IB DP Chemistry -S1.4.6 Avogadro's law - Study Notes - New Syllabus - 2026, 2027 & 2028
IB DP Chemistry – S1.4.6 Avogadro’s law- Study Notes – New Syllabus
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Structure 1.4.6 — Avogadro’s Law and Gas Volume Calculations
Structure 1.4.6 — Avogadro’s Law and Gas Volume Calculations
Avogadro’s Law
Avogadro’s Law states that:
“Equal volumes of all gases, under the same conditions of temperature and pressure, contain equal numbers of molecules.”
This implies that the volume of a gas is directly proportional to the number of moles of gas, provided pressure and temperature are constant:
\( V \propto n \)
This is useful for stoichiometry involving gases.
- You can use mole ratios from balanced equations to compare gas volumes directly.
- This works for any gas if they are at the same temperature and pressure.
Molar Volume (at STP):
At standard temperature and pressure (STP: 273 K and 1.00 × 10⁵ Pa), 1 mole of any gas occupies: 22.7 dm³
So:
- \( V = n \times 22.7 \)
- \( n = \frac{V}{22.7} \)
- Ensure volume is in dm³ for this shortcut to work.
Example
What volume of hydrogen gas (in dm³) is required to completely react with 100 cm³ of oxygen gas to form water, assuming all gases are measured at the same temperature and pressure?
\( 2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} \)
▶️Answer/Explanation
From the balanced equation: 2 volumes of H₂ react with 1 volume of O₂
So, if 100 cm³ of O₂ is used, H₂ required = \( 2 \times 100 = 200 \, \text{cm}^3 \)
Answer: 200 cm³ of hydrogen gas
Example
Calculate the volume of carbon dioxide produced at STP when 6.0 g of ethane (\( \text{C}_2\text{H}_6 \)) combusts completely.
\( 2\text{C}_2\text{H}_6 + 7\text{O}_2 \rightarrow 4\text{CO}_2 + 6\text{H}_2\text{O} \)
▶️Answer/Explanation
- Molar mass of \( \text{C}_2\text{H}_6 \) = 30 g/mol
- Moles of ethane = \( \frac{6.0}{30} = 0.200 \, \text{mol} \)
- From the equation, 2 mol \( \text{C}_2\text{H}_6 \) → 4 mol \( \text{CO}_2 \), so 1 mol \( \text{C}_2\text{H}_6 \) → 2 mol \( \text{CO}_2 \)
- CO₂ moles = \( 0.200 \times 2 = 0.400 \, \text{mol} \)
- Volume at STP = \( 0.400 \times 22.7 = 9.08 \, \text{dm}^3 \)
Answer: 9.08 dm³ of CO₂
Example
Ammonia is formed by the Haber process according to the equation:
\( \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \)
What volume of nitrogen is needed to produce 13.6 dm³ of ammonia at STP?
▶️Answer/Explanation
From the equation: 1 mol N₂ produces 2 mol NH₃ → 1 volume : 2 volumes
So, nitrogen required = \( \frac{13.6}{2} = 6.8 \, \text{dm}^3 \)
Answer: 6.8 dm³ of nitrogen gas