the expression in terms of π\pi until the final step — it often cancels or simplifies. Here we'll need a number, so approximate at the end.

2. Find the molecular volume of one mole of oxygen

One mole contains Avogadro's number of molecules, NA=6.022×1023 mol−1N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}.

So the total volume occupied by the molecules themselves (if packed without any gaps) is:

Vmolecular, mole=NA×Vmolecule=6.022×1023×1.4137×10−29 m3V_{\text{molecular, mole}} = N_A \times V_{\text{molecule}} = 6.022 \times 10^{23} \times 1.4137 \times 10^{-29}\ \text{m}^3

Compute:

Vmolecular, mole≈8.514×10−6 m3V_{\text{molecular, mole}} \approx 8.514 \times 10^{-6}\ \text{m}^3

Convert to litres (since molar volume is usually given in litres):

1 m3=1000 L⇒8.514×10−6 m3=8.514×10−3 L1\ \text{m}^3 = 1000\ \text{L} \quad \Rightarrow \quad 8.514 \times 10^{-6}\ \text{m}^3 = 8.514 \times 10^{-3}\ \text{L}

So the molecules themselves occupy about 8.58.5 millilitres per mole.

3. Find the actual volume occupied by one mole of oxygen at STP

At STP (Standard Temperature and Pressure: 0°C, 1 atm), one mole of any ideal gas occupies 22.4 litres. This is the molar volume:

Vactual, mole=22.4 L=2.24×10−2 m3V_{\text{actual, mole}} = 22.4\ \text{L} = 2.24 \times 10^{-2}\ \text{m}^3

Note

STP is defined as 273.15 K and 1 atm. The molar volume 22.4 L is an approximation; the more precise value is 22.414 L, but 22.4 L is standard for such problems.

4. Compute the fraction

The fraction ff is:

f=Vmolecular, moleVactual, mole=8.514×10−3 L22.4 L≈3.80×10−4f = \frac{V_{\text{molecular, mole}}}{V_{\text{actual, mole}}} = \frac{8.514 \times 10^{-3}\ \text{L}}{22.4\ \text{L}} \approx 3.80 \times 10^{-4}

Or in scientific notation:

f≈4×10−4f \approx 4 \times 10^{-4}

Watch out

A common mistake is to forget that the molecular volume uses the radius, not the diameter, in the sphere formula. Using dd directly in 43πr3\frac{4}{3}\pi r^3 without halving it gives an answer 8 times too large. Always halve the diameter first.

What this number means

A fraction of 4×10−44 \times 10^{-4} means that only 0.04% of the volume of oxygen gas at STP is actually occupied by the molecules themselves. The remaining 99.96% is empty space. This explains why gases are so compressible — you can squeeze them into a much smaller volume because the molecules have plenty of room to move closer together.

Important

This calculation assumes molecules are hard spheres. In reality, molecules are not rigid and have intermolecular forces, but the hard-sphere model gives a good order-of-magnitude estimate for the volume fraction.

✓Final answer

The fraction of molecular volume to actual volume for oxygen gas at STP is approximately 4×10−4\boxed{4 \times 10^{-4}}.

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